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Markus Antoni

Calculus with Curvilinear CoordinatesProblems and Solutions

Calculus with Curvilinear Coordinates

Markus Antoni

Calculus with CurvilinearCoordinatesProblems and Solutions

123

Markus AntoniInstitute of GeodesyUniversity of StuttgartStuttgart, Germany

ISBN 978-3-030-00415-6 ISBN 978-3-030-00416-3 (eBook)https://doi.org/10.1007/978-3-030-00416-3

Library of Congress Control Number: 2018954029

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https://doi.org/10.1007/978-3-030-00416-3

Preface

This book contains exercises and their solutions of the course ‘AdvancedMathematics’ in the GEOENGINE master program at the University of Stuttgart.

In this course, we recall and introduce the topics

• ordinary differential equations of second order and system of differentialequations,

• vector analysis in curvilinear coordinates, including integral theorems ofGreen, Stokes, and Gauß,

• partial differential equations and potential theory.

We present here only the vector analysis, which turned out to be a challenge formany students, due to the parametrization of objects and the different curvilinearcoordinate systems. The other courses of the program deal implicitly or explicitlywith several coordinate systems and coordinate transformations, e.g.,

• spherical coordinates: tachymetry, geodetic astronomy, gravity field repre-sentation, etc.,

• cylindrical coordinates: regional collocation, surveying in tunnels, TECcalculations, etc.,

• ellipsoidal coordinates: gravity field representation, map projection, statesurveying,

• modified torus coordinates: sampling and intersection of satellites tracks,• affine coordinates: misaligned observations in state surveying of nineteenth

century,• triaxial ellipsoidal coordinates, prolate or oblate spheroidal coordinates:

gravity field modeling of non-spherical celestial bodies.

Hence, we introduce the concept in terms of the tangential vectors of the frame. Inthis form, the procedures are valid for every orthogonal coordinate system, and wedo not need to restrict ourself to special cases.

v

Important remarks and conclusions are highlighted in the fashion of the fol-lowing statements:

• Vector analysis can be performed in different coordinate systems. An‘optimal’ system considers the symmetry (spherical, cylindrical, para-bolic, oblate spheroidal,…) of the problem, which reduces the calculationefforts.

• In case of ‘new’ coordinate systems, the reduction of work is not visible asthe frame vectors must be derived firstly.

In Chap. 1, we calculate the tangent vector and arc length of curves in Cartesianand curvilinear coordinates. We also present how to calculate the arc length in‘unknown’ coordinate systems and how to find a parameter representation from thealgebraic equation.

In Chap. 2, we consider the differential operators – gradient, curl, divergence –

acting on scalar and vector fields.In Chap. 3, we calculate the work for moving a unit mass in (non-)conservative

vector fields by the line integral or potential differences. The potential in arbitrarycurvilinear coordinates is determined from the corresponding vector field by inte-gration and differentiation.

In Chap. 4, we calculate the area of planar figures, the flux through a surface orvolume, and the circulation within a curved domain by the integral theorems ofvector analysis. In particular, we present the theorem of Green, Gauß, and Stokes.The benefit of the ‘optimal’ coordinate system is also demonstrated, for example inthe surface integrals over paraboloid and ellipsoid, where the field is given in themost adequate coordinate system.

Stuttgart, Germany Markus Antoni2018

vi Preface

Acknowledgements

The idea for this book arises due to discussions with the students in the masterprogram. I would like to honor the contributions of many alert students – inparticular my lab assistents – in the form of questions and critcial feedback. I wouldlike to thank my colleagues at the Institute of Geodesy at the University of Stuttgartfor their encouragement, support, and discussions contributing to this work. Inparticular, I would like to express my gratitude to Matthias Roth and Bramha DuttVishwakarma for reviewing the text. Remaining mistakes, typos or unclear state-ments are in the responsibility of the author. The thoughtful reader is cordiallyinvited to report these errors.

Last but not least, I hope that the well-disposed reader will find many chal-lenging and interesting exercises about differential geometry, vector analysis, andintegral theorems in this book. In particular, the presentation of questions andsolutions in arbitrary curvilinear coordinates might provide a new point of view tothe user.

Stuttgart, Germany Markus Antoni2018

vii

Contents

1 Arc Length and Tangent Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2 Differentiation of Field Quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . 39Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

3 Work, Line Integral and Potential . . . . . . . . . . . . . . . . . . . . . . . . . . 57Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

4 Integral Theorems of Vector Analysis . . . . . . . . . . . . . . . . . . . . . . . . 75Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

ix

Notation

e PlaneF Vector field in Cartesian coordinatesF Flux through surface or volume (index might denote the body)G Vector field in arbitrary curvilinear coordinatesjJj Jacobian determinantN Normal vector of a surfaceqi Single coordinate, e.g., q1 � r in spherical systemss Arc lengtht Parameter of the curveS Surface Sðu; wÞT Tangent vector of a curveVþ Upper part of surface/volume of Vivani’s figureZ Surface/volume of a cylinderg Additional parameter of a vector/scalar fieldW Parametric representation of a curveU Arbitrary scalar field, potentialR Surface/volume of a sphereRþ Surface/volume of a hemisphere (usually z� 0)X Circulation within a (curved) surface (index might denote

the body)fr; #; kg Spherical coordinatesfq;/; zg Cylindrical coordinatesfa; b; cg Arbitrary curvilinear coordinatesfhq1 ; hq2 ; hq3g Non-normalized tangential vector of the frame/‘frame vector’

f hq1 ; hq2 ; hq3g Normalized tangential vector of the frame/‘frame vector’

r� F, r� G Curl/rotation of the vector field F and G, respectivelydiv F, div G Divergence of the vector field F and G, respectivelyrF, rG Gradient of the vector field F and G, respectivelyfu;wg Parameters of a surface Sðu; wÞ

xi

List of Figures

Fig. 1.1 Approximation of a curve by a polygon . . . . . . . . . . . . . . . . . . . 2

Fig. 1.2 Closed curve WðtÞ ¼ 5 cos t � 2 cos5t2; 5 sin t � 2 sin

5t2

� �>. . . 12

Fig. 1.3 Square in polar coordinates. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14Fig. 1.4 Two Cardioid curves with a parametrization in polar

coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16Fig. 1.5 Closed figure: 3x2 þ jyj3 � y2 ¼ 0. . . . . . . . . . . . . . . . . . . . . . . . 17Fig. 1.6 Cassini ovals for c ¼ f1; 2; 5; 10g. . . . . . . . . . . . . . . . . . . . . . . . 20Fig. 1.7 Curve on the paraboloid x2 þ y2 � z ¼ 0 . . . . . . . . . . . . . . . . . . 22Fig. 1.8 Villarceau circle on a torus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25Fig. 1.9 Curve on a sphere:

W ¼�3 sin tþ sin 3t; 3 cos tþ cos 3t;

ffiffiffiffiffi12

psin t

�>. . . . . . . . . . . 28

Fig. 1.10 Spherical loxodrome with k0 ¼ 0 and c ¼ 8. . . . . . . . . . . . . . . . 30Fig. 1.11 Curve on the ellipsoid: W ¼ 10 sin t hq þ 8 cos t hz

with / ¼ 3t5

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

Fig. 1.12 Partial parabolic surfaces with c 2 ½0; p� and b ¼ 1and a ¼ 2, respectively . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

Fig. 1.13 Partial cardioid surfaces with c 2 ½0; p� and b ¼ 0:9and a ¼ 1:1, respectively . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

Fig. 3.1 Force field F ¼ 11þ y2

;�x

� �>and two different paths

between A and B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61Fig. 3.2 Folium of Descartes in the vector field F ¼ ð�y; xÞ> . . . . . . . . . 62Fig. 3.3 Visualization of Vivani’s figure by the intersection

of a (semi)-sphere and a cylinder . . . . . . . . . . . . . . . . . . . . . . . . 63Fig. 3.4 Intersection of the cylinder x2 þ y2 ¼ 1 and the

plane z ¼ 1� x2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

xiii

Fig. 4.1 Astroid curve with W ¼ cos3 t; sin3 t� �>

and its enclosedarea. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

Fig. 4.2 Folium of Descartes with x3 þ y3 � 3xy ¼ 0 . . . . . . . . . . . . . . . . 84Fig. 4.3 Lemniscate of Bernoulli with 2ðx2 � y2Þ ¼ ðx2 þ y2Þ2 . . . . . . . . 86Fig. 4.4 Example of a sector with two straight lines and a curved

line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87Fig. 4.5 Area enclosed by q ¼ cos n/ . . . . . . . . . . . . . . . . . . . . . . . . . . . 88Fig. 4.6 Geometrical center of the loop 3x2 þ y3 � y2 ¼ 0 . . . . . . . . . . . . 89Fig. 4.7 Geometrical center of a partial astroid figure . . . . . . . . . . . . . . . 91Fig. 4.8 Parabola x ¼ 0:5y2 and some of its pedal curves . . . . . . . . . . . . 94Fig. 4.9 Mathematical cylinder. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95Fig. 4.10 Torus with R ¼ 4 and r ¼ 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98Fig. 4.11 Vivani’s figure (upper part) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100Fig. 4.12 Mathematical cylinder where the bottom surface is defined

by the astroid of exercise 37. . . . . . . . . . . . . . . . . . . . . . . . . . . . 102Fig. 4.13 2D representation of cone, plane and Dandelin sphere . . . . . . . . 104Fig. 4.14 Cone, plane and one Dandelin sphere. . . . . . . . . . . . . . . . . . . . . 105Fig. 4.15 Ellipsoidal partial area

Op ¼ x 2 R3 : z ¼ 1

3

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi9� x2 � y2

p; jyj � 1

n o. . . . . . . . . . . . . . . 106

Fig. 4.16 Paraboloid: x2 þ y2 � 1 ¼ 2z . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108Fig. 4.17 Surface defined by the rotation of x ¼ z cos

ffiffiz3

paround

the z-axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110Fig. 4.18 Parabolic cylinder

S ¼ fx 2 R3 : y ¼ 4� z2; 0� x� 1; y� 0g . . . . . . . . . . . . . . . . 112

Fig. 4.19 Vivani’s surface (upper part) . . . . . . . . . . . . . . . . . . . . . . . . . . . 114Fig. 4.20 Hyperbolic partial areas of z ¼ xy in the domain x2 þ y2 � 1

and jxj þ jyj � 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115Fig. 4.21 Partial areas on paraboloid z ¼ x2 þ y2 . . . . . . . . . . . . . . . . . . . . 118Fig. 4.22 Partial cardioid surface with c 2 ½0; p�, b 2 ½0;1½

and a ¼ const . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

xiv List of Figures

Chapter 1Arc Length and Tangent Vector

Many engineers might face some of the following questions:

• What is the shortest flight connection between two airports?• Is the new roller coaster longer than the previous one?• How long will a new road be in a mountainous area?• When will the space probe reach the planet Mars?• Howmuchmaterial is required for a cable transport system or a suspension bridge?

These questions can be traced back to the problem of calculating the arc length of acurve, which is presented in this chapter.

Curves in the Cartesian Coordinate System

Definition: A curve is a one-dimensional geometrical object, which is defined by themapping of an interval I ∈ R into a set of well-defined points of the Euclidean spaceR

n . In this work, only continuous curves in two or three dimensions are considered.bA curve can be presented by (geometrical/physical) properties or a mathematicalexpression. In general, curves can be represented in

• explicit form (e.g. y(x) = √1 − x2),

• implicit form (e.g. x2 + y2 + x3 = sin y),• or a parametric form �(t) = (cos t, sin t, et cos t)�,but not every curve can be represented by all of these forms. In vector analysis, theparametric representation is preferred/required for calculations.Any parametric curve �(t) can be expressed in a Cartesian coordinate system

�(t) = (x(t), y(t), z(t))� = x(t) i + y(t) j + z(t) k (1.1)

© Springer Nature Switzerland AG 2019M. Antoni, Calculus with Curvilinear Coordinates,https://doi.org/10.1007/978-3-030-00416-3_1

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2 1 Arc Length and Tangent Vector

with the Cartesian base vectors

i = (1, 0, 0)� , j = (0, 1, 0)� , k = (0, 0, 1)�

and the components {x(t), y(t), z(t)}.Definition: The vector

T = �(t) =: limh→0

�(t + h) − �(t)

h= x(t) i + y(t) j + z(t) k (1.2)

is called the tangential vector or tangent vector of the curve �(t). The componentsare found by differentiating the expression of the curve w.r.t. the curve parameter t .Very often, this parameter is interpreted as time and so the notation x = dx

dt of math-ematical physics is used.

Arc Length

The arc length of a curve can be approximated by measuring a polygon which edgesthe curve (Fig. 1.1). The length of the polygon is calculated by

s =L−1∑

�=1

‖�(t�+1) − �(t�)‖.

In case of an equidistant parameter t�+1 = t� + �t , the expression will be simplified:

s =L−1∑

�=1

‖�(t� + �t) − �(t�)‖.

Fig. 1.1 Approximation of a curve by a polygon

1 Arc Length and Tangent Vector 3

The length s of the polygon approximates the true length s of the curve. The approx-imation will be improved for smaller increments �t and more data points:

s = lim�t=0,L→∞ s = lim

�t=0,L→∞

L−1∑

�=1

∥∥∥∥

�(t� + �t) − �(t�)

�t

∥∥∥∥ · �t = lim

�t=0,L→∞

L−1∑

�=1

‖T (t�)‖ · �t =

=τ∫

0

√T�Tdt. (1.3)

The formula above is independent of the coordinate system. In Cartesian coordinates,it can be expressed in the following:

s =τ∫

0

√T�Tdt =

τ∫

0

√x2(t) + y2(t) + z2(t)dt, (1.4)

where the argument (t) is often skipped for reason of compactness.

Alternative Coordinate Systems

The representation of a curve – and the further calculations – requires a set of linearindependent base vectors, but not necessarily a Cartesian coordinate system. Each

curve can be expressed in any alternative base vectors{hq1 , hq2 , hq3

}by

�(t) = ξ(t) hq1 + υ(t) hq2 + ζ(t) hq3 (1.5)

with the components {ξ(t), υ(t), ζ(t)} in the new coordinate system.

Definition: A new coordinate system is defined by the relationship

x = x(q1, q2, q3)

y = y(q1, q2, q3)

z = z(q1, q2, q3)

(1.6)

between the Cartesian coordinates (x, y, z) and the curvilinear coordinates(q1, q2, q3). The corresponding base vectors hqi (i = 1, 2, 3) are the tangential vec-tors of the curvilinear coordinate system. In the following text, these vectors will bedenoted as ‘frame vectors’ for compactness.

In this book, only orthogonal coordinate systems are considered in which the innerproduct of different base vectors is zero:

h�q1 hq2 = h

�q1 hq3 = h

�q2 hq3 = 0. (1.7)


‘Frame Vectors’

The ‘frame vectors’ are found by differentiation of the relation between the coordi-nates. To save space, the ‘frame vectors’ and their derivations are presented in thefollowing form:

hqi(1)= 1

‖hqi ‖

∂

⎛

⎝xyz

⎞

⎠

∂qi

(3)= 1

‖hqi ‖hqi

⇒ ‖hqi ‖ (2)=√h�qi hqi =: hqi i = 1, 2, 3.

(1.8)

In step (1), the relation (1.6) is differentiated w.r.t. the coordinate qi and the resultis divided by the (so far unknown) norm ‖hqi ‖. This norm is calculated in step (2)in the line below. In step (3), the normalized and simplified ‘frame vector’ is thenwritten down in the upper line.

Tangent Vector in Curvilinear Coordinates

According to the definition (1.2), the tangent vector in Cartesian representation isobtained by differentiating the components

T = �(t) = (x(t), y(t), z(t))� = x(t) i + y(t) j + z(t) k. (1.9)

In a curvilinear coordinate system, the norm of the ‘frame vectors’ hqi = ‖hqi ‖ andthe derivative of the coordinates qi are required. To derive the formula, the Cartesianexpression is re-ordered:

T = d

dt

{x(q1(t), q2(t), q3(t)) i + y(q1(t), q2(t), q3(t)) j + z(q1(t), q2(t), q3(t)) k

}=

=3∑

i=1

(∂x

∂qiqi

)i +

3∑

i=1

(∂y

∂qiqi

)j +

3∑

i=1

(∂z

∂qiqi

)k =

=3∑

i=1

qi

(∂x

∂qii + ∂y

∂qij + ∂z

∂qik)

=

=3∑

i=1

qi hqi =3∑

i=1

qi‖hqi ‖ hqi = q1‖hq1‖ hq1 + q2‖hq2‖ hq2 + q3‖hq3‖ hq3 .

(1.10)


The arc length s of a curve � in the interval [t1, t2] is calculated by integration:

s =t2∫

t1

√T�Tdt =

⎧⎪⎪⎨

⎪⎪⎩

t2∫

t1

√x2 + y2 + z2dt

t2∫

t1

√q21‖hq1‖2 + q2

2‖hq2‖2 + q23‖hq3‖2dt.

(1.11)

Notation

When the meaning is obvious from the context, the word ‘system’ might be omitted:

• ‘Express the tangent vector in spherical coordinates’, instead of ‘Express the tan-gent vector in the spherical coordinate system’.

• ‘Calculate the gradient in parabolic coordinates’ instead of ‘Calculate the gradientin the parabolic coordinate system’.

To avoid ‘subindices’ in the exercises, the notation of the coordinates {q1, q2, q3} ischanged by the following conventions:

• In the spherical coordinate system, the symbols {r, λ, ϑ} are replacing theqi -notation and the coordinate system is defined by

x = r cos λ sin ϑ

y = r sin λ sin ϑ (1.12)

x = r cosϑ

where r is called the radius, ϑ is the co-latitude and λ is the longitude. The ‘framevectors’ of the spherical system are obtained according to the given procedure by

hr = 1

‖hr‖

⎛

⎝cos λ sin ϑ

sin λ sin ϑ

cosϑ

⎞

⎠ =⎛

⎝cos λ sin ϑ

sin λ sin ϑ

cosϑ

⎞

⎠

⇒ ‖hr‖ =√sin2ϑ(cos2λ + sin2λ) + cos2ϑ = 1

hλ = 1

‖hλ‖

⎛

⎝−r sin λ sin ϑ

r cos λ sin ϑ

0

⎞

⎠ =⎛

⎝− sin λ

cos λ

0

⎞

⎠

⇒ ‖hλ‖ =√r2sin2ϑ(cos2λ + sin2λ) = r sin ϑ

hϑ = 1

‖hϑ‖

⎛

⎝r cos λ cosϑ

r sin λ cosϑ

−r sin ϑ

⎞

⎠ =⎛

⎝cos λ cosϑ

sin λ cosϑ

− sin ϑ

⎞

⎠

⇒ ‖hϑ‖ =√r2cos2ϑ(cos2λ + sin2λ) + r2sin2ϑ = r

(1.13)


The inverse mapping is given by

r = √x2 + y2 + z2

λ = arctany

x

ϑ = arccosz

r

(1.14)

where the longitude λ requires the investigation of the quadrant.

If a curve �(t) is defined by λ = λ(t), ϑ = ϑ(t) and r = r(t) in spherical coor-dinates, then the tangent vector is calculated by:

T =3∑

i=1

qi · ‖hi‖ hi = r hr + ϑ · r hϑ + λ · r sin ϑ hλ (1.15)

• In the cylindrical coordinate system

x = ρ cosφ

y = ρ sin φ

z = z(1.16)

the value ρ is known as radius or polar distance, and φ as polar angle or longitude.For better distinction, the z-coordinate is sometimes denoted by ζ as well. Theframe vectors are easy to find:

hρ = 1

‖hρ ‖

⎛

⎝cosφ

sin φ

0

⎞

⎠ =⎛

⎝cosφ

sin φ

0

⎞

⎠

⇒‖hρ ‖=√cos2 φ + sin2 φ = 1

hφ = 1

‖hφ ‖

⎛

⎝−ρ sin φ

ρ cosφ

0

⎞

⎠ =⎛

⎝− sin φ

cosφ

0

⎞

⎠

⇒‖hφ ‖=√

ρ2(cos2 φ + sin2 φ

) = ρ

hz =⎛

⎝001

⎞

⎠ = hz = k

(1.17)

and also the inverse mapping

ρ = √x2 + y2

φ = arctany

x,

where the angle requires again the consideration of the quadrant.


If the curve is expressed by ρ = ρ(t), φ = φ(t) and z = z(t), then the tangentvector is determined by

T =3∑

i=1

qi · ‖hi‖hi = ρ hρ + φ · ρ hφ + z hz (1.18)

If the geometry is limited to the plane z = 0, the corresponding system is alsoknown as polar coordinate system. In this case, very often the angle is used ascurve parameter, which leads to the tangent vector

T = ρ ′ hρ + ρ hφ (1.19)

with ρ ′ = dρdφ .

• For all other systems, the qi -notation is replaced by the coordinates α, β and γ .

Questions

In particular, the following problems are investigated in the exercises:

• How to calculate the arc length in Cartesian and curvilinear coordinates?• How to express the tangent vector in another coordinate system?• How to find a parametric representation for a curve?• How to verify that a curve is lying on a given surface?

Exercises

Curves in the Plane

1. Calculate the arc length of the curve

�(t) =(ln√1 + t2, arctan t

)�for 0 ≤ t ≤ 2.

2. Determine the arc length of the closed curve

�(t) = (5 cos t − 2 cos 5t2 , 5 sin t − 2 sin 5t

2

)�.

The curve contains multiple points, which must be considered for finding theinterval length.

3. Calculate the arc length of the curve, which is given in polar coordinates withthe radius

ρ(φ) = 1

| cosφ| + | sin φ| for φ ∈ [0, 2π ].


4. Use polar coordinates to parametrize the closed curve, which is given in theimplicit form

(2x + x2 + y2)2 = 4(x2 + y2)

and determine the arc length s.5. Given two points B1 = (1, 0)� and B2 = (−1, 0

)�in the xy-plane. The curve

C consist in all points X = (x, y)�, where the product of the Euclidean distancesXB1 · XB2 = c is constant.

(a) Find a parametric representation of the curve for c ∈ R+ by introducing

polar coordinates. Use the addition theorem of cosine for simplification.(b) Consider now the case c = 1 and express the tangent vector T in terms

of normalized polar coordinates { hr , hφ} and simplify the term√T�T

without calculating the arc length.

6. Given the implicit curve

3x2 + |y|3 − y2 = 0.

Find three different parametrizations and calculate for each of them the arclength.

Arc Length in Cartesian, Spherical and Cylindrical Coordinates

7. Verify, that the curve

�(t) =(sin t, sin2 t

2 , 12 (t − sin t cos t)

)�

is already parametrized w.r.t. its arc length.8. Given the curve

�(t) = t hρ + hφ + (t2 + 1) hz t ∈ R+0

in cylindrical coordinates with ρ = √t2 + 1 and φ = t − arctan t + π

2 .

(a) Determine the arc length s.(b) Verify, that the curve is lying on the paraboloid x2 + y2 − z = 0.(c) Re-write the curve in Cartesian representation with the arc length s as

parameter.

Exercises 9

9. Investigate the curve

�(t) = (2t3 − 2t, 4t2, t3 + t)�

t ∈ R+0 .

(a) Express the tangent vector T in spherical coordinates in terms of nor-malized ‘frame vectors’ hqi .

(b) Verify, that all points of the curve are lying on a quadric surface Q anddetermine its type and coefficients. The general form of a quadric is givenby

Q :3∑

i=1

3∑

k=1

aik xi xk + 23∑

i=1

bi xi + c = 0 (x = x1, y = x2, z = x3).

Conclusions based on previous results/geometry are strongly recom-mended!

10. Verify, that for every chosen parameter a ∈ [−1, 1] the Villarceau curves

�a =⎛

⎝ra + Ra cos t + √

R2 − r2√1 − a2 sin t

r√1 − a2 + R

√1 − a2 cos t − √

R2 − r2a sin tr sin t

⎞

⎠ , t ∈ [0, 2π ]

consist in circles on a torus (x2 + y2 + z2 + R2 − r2)2 − 4R2(x2 + y2) = 0with R > r .

11. Given the curve

� =⎛

⎝3 sin t + sin 3t3 cos t + cos 3t√

12 sin t

⎞

⎠ .

(a) Calculate the arc length for t ∈ [0, 2π ].(b) Prove, that the curve has a constant distance to the origin 0.

(c) Express the tangent vector via the formula T =3∑

i=1qi‖hqi ‖ hqi in spher-

ical coordinates and simplify the expressions. Explicit determination ofthe ‘frame vectors’ hqi is not necessary here.

12. A curve is called a ‘loxodrome’ �S(t) of the surface S, if the intersection anglesbetween the curve and the (orthogonal) parameter lines are constant. Derive arepresentation of a loxodrome on the unit sphere, which is parametrized w.r.t.longitude and co-latitude, and determine the arc length s.


Arc Length in Alternative and Modified Coordinate Systems

13. Given the curve � = (10 sin t) hρ + 8 cos t hz with φ = 3t5 in cylindrical coor-

dinates.

(a) Calculate the tangent vector T and the arc length s without using Carte-sian expressions.

(b) Determine the coordinates {α(t), β(t), γ (t)} and the (positive) scalingparameter p for this curve in (modified) oblate spheroidal coordinates

x = p cosh α sin β sin γ

y = p cosh α sin β cos γ

z = p sinh α cosβ

and express the tangent vector T in terms of non-normalized ‘framevectors’ without calculating hqi explicitly. All points of the curve are onthe surface α = const.

14. Calculate the arc length of the curve

� =(

2t√1 + t2

+ t2)

hα +(

2t2√1 + t2

− 1

)hβ

with β = 1, α = t and γ = π2 − t given in parabolic coordinates:

x = αβ cos γ, y = αβ sin γ, z = α2 − β2

2.

15. The relationship between Cartesian coordinates and a set of curvilinear coordi-nates (α, β, γ ) is given by

x = αβ

(α2 + β2)2cos γ

y = αβ

(α2 + β2)2sin γ

z = α2 − β2

2(α2 + β2)2.

(a) Determine the tangent vector T and the arc length s of the ‘meridian’with α = const. and γ = const. for β = [0, B].

(b) Verify that these meridians are cardioid curves which fulfill the equation

x2 + y2 + z2 = c(√

x2 + y2 + z2 + z)and determine c.

16. The curve is given by the coordinates α = 0, β = t and γ = t in the coordinatesystem

Exercises 11

x = √2

α(sin β − cosβ

) 1

cosh γ

y = √2

α(cosβ + sin β

) 1

cosh γ

z = √2

α+1tanh γ.

(a) Derive/note down all normalized ‘frame vectors’ of this system.(b) Calculate the arc length s.

Solutions

1.1. Arc Length of �(t) =(ln

√1 + t2, arctan t

)�

We differentiate the curve to obtain the tangent vector and its inner product:

T =(

t

1 + t2,

1

1 + t2

)�

T�T =(

t

1 + t2

)2

+(

1

1 + t2

)2

= 1

(1 + t2).

Based on the inner product we calculate the arc length:

s =2∫

0

√T�Tdt =

2∫

0

1√1 + t2

dt =[arsinh t

]2

0= arsinh 2 = ln

(2 + √

5)

.

1.2. Closed Curve: �(t) = (5 cos t − 2 cos 5t

2 , 5 sin t − 2 sin 5t2

)�

First, we have to determine the length of the interval so that the curve is closed. Weassume t0 = 0 as the starting point with the coordinates �(0) = (3, 0)�. The curvewill be closed when this point is reached again.Taking the square of the radius, we can reformulate the condition:

x2 + y2 =(5 cos t − 2 cos

5t

2

)2+(5 sin t − 2 sin

5t

2

)2 != 32 + 02

25(cos2 t + sin2 t

)+ 4

(cos2

5t

2+ sin2

5t

2

)− 20

(cos t cos

5t

2+ sin t sin

5t

2

)!= 9

−20 cos

(5t

2− t

)= −20 cos

3t

2= −20.


The curve can only be closed for t = 4π3 k, where k is an integer number. We try

the first multiples and find the curve to be closed for k = 3, which is equivalent tot = 4π .To ensure that it is not a multiple point on the curve, we calculate the tangent vector

T = (−5 sin t + 5 sin 5t2 , 5 cos t − 5 cos 5t

2

)�

with T (0) = T (4π) = (0, 0)�. Since both the coordinates and the tangential vectorare identical in this point, the curve is closed for the interval t ∈ [0, 4π ], which canbe seen also from Fig. 1.2.For the arc length we simplify the inner product:

T�T =(

−5 sin t + 5 sin5t

2

)2+(5 cos t − 5 cos

5t

2

)2=

= 25(cos2 t + sin2 t

)+ 25

(sin2

5t

2+ sin2

5t

2

)− 50

(sin t sin

5t

2+ cos t cos

5t

2

)=

= 50

[1 −

(sin t sin

5t

2+ cos t cos

5t

2

)]= 50

[1 − cos

3t

2

]

and use the trigonometric identity√1 − cosφ =

∣∣∣√2 sin φ

2

∣∣∣ for the arc length:

Fig. 1.2 Closed curve �(t) =(5 cos t − 2 cos 5t

2 , 5 sin t − 2 sin 5t2

)�(exercise 2)

Solutions 13

s =∫ √

T�Tdt =4π∫

0

5√2

√

1 − cos3t

2dt = 5

√2

4π∫

0

∣∣∣∣√2 sin

3t

2 · 2∣∣∣∣ dt =

= 10

⎛

⎜⎝

4π/3∫

0

sin3t

4dt −

8π/3∫

4π/3

sin3t

4dt +

4π∫

8π/3

sin3t

4dt

⎞

⎟⎠ = 80.

• In particular for an arc length containing sine and cosine expressions, onemust keep in mind:

√f (t)2 = | f (t)| �= f (t).

• In case of expected multiple points, the tangent vector in start and end pointmust be checked.

• The zero vector as tangential vector indicates that the parametrization hasa non-regular point (here: at t= 0), and the curve might have a cusp at thislocation.

1.3. Arc Length of ρ(φ) = 1| cosφ|+| sin φ| in Polar Coordinates

Instead of investigating four cases due to modulus, we prove the symmetry:

1

| cosφ| + | sin φ|τ=[0, π

2 ]=

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

1| cos τ |+| sin τ | = 1

cos τ+sin τφ ∈ [0, π

2

]

1| cos(τ+ π

2 )|+| sin(τ+ π2 )| = 1

|−sin τ |+| cos τ | φ ∈ [π2 , π

]

1| cos(τ+π)|+| sin(τ+π)| = 1

|−cos τ |+|−sin τ | φ ∈ [π, 3π2

]

1| cos(τ+3 π

2 )|+| sin(τ+3 π2 )| = 1

| sin τ |+|−cos τ | φ ∈ [ 3π2 , 2π].

If a point P = (x, y)� is on the curve, then the points P∗ = (±x, ±y)�

are on thecurve as well. Therefore, we can calculate the arc length in the first quadrant withτ ∈ [0, π

2

]andmultiply the result by 4. In addition, we use the trigonometric identity

cos x + sin x = √2 sin

(x + π

4

)and find the arc length:

s =2π∫

0

√(ρ′(φ))2 + (ρ(φ))2dφ = 4

π/2∫

0

√√√√√

(1√

2 sin(φ + π

4

)

)2+⎛

⎝−√2 cos

(φ + π

4

)

√22sin2

(φ + π

4

)

⎞

⎠

2

dφ =

= 4

π/2∫

0

√√√√

1√22sin2

(φ + π

4

)

√1 + cot2

(φ + π

4

)

︸︷︷︸√sin2

(φ+ π

4

)−1

dφ = 4

π/2∫

0

1√2

1

sin2(φ + π

4

)dφ =

= 4√2

[− cot

(φ + π

4

)]π/2

0= 4

√2.


Fig. 1.3 Square in polarcoordinates (exercise 3)

• The curve is a square in polar coordinates with the corner points{(±1, 0)�

,(0, ±1

)�}, which is shown in Fig. 1.3. Hence, Cartesian coordi-

nates would be better suited for this problem. The solution demonstrates thatthe arc length can be calculated in any coordinate system, but with differentamount of effort.

• Integration of triangles and squares in polar coordinates has a practical appli-cation for boundary elements. In this method, the (gravitational) potential ismodeled by single layers on the planar surface elements, and the integralsbecome singular for points on the element. One could overcome the singular-ity in some cases by introducing polar coordinates in the integrals.

• Recognizing (and proving) symmetry reduces the calculation effort. Ignoringsymmetries would lead to

ρ ′(φ) = cosφ sin φ

1| cosφ| − 1

| sin φ|(| cosφ| + | sin φ|)2

and four cases in the integration.

Solutions 15

1.4. Length of the Implicit Curve (2x + x2 + y2)2 = 4(x2 + y2)

We insert polar coordinates into the equation to find the radius

(2ρ cosφ + ρ2)2 = 4ρ2

ρ2 + 4ρ cosφ + 4 cos2 φ − 4 = 0

ρ = −2 cosφ ± 2.

Both solutions lead to a curve which fulfills the implicit equation. We simplify theintegrand

√T�T =

√ρ2 + ρ ′2 =

√(−2 cosφ ± 2)2 + (2 sin φ)2 =

=√4(cos2 φ + sin2 φ) + 4 ∓ 8 cosφ = √

8√1 ∓ cosφ =

=

⎧⎪⎪⎨

⎪⎪⎩

√8∣∣∣√2 sin φ

2

∣∣∣

√8∣∣∣√2 cos φ

2

∣∣∣

and calculate the two arc lengths:

s1 =2π∫

0

√8

∣∣∣∣√2 sin

φ

2

∣∣∣∣ dφ = 4

[−2 cos

φ

2

]2π

0

= 16,

s2 =2π∫

0

√8

∣∣∣∣√2 cos

φ

2

∣∣∣∣ dφ = 4

[2 sin

φ

2

]π

0

− 4

[2 sin

φ

2

]2π

π

= 16.

• A radius is usually assumed to be positive in polar coordinates. If negativevalues are obtained by calculations, then this solution should be investigatedcarefully. It can be either a artifact (e.g. exercise 43) or a real solution. Inthe later case it might be better to replace the word ‘radius’ by a neutralexpression like ‘curve parameter’.

• The two curves are known as Cardioids and they are visualized in Fig. 1.4together. In the geometrical definition, these curves are created by tracing amarked point on a circle, which is rolling around a fixed circle of the sameradius. Therefore, the curves belong to the family of epicycloids.

• In polar coordinates, we obtain two curves with different radii. There mightbe other parameterizations which describe only one curve with the total arclength s = 32.


Fig. 1.4 Two Cardioidcurves with aparametrization in polarcoordinates (exercise 4)

1.5. Different Parametrizations of 3x2 + | y|3 − y2 = 0

All terms occur either in squared formorwith themodulus, i.e. the curve is symmetricw.r.t x- and y-axis, which can be seen also in Fig. 1.5. Therefore, we can determinethe arc length only in the first quadrant and multiply the result by 4. In this concept,we can ignore the modulus for the following.

a. Parametrization in the Variable y

We use y as a variable and solve the equation

x = ± 1√3

√y2 − y3.

The derivative is then found by

dx

dy= (2 − 3y)y

2√3√y2 − y3

with the arc length

s = 4

1∫

0

√(dx

dy

)2

+ 1dy = 4

1∫

0

√√√√(

(2 − 3y)y

2√3√y2 − y3

)2

+ 12dy =

= 4

1∫

0

√(4y2 − 12y3 + 9y4) + 12(y2 − y3)

2√3√y2 − y3

dy =[4(y − 2)(y − 1)

√(4 − 3y2)√

3√1 − y(3y − 4)

]1

0

= 8√3.

Solutions 17

Fig. 1.5 Closed figure:3x2 + |y|3 − y2 = 0(exercise 5)

In this particular case, the integration is performed and evaluated withMathematica.

b. Rational Parametrization by x = y · tTheorigin0 = (0, 0)� is lying on the curve.Hence,we can try a rational parametriza-tion x = y · t

3y2t2 + y3 − y2 = 0

y = 1 − 3t2

⇒ x = t − 3t3.

In the first quadrant, the coordinates {x, y} are both positive and so we find the

interval t ∈[0, 1√

3

]and the arc length

s = 4∫ √

x2 + y2 dt = 4

1√3∫

0

√(1 − 9t2)2 + (−6t)2dt = 4

1√3∫

0

√(9t2 + 1)2dt = 8√

3.

c. Rational Parametrization by y = x · tA second rational parametrization is found by y = x · t


3x2 + x3t3 − x2t2 = 0

x = t2 − 3

t3= 1

t− 3

t3

⇒ y = 1 − 3

t2

with the inner product

T�T =(

− 1

t2+ 9

t4

)2

+(6

t3

)= 1

t4t4

t4− 18

t6t2

t2+ 81

t8+ 36

t6t2

t2= (t2 + 9)2

t8

and the arc length

s = 4

∞∫

0

√T�Tdt = 4

∞∫

0

t2 + 9

t4dt = 4

[−1

t− 3

t3

]∞√3

= 8√3.

Below we sketch some more parametric representations without calculations:

1. A rational parametrization is possible for every rational point of the curve, inparticular for

(0, 0

)�,(0, ±1

)�and

(± 29 , ± 2

9

)�.

2. By introducing polar coordinates we obtain

3ρ2 cos2 φ + ρ3| sin φ|3 − ρ2 sin2 φ = 0

⇒ ρ(φ) = 1

| sin φ|3(sin2 φ − 3 cos2 φ

)= 4 sin2 φ − 3

sin3 φ.

3. We can also solve the algebraic equation of order 3 in a closed form:

y(x) = 1

3

⎛

⎝3√

−81x2 + 9√81x4 − 4x2 + 23√2

+3√2

3√

−81x2 + 9√81x4 − 4x2 + 2

+ 1

⎞

⎠ .

• A curve might have different parametric representations and the calculationeffort for arc length (or line integral, area, geometrical center…) can differsignificantly.

• It should be pointed out that not every implicit curve has a parametric repre-sentation.

• The idea of the rational parametrization exploits the relationship between a(common) curve point with rational coordinates and a family of straight linespassing through the common point and the other points of the curve. Thisparametrization often leads to simpler integrals than a polar representation.

Solutions 19

1.6. Constant Product of Euclidean Distances

a. Parametrization in Polar Coordinates

We insert polar coordinates into the constant product condition of the Euclideandistances to the points B1 = (1, 0)� and B2 = (−1, 0

)�and get a equation for the

radius ρ:

√(x − 1)2 + y2

√(x + 1)2 + y2 = c

((ρ cosφ − 1)2 + ρ2 sin2 φ

)((ρ cosφ + 1)2 + ρ2 sin2 φ

)= c2

(ρ2 − 2ρ cosφ + 1

)(ρ2 + 2ρ cosφ + 1

)= c2

ρ4 + 2ρ2 − 4ρ2 cos2 φ + 1 = c2.

By using the trigonometric identity (− cos 2φ) = (−2 cos2 φ + 1), we get

ρ4 − 2ρ2 cos 2φ − (c2 − 1) = 0.

The squared radius is determined from the quadratic equation in the variable ρ2:

ρ2 = cos 2φ ±√cos2 2φ − (1 − c2),

⇒ ρ(φ) = ±√cos 2φ ±

√cos2 2φ − (1 − c2).

b. Tangent Vector for c = 1

In the case c = 1, we can eliminate the inner root (for cos 2φ ≥ 0)

ρ(φ) = ±√2√cos 2φ

ρ ′ = ±√2

−2 sin 2φ

2√2 cos 2φ

= ∓√2

sin 2φ√cos 2φ

and obtain with the formula (1.19) the tangent vector

T = ∓√2

sin 2φ√cos 2φ

hρ ± √2√cos 2φ hφ

√T�T =

√

2sin2 2φ

cos 2φ+ 2(cos 2φ) = √

2

√sin2 2φ + cos2 2φ

cos 2φ= √

21√

cos 2φ.

The parametrization is valid for cos 2φ ≥ 0, which leads to 3 intervals[0, π

4

],[

3π4 , 5π

4

], and

[7π4 , 2π

]for the curve parameter φ.


Fig. 1.6 Cassini ovals for c = {1, 2, 5, 10} (exercise 6)

• The curves are known as Cassini ovals in the general case for c ∈ R andfour of the curves are visualized in Fig. 1.6.

• For c = 1 we get the Lemniscate of Bernoulli. Its arc length leads to the

elliptic integral of Gauß: s = 4√2

1∫

0

dt√1−t4

= 2√2� .

1.7. Parametrization of a Curve w.r.t. its Arc Length:

�(t) =(sin t, sin2 t

2 , 12(t − sin t cos t)

)�

A curve is parametrized w.r.t. its arc length, if the norm of the tangent vector ‖T‖ isequal to one for all points on the curve:

Solutions 21

T = (cos t, sin t cos t, 12 (1 − cos2 t + sin2 t)

)� = (cos t, sin t cos t, sin2 t)�

‖T‖2 = cos2 t + sin2 t cos2 t + sin4 t = cos2 t + sin2 t (cos2 t + sin2 t) = 1 ��

• The parametrization w.r.t. the arc length leads to easier expressions for furtherquantities in differential geometry like the curvature of a curve. In vectoranalysis, the re-parametrization is usually not performed.

1.8. Curve on a Paraboloid: �(t) = t hρ + hφ + (t2 + 1) hz

a. Arc Length

We calculate the derivatives of the coordinates

ρ = √t2 + 1 ⇒ ρ = t√

1+t2

φ = t − arctan t + π2 ⇒ φ = 1 − 1

1+t2

z = t2 + 1 ⇒ z = 2t

and the tangent vector

T = t√t2 + 1

hρ +(1 − 1

1 + t2

)√t2 + 1 hφ + 2t hz

T�T = t2

(t2 + 1)+ t4

1 + t2+ 4t2 = 5t2.

Hence, we obtain the arc length

s =∫ √

T�Tdt =∫ √

5t2dt =√5

2t2.

b. On a Paraboloid

• To prove, whether a curve is lying on a certain surface, we have to insert itscomponents into the surface equation and achieve a true expression.

In this case, we obtain

x2 + y2 − z = (t cos t − sin t)2 + (t sin t + cos t)2 − (t2 + 1) = (t2 + 1) − (t2 + 1) = 0,

so that all points are lying on the paraboloid (cf. Fig. 1.7).


Fig. 1.7 Curve on the paraboloid x2 + y2 − z = 0 (exercise 8)

c. Re-write w.r.t. Arc Length

For the parametrization w.r.t. the arc length, we invert the relation s =√52 t2 and

insert t =√

2s√5into the original curve:

�(s) =

⎛

⎜⎜⎜⎜⎝

√2s√5cos√

2s√5

− sin√

2s√5√

2s√5sin√

2s√5

+ cos√

2s√5

(√2s√5

)2 + 1

⎞

⎟⎟⎟⎟⎠

.

1.9. Curve �(t) = (2t3 − 2t, 4t2, (t3 + t)

)�Lying on a

Quadric

a. Spherical Coordinates

The relation between the spherical and the Cartesian coordinate system provide us:

r2 = (2t3 − 2t)2 + (4t2)2 + (t3 + t)2 = 5t6 + 10t4 + 5t2 == 5t2(t4 + 2t2 + 1) = 5t2(t2 + 1)2

⇒ r = t√5(t2 + 1) = √

5(t3 + t), ϑ = arccost3 + t

t√5(t2 + 1)

= arccos1√5.

λ = arccot2t3 − 2t

4t2= arccot

t2 − 1

2t.

Solutions 23

The longitude might by expressed by arctan-function as well, the arccot-function ischosen here to avoid the problem for t = ±1.The derivatives of the coordinates are given by

r = √5(3t2 + 1),

ϑ = 0,

λ = − 1

1 +(t2−12t

)22t (2t) − 2(t2 − 1)

4t2= − 4t2

4t2 + t4 − 2t2 + 1

2(t2 + 1)

4t2= − 2

1 + t2,

which leads to the tangent vector

T = √5(3t2 + 1) hr + √

5(t3 + t)

√

1 − 1

5

−2

1 + t2hλ = √

5(3t2 + 1) hr − 4t hλ

in spherical coordinates.

b. Quadric

The quadric form

Q :3∑

i=1

3∑

k=1

aik xi xk + 23∑

i=1

bi xi + c = 0 (x = x1, y = x2, z = x3)

has at a first glance up to 13 unknowns {aik, bk, c}. From linear algebra it is knownthat the matrixA = [aik] is symmetric, which reduces the problems to 10 unknowns.Now we have a closer look on the geometry:

• We find c = 0 as the point 0 = (0, 0, 0)� is lying on the curve.• The curve is on a circular cone because of the constant co-latitude for all points:

– There are no mixed terms (aik)i �=k due to circular symmetry.– In circular cones we get also a11 = a22.– The curve is centered around the origin and so there are no translation terms:bi = 0 for i = 1, 2, 3.

The combination of these facts leads to the reduced quadric form:

a11x2 + a11y

2 + a33z2 = 0

with only two unknowns. We insert one point of the curve e.g. �(1) = (0, 4, 2)�,to find a relation between a11 and a33:

a11(02 + 42) + a33(2

2) = 16a11 + 4a33 = 0

⇒ a33 = −4a11.


As a last check, we insert now the parameter form of the curve, to make sure, thatall points are on the cone:

x2 + y2 − 4z2 = a11[(2t3 − 2t)2 + (4t2)2

]− 4a11

[(t3 + t)2

]=

= a11[4t6 − 8t4 + 4t2 + 16t4 − 4t6 − 8t4 − 4t2

]= 0. ��

• The study of quadratic form or quadric is usually performed in courses onlinear algebra. The type is determined by the sign of the eigenvalues. Viatranslation, rotation, and re-scaling the so-called normal form of the quadricform is achieved.

• Prominent examples of the quadratic form in 3D space are spheres, ellipsoids,cones, cylinder, paraboloids and hyperboloids.

1.10. Villarceau Circles

The investigation of the Villarceau circles is carried out in the following steps:

1. We verify that all points are lying on a torus with the given equation.2. We prove that all points are lying on a sphere by calculating the curvature.3. We prove that the curve is planar.

Lying on a Torus

With the abbreviations σ := √R2 − r2 and τ = √

1 − a2 we first calculate the termρ2 = x2 + y2, which occurs twice in the torus equation:

ρ2 = r2a2 + R2a2 cos2 t + σ 2τ2 sin2 t + 2ra2R cos t + 2raστ sin t + 2Raστ cos t sin t++ r2τ2 + R2τ2 cos2 t + σ 2a2 sin2 t + 2rτ2R cos t − 2rτσa sin t − 2Rσaτ cos t sin t =

= r2(a2 + 1 − a2)+R2(a2 + 1 − a2) cos2 t+σ 2(1 − a2 + a2) sin2 t+2r R(a2 + 1 − a2) cos t == (r cos t + R)2.

In the first sum, we add the term (z2 + R2 − r2)

ρ2 + z2 + R2 − r2 = (r2 cos2 t + 2r R cos t + R2) + r2 sin2 t + R2 − r2 == 2r R cos t + 2R2

and insert the result into the torus equation:

(ρ2 + z2 + R2 − r2)2 − 4R2(ρ2) = (2r R cos t + 2R2)2 − 4R2(r cos t + R)2 = 0. ��

Solutions 25

Fig. 1.8 Villarceau circle on a torus (exercise 10)

As this equation is fulfilled for all parameter t and every choice of a, every point ofthese curves is lying on a torus (Fig. 1.8).

Radius of the Circle

To verify, that a curve is lying on a sphere, we can either consider the equation

(x − mx )2 + (y − my)

2 + (z − mz)2 = r2

with the unknown center M = (mx , my, mz)�

and the radius r , or we demonstratethat the curvature

κ = ‖� × �‖‖�‖3

is constant and positive. In the latter method, we don’t have to determine the center.The cross product of first and second derivatives of the curve delivers

� × � =⎛

⎝−Ra sin t + σ

√1 − a2 cos t

−R√1 − a2 sin t − σa cos t

r cos t

⎞

⎠×⎛

⎝−Ra cos t − σ

√1 − a2 sin t

−R√1 − a2 cos t + σa sin t

−r sin t

⎞

⎠ =

=⎛

⎝R√1 − a2r

−Rar−Rσ

⎞

⎠

which leads to the curvature

κ =√

(1 − a2)r2R2 + a2R2r2 + R2(σ 2)√

(R2(a2 + 1 − a2) sin2 t + σ 2(1 − a2 + a2) cos2 t + r2 cos2 t)3 = R2

R3.


The curve has a positive constant curvature, so all the points are lying on a spherewith radius 1

κ= R.

Lying in a Plane

Finally we prove that all points are lying in a plane, which is depending on theparameter a.We select three points on the curve, and use their vectors to span up the plane E . Forsimplicity we choose points where either sine or cosine-terms vanish:

�(0) =⎛

⎝ra + Ra

r√1 − a2 + R

√1 − a2

0

⎞

⎠ �(π) =⎛

⎝ra − Ra

r√1 − a2 − R

√1 − a2

0

⎞

⎠

�(0.5π) =⎛

⎝ra + σ

√1 − a2

r√1 − a2 − σa

r

⎞

⎠ .

These points are used to find the normal vector of the plane

N =(�(π) − �(0)

)×(�(0.5π) − �(0)

)=

=⎛

⎝−2Ra

−2R√1 − a2

0

⎞

⎠×⎛

⎝σ√1 − a2 − Ra

−σa − R√1 − a2

r

⎞

⎠ = 2R

⎛

⎝−r

√1 − a2

raσ

⎞

⎠

which leads to the plane equation

E :[

− r√1 − a2

]x +

[ra]y +

[σ]z = const.

Now we want to ensure that all points are in the plane which is spanned by these 3vectors. Hence we insert the curve components into the plane equation:

[− r√1 − a2

] (ra + Ra cos t + σ

√1 − a2 sin t

)+

+[ra] (

r√1 − a2 + R

√1 − a2 cos t − σa sin t

)+[σr](

sin t)

== −rσ(1 − a2) sin t − a2rσ sin t + σr sin t = 0. ��

The curve is on a sphere and a plane, hence, the points are lying on a circle.

Solutions 27

1.11. Curve on a Sphere:

� =(3 sin t + sin 3t, 3 cos t + cos 3t,

√12 sin t

)�

a. Arc Length

We calculate the tangent vector and simplify the inner product by trigonometricidentities:

T =⎛

⎝3 cos t + 3 cos 3t−3 sin t − 3 sin 3t√

12 cos t

⎞

⎠

T�T = 9(cos2 t+ sin2 t)+9(cos2 3t + sin2 3t)+18 cos t cos 3t+18 sin t sin 3t+12 cos2 t==18+18 cos 2t+12 cos2 t=18 cos2 t+18 sin2 t+18 cos2 t − 18 sin2 t+12 cos2 t == 48 cos2 t.

For the arc length, we must consider the modulus

s =2π∫

0

√48 cos2 tdt = 4

√3

2π∫

0

| cos t |dt = 4√3

([sin t

]π/2

0−[sin t

]3π/2

π/2+[sin t

]2π

3π/2

)=

= 16√3.

b. Distance to the Origin

The distance to the origin is equivalent to the radius in spherical coordinates:

r2 = 9(sin2 t + cos2 t) + (sin2 3t + cos2 3t) + 6 sin t sin 3t + 6 cos t cos 3t︸︷︷︸6 cos(3t−t)

+12 sin2 t =

= 10 + 6 cos 2t + 12 sin2 t = 10 + 6 cos2 t − 6 sin2 t + 12 sin2 t == 10 + 6(cos2 t + sin2 t) = 16.

All points of the curve are lying on a sphere around the origin 0 with the radiusr = √

16 = 4 (cf. Fig. 1.9).

c. Tangent Vector in Spherical Coordinates

The tangent vector in spherical coordinates is given by formula (1.15). In the previousstep, we found already r = 4 and r = 0. We calculate the co-latitude ϑ by

ϑ = arccosz

r= arccos

√12 sin t

4= arccos

√3 sin t

2

⇒ ϑ = − 1√

1 −(√

3 sin t2

)2

√3 cos t

2= −

√3 cos t

√4 − 3 sin2 t


Fig. 1.9 Curve on a sphere: � =(3 sin t + sin 3t, 3 cos t + cos 3t,

√12 sin t

)�(exercise 11)

and consider sin ϑ = √1 − cos2 ϑ =

√1 − 3 sin2 t

4 .

For the longitude, we differentiate the expression λ = arctan yx in general:

λ = 1

1 + y2

x2

x y − yx

x2= x y − yx

x2 + y2.

It might be necessary to add ±π to the longitude for the correct quadrant, but thiswill not affect the derivative λ. We insert the components of the curve � into the lastformula

λ = (3 sin t + sin 3t)(−3 sin t − 3 sin 3t) − (3 cos t + cos 3t)(3 cos t + 3 cos 3t)

10 + 6 cos 2t=

= −12 + 12 sin t sin 3t + 12 cos t cos 3t

10 + 6 cos 2t= −6 + 6 cos 2t

5 + 3 cos 2t

and obtain the tangent vector

T = −6 + 6 cos 2t

5 + 3 cos 2t4

√

1 − 3 sin2 t

4hλ −

√3 cos t

√4 − 3 sin2 t

4 hϑ .

Solutions 29

1.12. Loxodrome on the Sphere

The angle ψ(t) between two curves �1(t) and �1(t) is determined by the innerproduct of the normalized tangent vectors

1

‖�1‖�1

� · 1

‖�2‖�2 = cosψ(t).

In the following, we present the calculation in spherical and Cartesian coordinates tohighlight again the advantage of using adequate coordinates. (Even more adequateis the introduction of cylindrical coordinates, to find the loxodrome curves of allrotational surfaces.)Every curve on a unit sphere can be re-written in the form

�1 = hr =⎛

⎝cos λ(t) sin ϑ(t)sin λ(t) sin ϑ(t)

cosϑ(t)

⎞

⎠

with differentiable functions λ(t) and ϑ(t). The condition of a loxodrome providesa relation between these two functions. Hence, we choose one of them to be theparameter of the curve, e.g. ϑ(t) := t .We obtain the tangent vector

�1 = T = λ sin t hλ + hϑ =⎛

⎝−λ sin λ sin t + cos λ cos tλ cos λ sin t + sin λ cos t

− sin t

⎞

⎠ .

The parameter line λ = t of constant co-latitude ϑ = ϑ0 is a circle

�2 =⎛

⎝cos t sin ϑ0

sin t sin ϑ0

cosϑ0

⎞

⎠ = hr

with the tangent vector

�2 = sin ϑ0 hλ =⎛

⎝− sin t sin ϑ0

cos t sin ϑ0

0

⎞

⎠ .

We calculate the inner product

1

‖�1‖�1

� · 1

‖�2‖�2 = 1

√λ2 sin2 t + 1

(λ sin t hλ + hϑ

)� · 1√sin2ϑ0

sin ϑ0 hλ =

= λ sin t√

λ2 sin2 t + 1.


The intersection angle is constant for

cosψ = λ sin t√

λ2 sin2 t + 1= const.,

which provides the relation

λ = c

sin t

⇒ λ = c∫

1

sin tdt = c

∫1 + u2

2u

2

1 + u2du = c ln

∣∣∣∣tan

t

2

∣∣∣∣+ λ0,

using the substitution ofWeierstraß (u = tan t2 ) for the integration. Hence, we obtain

the expression

�� =⎛

⎝cos(c ln∣∣tan t

2

∣∣+ λ0)sin t

sin(c ln∣∣tan t

2

∣∣+ λ0)sin t

cos t

⎞

⎠

with c√c2+1

= cosψ for the spherical loxodrome. The arc length can be read fromthe previous nominator

s =∫ √

T�Tdt =∫ √

λ2 sin2 t + 1dt =√c2 + 1t.

• A loxodrome on the sphere is visualized in Fig.1.10 and by straight lines inthe Mercator projection.

Fig. 1.10 Sphericalloxodrome with λ0 = 0 andc = 8 (exercise 12)

Solutions 31

• There are other parametric representations due to different reasons:

– other spherical coordinates, e.g. latitude instead of co-latitude,– alternative solution of the integral

∫1

sin t dt ,– or by using the longitude λ := t as curve parameter.

1.13. Arc Length of � = (10 sin t) hρ + 8 cos t hz

a. Arc Length

Based on the expression in cylindrical coordinates, we can identify:

ρ = 10 sin t ⇒ ρ = 10 cos t,

φ = 3t5 ⇒ φ = 3

5 ,

z = 8 cos t ⇒ z = −8 sin t.

We determine the tangent vector

T = 10 cos t · 1 hρ + 3

5· 10 sin t hρ − 8 sin t · 1 hz

T�T = 100 cos2 t + 36 sin2 t + 64 sin2 t = 100

and the arc length

s =∫ √

100dt = 10t.

b. Oblate Spheroidal Coordinates

We have to compare the Cartesian expression (based on the cylindrical system) andthe oblate spherical coordinates:

10 sin t cos3t

5!= p cosh α sin β sin γ,

10 sin t sin3t

5!= p cosh α sin β cos γ,

8 cos t!= p sinh α cosβ.

Based on the (squared) polar distance

x2 + y2 = 100 sin2 t

(sin

3t

5+ cos

3t

5

)!= p2 cosh2 α sin β2

(sin2 γ + cos2 γ

),


Fig. 1.11 Curve on theellipsoid:� = 10 sin t hρ + 8 cos t hzwith φ = 3t

5 (exercise 13)

we find the coordinates γ = π2 − 3t

5 and β = t . The last conditions p cosh α = 10and p sinh α = 8 are fulfilled by

tanh α = 4

5⇒ α = artanh

4

5,

p = 10

cosh α⇒ p = 6.

The values α and p are constant for this curve. Hence, all points are lying completyon the ellipsoid (cf. Fig. 1.11), which is defined by these parameters.The tangent vector in terms of non-normalized ‘frame vectors’ is

T = 1hβ − 3

5hγ .

• oblate spheroidal coordinates:The coordinate surface for fixed values of {p, α} is a rotational ellipsoid:

x2 + y2

p2 cosh2 α+ z2

p2 sinh2 α= sin2 β + cos2 β = 1.

The value of α is responsible for the ratio between semi-major and semi-minor axes, while p provides an up- or downscaling of the complete figure.In a similar way, constant values of {p, β} lead to one-sheeted hyperboloids

x2 + y2

p2 sin2 β− z2

p2 cos2 β= cosh2 α − sinh2 α = 1.

• Compared to other textbooks, the sine- and cosine terms are interchanged,which doesn’t affect the principle.

Solutions 33

1.14. Arc Length in Parabolic Coordinates

‘Frame Vectors’ of Parabolic Coordinates

We differentiate the relationship between Cartesian and parabolic coordinates toobtain the ‘frame vectors’ hqi :

hα = 1

‖hα‖

⎛

⎝β cos γ

β sin γ

α

⎞

⎠ = 1√

β2 + α2

⎛

⎝β cos γ

β sin γ

α

⎞

⎠

⇒ ‖hα‖ =√

β2 + α2,

hβ = 1

‖hβ‖

⎛

⎝α cos γ

α sin γ

−β

⎞

⎠ = 1√

β2 + α2

⎛

⎝α cos γ

α sin γ

−β

⎞

⎠

⇒ ‖hβ‖ =√

β2 + α2,

hγ = 1

‖hγ ‖

⎛

⎝−αβ sin γ

αβ cos γ

0

⎞

⎠ =⎛

⎝− sin γ

cos γ

0

⎞

⎠

⇒ ‖hγ ‖ = βα.

Based on the expression β = 1, α = t and γ = π2 − t for the curve, we obtain the

tangent vector

T = 1‖hα‖ hα + 0 hβ − 1‖hγ ‖ hγ

T�T = 12(√

t2 + 12)2 + 12

(1t)2 = 2t2 + 1

and the corresponding arc length

s =∫ √

2t2 + 1 dtt= sinh u√

2==∫ √

2

√1

2

(1 + sinh2 u

) 1√2cosh udu =

= 1√2

∫cosh2 udu = 1√

2

sinh u cosh u − u

2=

= 1√2

sinh u√1 + sinh2 u − u

2= 1√

2

√2t

√1 + 2t2 − arsinh

√2t

2.


Fig. 1.12 Partial parabolicsurfaces with γ ∈ [0, π ] andβ = 1 and α = 2,respectively (exercise 14)

• parabolic coordinates:The coordinate surfaces consist of twopairs ofconfocal paraboloids,whichare rotated around the z-axis, and the half planes with constant value of γ .For a constant value of β, the surface

2z = x2 + y2

β2− β2

describes a confocal paraboloid that open upwards. For a constant value ofα, the surface

2z = − x2 + y2

α2+ α2

describes a confocal paraboloid that open downwards (cf. Fig.1.12).

1.15. Meridian of Cardioid Coordinates

a. Arc Length of the Meridian

Tangential Vectors of the Frame

We differentiate the fractions separately w.r.t. the coordinate α:

Solutions 35

∂

∂α

{β

(α)

(α2 + β2)2

}= β

(α2 + β2)2 − α2(α2 + β2)(2α)

(α2 + β2)4= (−3α2β + β3)

(α2 + β2)3

∂

∂α

{1

2

(α2 − β2)

(α2 + β2)2

}

= 1

2

(α2 + β2)2(2α) − (α2 − β2)2(α2 + β2)(2α)

(α2 + β2)4=(

(−α3 + 3β2α)

(α2 + β2)3

)

.

The derivatives w.r.t. the coordinate β are recognizable the same, apart from a minussign.For the ‘frame vectors’ we get

hα = 1

‖hα‖

⎛

⎜⎜⎝

(−3α2β+β3)

(α2+β2)3cos γ

(−3α2β+β3)

(α2+β2)3sin γ

(−α3+3β2α)

(α2+β2)3

⎞

⎟⎟⎠ =

⎛

⎜⎜⎜⎜⎝

(−3α2β+β3)√(α2+β2)3

cos γ

(−3α2β+β3)√(α2+β2)3

sin γ

(−α3+3β2α)√(α2+β2)3

⎞

⎟⎟⎟⎟⎠

⇒ ‖hα‖ =√(

(−3α2β + β3)

(α2 + β2)3

)2

+(

(−α3 + 3β2α)

(α2 + β2)3

)2

=√

1

(α2 + β2)3,

hβ = 1

‖hβ‖

⎛

⎜⎜⎝

(α3−3β2α)

(α2+β2)3cos γ

(α3−3β2α)

(α2+β2)3sin γ

(−3α2β+β3)

(α2+β2)3

⎞

⎟⎟⎠ =

⎛

⎜⎜⎜⎜⎝

(α3−3β2α)√(α2+β2)3

cos γ

(α3−3β2α)√(α2+β2)3

sin γ

(−3α2β+β3)√(α2+β2)3

⎞

⎟⎟⎟⎟⎠

⇒ ‖hβ‖ =√(−(−α3 + 3β2α)

(α2 + β2)3

)2

+(

(−3α2β + β3)

(α2 + β2)3

)2

= ‖hα‖,

hγ = 1

‖hγ ‖

⎛

⎜⎝

− αβ

(α2+β2)2sin γ

αβ

(α2+β2)2cos γ

0

⎞

⎟⎠ =

⎛

⎝− sin γ

cos γ

0

⎞

⎠

⇒ ‖hγ ‖ = αβ

(α2 + β2)2.

Arc Length

Themeridian in β-direction is the curve with a constant value of α and γ which leadsto the tangent vector

T = 0hα + hβ + 0hγ = 1√

(α2 + β2)3hβ.


For the corresponding arc length, we find

s =B∫

0

√1

(α2 + β2)3dβ

β=α sinh t==arsinh B

α∫

0

1√

α2 + α2 sinh2 t3 α cosh tdt =

arsinh Bα∫

0

α cosh t

α3 cosh3 tdt =

=[1

α2tanh t

]arsinh Bα

0=[

α sinh t

α2α√1 + sinh2 t

]arsinh Bα

0

= B

α2√

(α2 + B2).

b. Cardioid

We calculate the terms

x2 + y2 + z2 = 4α2β2

4(α2 + β2)4+ α4 − 2α2β2 + β4

4(α2 + β2)4= (α2 + β2)2

4(α2 + β2)4= 1

4(α2 + β2)2

√(x2 + y2 + z2) + z = 1

2(α2 + β2)+ α2 − β2

2(α2 + β2)2=

= (α2 + β2)

2(α2 + β2)2+ α2 − β2

2(α2 + β2)2= 2α2

2(α2 + β2)2

of the cardioid equation:

(x2 + y2 + z2) = c(√

x2 + y2 + z2 + z)

which leads to the conclusion c = 14α2 .

• cardioid coordinates:The coordinate surfaces consist of two pairs of cardioids, which are rotatedaround the z-axis, and the half planeswith constant value ofγ .Fora constantvalue of α, the surface

(x2 + y2 + z2) = 1

4α2

(√x2 + y2 + z2 + z

)

describes a cardioid intersecting the positive z-axis. For a constant valueof β, the surface

(x2 + y2 + z2) = 1

4β2

(√x2 + y2 + z2 − z

)

describes a cardioid intersecting the negative z-axis (cf. Fig.1.13).

Solutions 37

Fig. 1.13 Partial cardioidsurfaces with γ ∈ [0, π ] andβ = 0.9 and α = 1.1,respectively (exercise 15)

1.16. Curve on a Sphere

a. ‘Frame Vectors’ of Modified Spherical Coordinates

We differentiate the relationship between Cartesian and the new curvilinear coordi-nates to obtain the ‘frame vectors’ hqi :

hα = 1

‖hα‖ · √2

αln

√2

⎛

⎜⎝

(sin β − cosβ) 1cosh γ

(sin β + cosβ) 1cosh γ√

2 tanh γ

⎞

⎟⎠ = 1√

2

⎛

⎜⎝

(sin β − cosβ) 1cosh γ

(sin β + cosβ) 1cosh γ√

2 tanh γ

⎞

⎟⎠

⇒ ‖hα‖ =(√

2α+1

ln√2),

hβ = 1

‖hβ‖ · √2

α

⎛

⎜⎜⎝

(cosβ + sin β

)1

cosh γ(cosβ − sin β

)1

cosh γ

0

⎞

⎟⎟⎠ = 1√

2

⎛

⎝cosβ + sin β

cosβ − sin β

0

⎞

⎠

⇒ ‖hβ‖ = √2

( √2

α

cosh γ

)

,


hγ = 1

‖hγ ‖ · √2

α

⎛

⎜⎜⎝

−(sin β − cosβ)sinh γ

cosh2 γ

−(sin β + cosβ)sinh γ

cosh2 γ√2 1cosh2 γ

⎞

⎟⎟⎠ = 1√

2

⎛

⎜⎜⎝

−(sin β − cosβ)sinh γ

cosh γ

−(sin β + cosβ)sinh γ

cosh γ√2 1cosh γ

⎞

⎟⎟⎠

⇒ ‖hγ ‖ = √2

( √2

α

cosh γ

)

.

b. Arc Length

Based on the expression α = 0, β = t and γ = t for the curve, we calculate thetangent vector

T = ‖hβ‖ hβ + ‖hγ ‖ hγ =√2

cosh t( hβ + hγ )

T�T =( √

2

cosh t

)2

+( √

2

cosh t

)2

= 4

cosh2 t

and the corresponding arc length

s =∫ √

4

cosh2 tdt = 2

∫2

u + 1u

du

u= 4 arctan et .

• The result of x2 + y2 + z2 =(√

2α)2 (

(2 sin2 β+2 cos2 β)

cosh2 γ+ 2 tanh2 γ

)= 2α+1

is independent of the coordinates {γ, β}. Hence, we have a spherical sur-face for constant values of α. In other words, it is a modified spherical system,where α is responsible for the distance to the origin and β for the ‘longitude’

Chapter 2Differentiation of Field Quantities

Vector analysis deals not only with geometry (like arc length, surface or volume), butalso with field quantities and differential operators. Geometry and field quantity willbe combined for evaluating line integrals and integral theorems in Chaps. 3 and 4.

Definition: From a simple point of view, a scalar or vector field assigns a scalar orvector quantity to every point in an Euclidean space Rn .

Examples for every day scalar fields are temperature, pressure or density of theair, while wind velocity or gravitational attraction are vector fields. A differentialoperator – in this context – is acting on a field quantity to obtain related field quantitiesoffering another point of view.In physical geodesy for example, the gravitational field of the Earth is representedas a scalar field, which is called the gravitational potential. A differentiation of thepotential is performed in several cases:

• For orbit simulation, the gravity is calculated as the gradient of the potential.• Actual and future satellite missions are not observing potential differences but thegradient or the tensor of the gravity field in a (rotating) coordinate system (of thesatellite) or its projection in certain directions.

Gradient of a Scalar Field

Definition: The gradient of a scalar field �(x, y, z) is defined in Cartesian coordi-nates by

∇� = grad �(x, y, z) = ∂�

∂xi + ∂�

∂yj + ∂�

∂zk. (2.1)


39


https://doi.org/10.1007/978-3-030-00416-3_3

https://doi.org/10.1007/978-3-030-00416-3_4

40 2 Differentiation of Field Quantities

For a scalar field in curvilinear coordinates �(q1, q2, q3) the chain rule must beconsidered:

∂�

∂qi= ∂�

∂x

∂x

∂qi+ ∂�

∂y

∂y

∂qi+ ∂�

∂z

∂z

∂qi.

The derivatives ∂(x,y,z)∂qi

can be interpreted as the (non-normalized) ‘frame vectors’:

⎛⎜⎜⎝

∂�∂q1∂�∂q2∂�∂q3

⎞⎟⎟⎠ =

⎛⎜⎜⎝

∂(x,y,z)∂q1

∂(x,y,z)∂q2

∂(x,y,z)∂q3

⎞⎟⎟⎠

⎛⎜⎝

∂�∂x∂�∂y∂�∂z

⎞⎟⎠ =

⎛⎜⎝h�q1

h�q2

h�q3

⎞⎟⎠

⎛⎜⎝

∂�∂x∂�∂y∂�∂z

⎞⎟⎠ .

The equation is inverted

⎛⎜⎝

∂�∂x∂�∂y∂�∂z

⎞⎟⎠ =

(hq1

‖hq1‖ ,hq2

‖hq2‖ ,hq3

‖hq3‖)⎛⎜⎜⎝

∂�∂q1∂�∂q2∂�∂q3

⎞⎟⎟⎠

to obtain the gradient in curvilinear (orthogonal) coordinates:

∇� =3∑

i=1

1

‖hqi ‖∂�

∂qihqi . (2.2)

In particular, we find for cylindrical coordinates

∇� = ∂�

∂ρhρ + 1

ρ

∂�

∂φhφ + ∂�

∂zhz (2.3)

and for spherical coordinates

∇� = ∂�

∂rhr + 1

r

∂�

∂ϑhϑ + 1

r sin ϑ

∂�

∂λhλ. (2.4)

The gradient of a scalar field � is a vector field, which points in the direction of thebiggest increase rate of the scalar field. The corresponding scalar field is called thepotential for some applications.

Notation

A vector field in the Cartesian coordinate system is noted down by

F(x, y, z) =⎛⎜⎝Fx (x, y, z)

Fy(x, y, z)

Fz(x, y, z)

⎞⎟⎠ = Fx (x, y, z) i + Fy(x, y, z) j + Fz(x, y, z) k

while the expression in a curvilinear coordinate system always requires the tangentialvectors of the frame:

2 Differentiation of Field Quantities 41

G(α, β, γ ) = Gα(α, β, γ ) hα + Gβ(α, β, γ ) hβ + Gγ (α, β, γ ) hγ

For a compact presentation, the arguments are often neglected in the following.

Divergence of a Vector Field

Definition: The mapping of a vector field F=Fx i + Fy j + Fz k – given in3-dimensional Cartesian coordinates – onto the scalar quantity

div F = ∂Fx (x, y, z)

∂x+ ∂Fy(x, y, z)

∂y+ ∂Fz(x, y, z)

∂z

is called the divergence of the field F.The divergence can be derived form the gradient for each orthogonal coordi-nate system by using the ‘frame vectors’ (cf. exercise 25 and 23). In particu-lar, the divergence is calculated in cylindrical coordinates with the vector fieldG = Gρ hρ + Gφ hφ + Gz hz by

div G = 1

ρ

∂{ρGρ}∂ρ

+ 1

ρ

∂{Gφ}∂φ

+ ∂{Gz}∂z

, (2.5)

and in sphercical coordinates with the field G = Gr hr + Gϑ hϑ + Gλ hλby

div G = 1

r2∂

∂r

{r2Gr

}+ 1

r sin ϑ

∂

∂ϑ

{Gϑ sin ϑ

}+ 1

r sin ϑ

∂

∂λ

{Gλ

}. (2.6)

The divergence of a vector field is a scalar field describing the ‘density of sources’of the vector field. Locations with a positive divergence are known as sources, whilenegative values are known as sinks. In case of the gravity field, the divergence ispositive within a mass and zero outside the mass.

Curl of a Vector Field

Definition: The mapping of a vector field F=Fx i + Fy j + Fz k – given in3-dimensional Cartesian coordinates – onto another vector field

∇ × F = curl F = det

⎛⎝

i j k∂∂x

∂∂y

∂∂z

Fx (x, y, z) Fy(x, y, z) Fz(x, y, z)

⎞⎠ =

=(

∂Fz

∂y− ∂Fy

∂z

)i +

(∂Fx

∂z− ∂Fz

∂x

)j +

(∂Fy

∂x− ∂Fx

∂y

)k

is called the curl of the field F.1

1The notation via determinants should be seen as a formal concept only, as the elements are not ofthe same type here.


The general expression for the curl in curvilinear coordinates can be found in a formaldeterminant

∇ × G = curl G = 1

hαhβhγ

det

⎛⎝hα hα hβ hβ hγ hγ

∂∂α

∂∂β

∂∂γ

hαGα hβGβ hγGγ

⎞⎠ , (2.7)

which is presented and discussed in exercises 24. In particular, one can obtain thecurl in spherical coordinates by

∇ × G =(

1

r sin ϑ

∂Gr

∂λ− 1

r

∂{rGλ}∂r

)hϑ +

(1

r

∂{rGϑ }∂r

− 1

r

∂Gr

∂ϑ

)hλ+ (2.8)

+ 1

r sin ϑ

(∂{Gλ sin ϑ}

∂ϑ− ∂Gϑ

∂λ

)hr ,

and in cylindrical coordinates by

∇ × G =(1

ρ

∂Gz

∂φ− ∂Gφ

∂z

)hρ +

(∂Gρ

∂z− ∂Gz

∂ρ

)hφ +

(1

ρ

∂{ρGφ}∂ρ

− 1

ρ

∂Gρ

∂φ

)hz .

(2.9)

The curl of a vector field provides information about the infinitesimal rotation at eachpoint. The result is again a vector field, where the direction is parallel to the axis ofrotation and the norm describes the ‘magnitude’ of rotation.

In most exercises of this book, a vector field in Cartesian coordinates is denoted byF = F(x, y, z) and in curvilinear coordinates by G = G(α, β, γ ). This is just forconvenience of the reader.

Conservative Vector Fields

• According toHelmholtz decomposition every ‘well behaved’ vector fieldcan be decomposed into a curl-free and a divergence-free part.

• A curl-free vector field is called a conservative (vector) field.• Each conservative vector field F or G has a corresponding scalar potential

� with the relation F = ∇� and G = ∇�, respectively. The quantity � iscalled potential.

• The gravity field is a conservative vector field and the divergence outsidethe body/mass is zero.

Questions

In particular, the following problems are investigated in the exercises:

• How to calculate the gradient, the curl and the divergence in Cartesian, sphericaland cylindrical coordinates?

2 Differentiation of Field Quantities 43

• How to express a vector field in another coordinate system?• How to derive the gradient and the divergence in a (unknown) curvilinear coordi-nate system?

Exercises

17. Determine the gradient and the curl of the gradient for the scalar field

� = r cos λ + cos(2λ + ϑ) − 10 sin λ

in spherical coordinates.18. Calculate the curl of the vector field

G(ρ, φ, z) = − cos2 φ tanh ρ hρ + sin 2φ ln ρ

√cosh ρ

cosh zhφ + cos2 φ tanh z hz .

19. Determine the divergence and the curl of the vector field

G = 1

r2hr − cos λ sin ϑ hϑ + sin 2ϑ sin λ hλ

w.r.t. spherical coordinates.20. Convert the field G of the previous exercise also into cylindrical coordinates

(φ, ρ, z) with φ ≡ λ and calculate the divergence. To avoid ambiguities, thesolution can be restricted to the upper space z ≥ 0.

21. Derive the divergence of the vector field

F = x2y

(x2 + y2)2i +

√x2 + y2 j + z(1 − x2 − y2) k

in cylindrical coordinates.22. A new set of coordinates is defined by the relationship

x = α

α2 + β2, y = β

α2 + β2, z = ζ.

Determine the gradient in this system for an arbitrary function �.23. Derive the divergence formula in spherical coordinates by inserting an arbitrary

vector field into its gradient.24. The curl in orthogonal coordinates is given by the formal determinant

curl G = 1

hαhβhγ

det

⎛⎝hα hα hβ hβ hγ hγ

∂∂α

∂∂β

∂∂γ

hαGα hβGβ hγGγ

⎞⎠ .


Evaluate the formula for spherical coordinates with α ≡ r , β ≡ λ and γ ≡ ϑ

and explain the difference to the known result∇ × G presented in formula (2.8).25. Parabolic coordinates are given by the relationship


2.

(a) Determine the ‘frame vectors’ hqi and verify that they are orthogonal.(b) Derive the expression of the divergence in this system.

26. Determine a general expression for a vector field whose curl is given by

∇ × G = − sin λ hϑ − cosϑ cos λ hλ + cot ϑ

r2hr

in spherical coordinates. Investigate then the two cases Gλ = 1r and Gλ = 0.

Solutions

2.17. Gradient and Curl of the Gradient for a Scalar Field

We insert the scalar field � into formula (2.4) to get the gradient

∇� = cos λ hr + 1

r

(− sin(2λ + ϑ)

)hϑ +

(− 2 sin(2λ + ϑ) − 10 cos λ − r sin λ

)

r sin ϑhλ.

The curl of the gradient is found by

∇ × (∇�) =

=⎛⎝ 1

r sin ϑ

∂{cos λ

}

∂λ− 1

r

∂{r 1r sin ϑ

(− 2 sin(2λ + ϑ) − 10 cos λ − r sin λ

)}

∂r

⎞⎠hϑ+

+⎛⎝1

r

∂{r 1r (− sin(2λ + ϑ))

}

∂r− 1

r

∂{cos λ

}

∂ϑ

⎞⎠hλ+

+ 1

r sin ϑ

⎛⎝∂

{1r

(−2 sin(2λ + ϑ) − 10 cos λ − r sin λ

)}

∂ϑ−

∂{1r

(−sin(2λ + ϑ)

)}

∂λ

⎞⎠hr =

=(

1

r sin ϑ(− sin λ) − 1

r

1

sin ϑ

(− sin λ

))hϑ + 0 hλ+

+ 1

r sin ϑ

(1

r

(− 2 cos(2λ + ϑ)

)− 1

r

(− 2 cos(2λ + ϑ)

))hr = 0.

Solutions 45

• The curl of a field is a vector and so the answer ‘∇ × (∇�) = 0’ is wrong.Either one of the ‘frame vectors’ or the null vector must appear here.

• The vector field has a potential, namely �, hence it must be curl-free. Thesecond calculation is not necessary at all.

2.18. Curl in Cylindrical Coordinates

We apply the rules of logarithm to remove the roots

G(ρ, φ, z) = − cos2 φ tanh ρ hρ + sin 2φ ln ρ

√cosh ρ

cosh zhφ + cos2 φ tanh z hz =

= − cos2 φ tanh ρ hρ +sin 2φ

(− 1

ρln cosh z+ 1

ρln cosh ρ

)hφ +cos2 φ tanh z hz

and insert the vector field into the curl formula (2.9):

∇ × G =⎡⎣ 1

ρ

∂{cos2 φ tanh z

}

∂φ−

∂{sin 2φ

(− 1

ρln cosh z + 1

ρln cosh ρ

)}

∂z

⎤⎦ hρ+

+⎡⎣ ∂

{− cos2 φ tanh ρ

}

∂z−

∂{cos2 φ tanh z

}

∂ρ

⎤⎦ hφ+

+ 1

ρ

⎡⎣ ∂

{ρ sin 2φ

(− 1

ρln cosh z + 1

ρln cosh ρ

)}

∂ρ−

∂{

− cos2 φ tanh ρ}

∂φ

⎤⎦ hz =

=[1

ρ

(− 2 cosφ sin φ tanh z

)+

(sin 2φ

1

ρtanh z

)]hρ + 0 hφ+

+ 1

ρ

[sin 2φ tanh ρ + (−2 cosh φ sin φ) tanh ρ

]hz = 0 hz = 0.

2.19. Divergence and Curl in Spherical Coordinates

We insert the components of the vector field G into the formula of the divergence

div G = 1

r2∂

∂r

{r2

1

r2

}+ 1

r sin ϑ

∂

∂ϑ

{− sin ϑ cos λ sin ϑ

}+ 1

r sin ϑ

∂

∂λ

{sin 2ϑ sin λ

}=

= 0 − 1

r sin ϑ(sin 2ϑ cos λ) + 1

r sin ϑ(sin 2ϑ cos λ) = 0


and the curl

∇ × G =

=⎛⎝ 1

r sin ϑ

∂{

1r2

}

∂λ− 1

r

∂{r(sin 2ϑ sin λ)

}

∂r

⎞⎠hϑ +

⎛⎝1

r

∂{

− r cos λ sin ϑ}

∂r− 1

r

∂{

1r2

}

∂ϑ

⎞⎠hλ+

+ 1

r sin ϑ

⎛⎝∂

{(sin 2ϑ sin λ) sin ϑ

}

∂ϑ−

∂{

− cos λ sin ϑ}

∂λ

⎞⎠hr =

=(0 − sin 2ϑ sin λ

r

)hϑ +

(− cos λ sin ϑ

r− 0

)hλ+

+ 1

r sin ϑ

(2 cos 2ϑ sin ϑ sin λ + sin 2ϑ sin λ cosϑ − sin λ sin ϑ

)hr =

=(

− sin 2ϑ sin λ

r

)hϑ +

(− cos λ sin ϑ

r

)hλ + sin λ

r3 cos 2ϑ hr .

We can conclude that this vector field is source-free but not curl-free. The vectorfield G will be used again in the later exercises 20 and 29 due to these properties.

2.20. Re-writing from Spherical to Cylindrical Coordinates

First, we express the spherical and cylindrical ‘frame vectors’ in matrix notation:

⎛⎜⎝

hrhλ

hϑ

⎞⎟⎠ =

⎛⎝cos λ sin ϑ sin λ sin ϑ cosϑ

− sin λ cos λ 0cos λ cosϑ sin λ cosϑ − sin ϑ

⎞⎠

⎛⎜⎝

ijk

⎞⎟⎠ ,

⎛⎜⎝

hρ

hλ

hz

⎞⎟⎠ =

⎛⎝

cos λ sin λ 0− sin λ cos λ 0

0 0 1

⎞⎠

⎛⎜⎝

ijk

⎞⎟⎠ .

Then, we solve the second equation for the Cartesian base vectors and find therelationship:

⎛⎜⎝

hrhλ

hϑ

⎞⎟⎠ =

⎛⎝cos λ sin ϑ sin λ sin ϑ cosϑ

− sin λ cos λ 0cos λ cosϑ sin λ cosϑ − sin ϑ

⎞⎠

⎛⎜⎝cos λ hρ − sin λ hλ

sin λ hρ + cos λ hλ

hz

⎞⎟⎠

=⎛⎜⎝sin ϑ hρ + cosϑ hz

hλ

cosϑ hρ− sin ϑ hz

⎞⎟⎠ = T

⎛⎜⎝

hρ

hλ

hz

⎞⎟⎠ .

Solutions 47

• The relation between two sets of orthonormal coordinates systems can con-sist only in translations, rotations and reflections. The determinant of thetransformation matrix is always detT = ±1, where a negative sign indi-cates reflection or incorrect order of the ‘frame vectors’.

• The relation between spherical and cylindrical coordinates is only a rotationabout the co-latitude.

The relation between the coordinates is

r =√

ρ2 + z2,

cosϑ = z√ρ2 + z2

,

sin ϑ = ±√1 − cos2 ϑ = ±

√1 − z2

ρ2 + z2= ± ρ√

ρ2 + z2.

The minus sign in the sine-term can be neglected as the co-latitude is limited byϑ ∈ [0, π ].Now, we can re-write the vector field into cylindrical coordinates and simplify theexpressions2:

G = 1

r2

(sin ϑ hρ + cosϑ hz

)− cos λ sin ϑ

(cosϑ hρ− sin ϑ hz

)+ sin 2ϑ sin λ hλ =

=(

1

r2− cos λ cosϑ

)sin ϑ hρ +

(1

r2cosϑ+ cos λ sin2 ϑ

)hz + (2 sin ϑ cosϑ) sin λ hλ =

=(

ρ√

ρ2 + z23 − cos λ

zρ

ρ2 + z2

)hρ +

(z

√ρ2 + z2

3 + cos λρ2

ρ2 + z2

)hz+

(2

ρz√

ρ2 + z22

)sinλ hλ.

By inserting the vector field into the formula of the divergence in cylindrical coor-dinates and with λ = φ, we obtain

div G= 1

ρ

∂

{ρ2√

ρ2+z23 −cos λ

zρ2

ρ2+z2

}

∂ρ+1

ρ

∂

{2 ρz

ρ2+z2sin λ

}

∂λ+

∂

{z√

ρ2+z23 + cos λ

ρ2

ρ2+z2

}

∂z=

= 2√

ρ2 + z23 − 3ρ2

√ρ2 + z2

(ρ2 + z2)3− z cos λ

2(ρ2 + z2) − 2ρ2

(ρ2 + z2)2+

(2z

ρ2 + z2cos λ

)+

+√

ρ2 + z23 − 3z2

√ρ2 + z2

(ρ2 + z2)3+ cos λ

−ρ22z

(ρ2 + z2)2=

= cos λ−2ρ2z − 2z3+2ρ2z + 2zρ2 + 2z3−2zρ2

(ρ2 + z2)2+ 2r3 − 3(ρ2 + z2)r + r3

(ρ2 + z2)3= 0.

2The first 2 lines of this equation should be handled with care, as we are mixing temporarilycoordinates of different systems.


• The divergence of a field is independent from the coordinate system. Hence,the result is the same as in the previous exercise.

2.21. Divergence in Cylindrical and Cartesian Coordinates

Re-writing the Vector Field

In the first method, we re-write the field

F = x2y

(x2 + y2)2i +

√x2 + y2 j + z(1 − x2 − y2) k

in cylindrical coordinates. Based on exercise 20, we have the relation⎛⎜⎝

hρ

hφ

hz

⎞⎟⎠ =

⎛⎝

cosφ sin φ 0− sin φ cosφ 0

0 0 1

⎞⎠

⎛⎜⎝

ijk

⎞⎟⎠ ,

which is equivalent to

i = cosφ hρ − sin φ hφ

j = sin φ hρ + cosφ hφ

k = hz .

We insert x = ρ cosφ and y = ρ sin φ into the field

F = ρ3 cos2 φ sin φ

ρ4

(cosφ hρ − sin φ hφ

)+ ρ

(sin φ hρ + cosφ hφ

)+ z(1 − ρ2) hz =

=(1

ρcos3 φ sin φ + ρ sin φ

)hρ +

(− 1

ρcos2 φ sin2 φ + ρ cosφ

)hρ + z(1 − ρ2) hz

and obtain the divergence

div F =

= 1

ρ

∂{ρ(1ρ cos3 φ sin φ+ρ sin φ

)}

∂ρ+ 1

ρ

∂{(

− 1ρ cos2 φ sin2 φ+ρ cosφ

)}

∂φ+

∂{z(1−ρ2)

}

∂z=

= 2 sin φ + 1

ρ

(− 1

ρ(−2 cosφ sin3 φ + 2 cos3 φ sin φ) − ρ sin φ

)+ (1 − ρ2) =

= sin φ − 1

ρ2sin 2φ cos 2φ + (1 − ρ2).

Independence of Coordinates

In the second method, we consider that the divergence is independent of the chosencoordinate system. Hence, we can calculate the quantity in Cartesian coordinates

Solutions 49

div F = (x2 + y2)2(2xy) − x2y2(x2 + y2)(2x)

(x2 + y2)4+ y√

x2 + y2+ (1 − x2 − y2) =

= 2xy

(x2 + y2)2− 4x3y

(x2 + y2)3+ y√

x2 + y2+ (1 − x2 − y2)

and later insert cylindrical coordinates

div F = 2ρ2 cosφ sin φ

ρ4− 4ρ4 cos3 φ sin φ

ρ6+ ρ sin φ

ρ+ (1 − ρ2) =

= sin φ + 1

ρ2sin 2φ(1 − 2 cos2 φ) + (1 − ρ2).

Due to the identity 1 − 2 cos2 φ = − cos 2φ, the two results of the divergence areequal.

• The divergence of a field is independent of the coordinate system. If theexpression of the vector field is not necessary in curvilinear coordinates forlater steps, then the divergence can be calculated in the Cartesian systemas well.

2.22. Gradient in a New Coordinate System

We differentiate the relationship between Cartesian and the new curvilinear coordi-nates to obtain the ‘frame vectors’ hqi :

hα = 1

‖hα‖

⎛⎜⎜⎝

(α2+β2)−2α2

(α2+β2)2

−2αβ

(α2+β2)2

0

⎞⎟⎟⎠ =

⎛⎜⎜⎝

β2−α2

(α2+β2)

−2αβ

(α2+β2)

0

⎞⎟⎟⎠

⇒ ‖hα‖ =√

(β2 − α2)2 + (4α2β2)

(α2 + β2)4= 1

α2 + β2,

hβ = 1

‖hβ‖

⎛⎜⎜⎝

−2αβ

(α2+β2)2

(α2+β2)−2β2

(α2+β2)2

0

⎞⎟⎟⎠ =

⎛⎜⎜⎝

−2αβ

(α2+β2)

α2−β2

(α2+β2)

0

⎞⎟⎟⎠

⇒ ‖hβ‖ =√

(−β2 + α2)2 + (4α2β2)

(α2 + β2)4= 1

α2 + β2

hζ = hζ =⎛⎝001

⎞⎠


Based on the ‘frame vectors’, we can express the gradient in this system:

∇� = (α2 + β2)

(∂�

∂αhα + ∂�

∂βhβ

)+ ∂�

∂ζhζ .

2.23. Divergence in Spherical Coordinates

The ‘frame vectors’ of the spherical coordinate system are presented already in(1.13):

hr =⎛⎝cos λ sin ϑ

sin λ sin ϑ

cosϑ

⎞⎠ , hλ =

⎛⎝

− sin λ

cos λ

0

⎞⎠ , hϑ =

⎛⎝cos λ cosϑ

sin λ cosϑ

− sin ϑ

⎞⎠

with the gradient

∇� = ∂�

∂rhr + 1

r sin ϑ

∂�

∂λhλ + 1

r

∂�

∂ϑhϑ .

We consider the gradient as a formal vector operator acting on an arbitrary vectorfield G = Gr hr + Gλ hλ + Gϑ hϑ . For simplicity, we skip the transpose symbolsin this question.

div G =[

∂

∂rhr + 1

r sin ϑ

∂

∂λhλ + 1

r

∂

∂ϑhϑ

]·[Gr hr + Gλ hλ + Gϑ hϑ

]=

=∂{Gr hr + Gλ hλ + Gϑ hϑ

}

∂rhr + 1

r sin ϑ

∂{Gr hr + Gλ hλ + Gϑ hϑ

}

∂λhλ+

+ 1

r

∂{Gr hr + Gλ hλ + Gϑ hϑ

}

∂ϑhϑ =

=[

∂Gr

∂rhr hr + Gr

∂ hr∂r

hr + Gλ∂ hλ

∂rhr + Gϑ

∂ hϑ

∂rhr

]+

+ 1

r sin ϑ

[∂Gλ

∂λhλ hλ + Gλ

∂ hλ

∂λhλ + Gr

∂ hr∂λ

hλ + Gϑ∂ hϑ

∂λhλ

]+

+ 1

r

[∂Gϑ

∂ϑhϑ hϑ + Gϑ

∂ hϑ

∂ϑhϑ + Gλ

∂ hλ

∂ϑhϑ + Gϑ

∂ hr∂ϑ

hϑ

].

Due to orthogonality, we find hqi∂ hqi∂qi

= 0 for every ‘frame vector’ and coordinate

qi . In the spherical system, we get also ∂ hλ

∂r = ∂ hλ

∂ϑ= ∂ hϑ

∂r = ∂ hr∂r = 0 and the inner

products

https://doi.org/10.1007/978-3-030-00416-3_1

Solutions 51

∂ hr∂λ

hλ =⎛⎝

− sin λ sin ϑ

cos λ sin ϑ

0

⎞⎠

⎛⎝

− sin λ

cos λ

0

⎞⎠ = sin ϑ

∂ hϑ

∂λhλ =

⎛⎝

− sin λ cosϑ

cos λ cosϑ

0

⎞⎠

⎛⎝

− sin λ

cos λ

0

⎞⎠ = cosϑ

∂ hr∂ϑ

hϑ =⎛⎝cos λ cosϑ

sin λ cosϑ

− sin ϑ

⎞⎠

⎛⎝cos λ cosϑ

sin λ cosϑ

− sin ϑ

⎞⎠ = 1.

Inserting into the formal product we conclude

∇ · G =[

∂Gr

∂r

]+ 1

r sin ϑ

[∂Gλ

∂λ+ Gr sin ϑ + Gϑ cosϑ

]+ 1

r

[∂Gϑ

∂ϑ+ Gr

]=

=[

∂Gr

∂r

]+ 1

r sin ϑ

∂Gλ

∂λ+

[Gr

r+ Gϑ

cot ϑ

r

]+ 1

r

[∂Gϑ

∂ϑ+ Gr

]=

=[1

r2r2

∂Gr

∂r+ 2

rGr

]+ 1

r sin ϑ

∂Gλ

∂λ+ 1

r sin ϑ

[Gϑ cosϑ + sin ϑ

∂Gϑ

∂ϑ

]=

= 1

r2∂

∂r

{r2Gr

}+ 1

r sin ϑ

∂

∂ϑ

{Gϑ sin ϑ

}+ 1

r sin ϑ

∂

∂λ

{Gλ

}= div G,

where the last line is equal to the known divergence formula.

2.24. Curl Calculation via Formal Determinant

We insert spherical coordinates into the formal determinant, write down the resultfor the moment as ∇ × G and simplify the expression

∇ × G = 1

sin ϑrhr

(∂{Gϑ }

∂λ− ∂{sin ϑGλ}

∂ϑ

)+ 1

rhλ

(∂{Gr }∂ϑ

− ∂{rGϑ }∂r

)+

+ hϑ

(1

r

∂{rGλ}∂r

− 1

r sin ϑ

∂{Gr }∂λ

).

When we compare our result with

∇ × G =(

1

r sin ϑ

∂Gr

∂λ− 1

r

∂{rGλ}∂r

)hϑ +

(1

r

∂{rGϑ }∂r

− 1

r

∂Gr

∂ϑ

)hλ+

+ 1

r sin ϑ

(∂{Gλ sin ϑ}

∂ϑ− ∂Gϑ

∂λ

)hr ,


we would conclude ∇ × G = −∇ × G. In fact, the formal determinant can provideeither the known curl or its negative expression, depending on the chosen order of thecoordinates. If the coordinates are used in the correct order of a right-handed-system– in the spherical case {r, ϑ, λ} –, we obtain standard expressions.

2.25. Parabolic Coordinates

a. Orthogonality

The ‘frame vectors’ and their norm are presented already in exercise 14. To verifyorthogonality, we evaluate the products of the 3 (non-normalized) ‘frame vectors’:

h�α hβ = (

β cos γ, β sin γ, α)⎛⎝

α cos γ

α sin γ

−β

⎞⎠ = βα(cos2 γ + sin2 γ − 1) = 0,

h�α hγ = (

β cos γ, β sin γ, α)⎛⎝

− sin γ

cos γ

0

⎞⎠ = β(− cos γ sin γ + sin γ cos γ ) = 0,

h�β hγ = (

α cos γ, α sin γ, −β)⎛⎝

− sin γ

cos γ

0

⎞⎠ = α(− cos γ sin γ + sin γ cos γ ) = 0.

As all products vanish, the 3 vectors form an orthogonal triad and the normalizationwill not change this property.

b. Divergence

Based on the ‘frame vectors’, we can write the gradient in this system:

∇� = 1

hα

∂�

∂αhα + 1

hβ

∂�

∂βhβ + 1

hγ

∂�

∂αhγ .

An arbitrary vector field has the representationG = Gα hα + Gβ hβ + Gγ hγ . Anal-ogous to exercise 23, we obtain the divergence by simplification of the product(∇ · G) while omitting the transpose symbols:

Solutions 53

div G= 1

hα

∂{Gα hα + Gβ hβ + Gγ hγ

}

∂αhα + 1

hβ


}

∂βhβ+

+ 1

hγ


}

∂γhγ =

= 1

hα

(∂Gα

∂α+ Gβ

∂ hβ

∂αhα

)+ 1

hβ

(∂Gβ

∂β+ Gα

∂ hα

∂βhβ

)+

+ 1

hγ

(∂Gγ

∂γ+ Gα

∂ hα

∂γhγ + Gβ

∂ hβ

∂γhγ

)=

= 1√α2 + β2

(∂Gα

∂α+ Gβ

β

β2 + α2 + ∂Gβ

∂β+ Gα

α

β2 + α2

)+

+ 1

αβ

(∂Gγ

∂γ+ Gα

β√β2 + α2

+ Gβα√

β2 + α2

).

In the last step, we considered hα = hβ , the obvious relation∂ hγ

∂α= ∂ hγ

∂β= 0 and the

products

∂ hβ

∂αhα =

⎡⎣ −α

(√

β2 + α2)3

⎛⎝α cos γ

α sin γ

−β

⎞⎠+ 1√

β2 + α2

⎛⎝cos γ

sin γ

0

⎞⎠⎤⎦ 1√

β2 + α2

⎛⎝β cos γ

β sin γ

α

⎞⎠= β

β2 + α2 ,

∂ hα

∂βhβ =

⎡⎣ −β

(√

β2 + α2)3

⎛⎝β cos γ

β sin γ

α

⎞⎠+ 1√

β2 + α2

⎛⎝cos γ

sin γ

0

⎞⎠⎤⎦ 1√

β2 + α2

⎛⎝α cos γ

α sin γ

−β

⎞⎠= α

β2 + α2 ,

∂ hα

∂γhγ = 1√

β2 + α2

⎛⎝−β sin γ

β cos γ

0

⎞⎠

⎛⎝− sin γ

cos γ

0

⎞⎠ = β√

β2 + α2,

∂ hβ

∂γhγ = α√

β2 + α2.

In principle, the above formula provides already the divergence in parabolic coordi-nates. Nevertheless we reformulate to compare our result with the version in text-books. First, we sort the terms w.r.t. the component Gα and identify (with pre-knowledge of the final result):


1√α2 + β2

(∂Gα

∂α+ Gα

α

β2 + α2

)+ 1

αβ

(Gα

β√β2 + α2

)=

= 1

α2 + β2

[√α2 + β2

∂Gα

∂β+ Gα

α√α2 + β2

+ 1

α

√α2 + β2Gα

]=

= 1

α2 + β2

[√α2 + β2

∂Gα

∂β+ Gα

∂√

α2 + β2

∂α+ 1

α

∂α

∂α

√α2 + β2Gα

]=

= 1

α2 + β2

1

α

∂{α√

α2 + β2Gα

}

∂α.

After performing the same steps for Gβ , we obtain the divergence

div G = 1

α2 + β2

⎡⎣ 1

α

∂{α√

α2 + β2Gα

}

∂α+ 1

β

∂{β√

α2 + β2Gβ

}

∂β

⎤⎦ + 1

αβ

∂Gγ

∂γ

in parabolic coordinates.

• For all orthogonal coordinate systems, the divergence can be calculated by

div G = (∇ · G) = 1

hαhβhγ

[∂

∂α

{Gαhβhγ

}+ ∂

∂β

{Gβhαhγ

}+ ∂

∂γ

{Gγ hαhβ

}]

after finding the ‘frame vectors’ and their norm.

2.26. Identification of a Vector Field from its Curl

We compare the general expression

∇ × G =(

1

r sin ϑ

∂Gr

∂λ− 1

r

∂{rGλ}∂r

)hϑ +

(1

r

∂{rGϑ }∂r

− 1

r

∂Gr

∂ϑ

)hλ+

+ 1

r sin ϑ

(∂{Gλ sin ϑ}

∂ϑ− ∂Gϑ

∂λ

)hr

with the given curl

∇ × G = −sin λhϑ − cosϑ cos λhλ + cot ϑ

r2hr .

Solutions 55

In all ‘components’, the derivatives w.r.t. two different variables sum up. Therefore,we assume a linear combination with the weight μ and (1 − μ) for the hr -terms:

1

r sin ϑ

(∂{Gλ sin ϑ}

∂ϑ− ∂Gϑ

∂λ

)hr

!= cot ϑ

r2hr ,

∂{Gλ sin ϑ}∂ϑ

!= (μ)cosϑ

r,

−∂Gϑ

∂λ

!= (1 − μ)cosϑ

r.

By integration, we find the components

Gϑ = −λ(1 − μ)cosϑ

r+ cϑ(ϑ, r),

Gλ =(

μsin ϑ

r+ cλ(λ, r)

)1

sin ϑ= μ

r+ cλ(λ, r)

sin ϑ.

The terms cϑ(ϑ, r) and cλ(λ, r) are ‘integration constants’ which depend on twovariables, but not on the third one. Using the expression of Gλ, we compare thehϑ -terms:

(1

r sin ϑ

∂Gr

∂λ− 1

r

∂{rGλ}∂r

)hϑ

!= −sin λhϑ ,

1

r sin ϑ

∂Gr

∂λ

!= −(ν) sin λ,

−1

r

∂{rGλ}∂r

= −1

r

1

sin ϑ

∂{rcλ(λ, r)}∂r

!= −(1 − ν) sin λ.

We conclude from the last line ν = 1 and cλ(λ, r) = gλ(λ)

r . For the term Gr , weintegrate with another ‘integration constant’

Gr = r sin ϑ cos λ + cr (r, ϑ).

The last condition is found by the equation

(1

r

∂{rGϑ }∂r

− 1

r

∂Gr

∂ϑ

)hλ

!=−cosϑ cos λ hλ

(∂{−λ(1 − μ) cosϑ + rcϑ(ϑ, r)}

∂r− ∂{r sin ϑ cos λ + cr (r, ϑ)}

∂ϑ

)!=−r cosϑ cos λ

cϑ(r, ϑ) + r∂cϑ(r, ϑ)

∂r− r cosϑ cos λ − ∂cr (r, ϑ)

∂ϑ= −r cosϑ cos λ,

cϑ(r, ϑ) + r∂cϑ(r, ϑ)

∂r− ∂cr (r, ϑ)

∂ϑ= 0.


Hence, we obtain the corresponding vector field

G =(

μ

r+ gλ(λ)

r sin ϑ

)hλ + (

r sin ϑ cos λ + cr (r, ϑ))hr+

+(

−λ(1 − μ)cosϑ

r+ cϑ(ϑ, r)

)hϑ

with 3 unknown functions and the condition cϑ(r, ϑ) + r ∂cϑ (r,ϑ)

∂r − ∂cr (r,ϑ)

∂ϑ= 0.

Case 1: Gλ = 1r

Based on Gλ = 1r , we determine μ = 1 and the vector field

G1 =(1

r

)hλ + (

r sin ϑ cos λ + cr (r, ϑ))hr + (

cϑ(ϑ, r))hϑ .

Case 2: Gλ = 0

Based on Gλ = 0 we determine μ = 0 and the field

G2 = (r sin ϑ cos λ + cr (r, ϑ)

)hr +

(−λ

cosϑ

r+ cϑ(ϑ, r)

)hϑ .

Chapter 3Work, Line Integral and Potential

The work or energy W , which is necessary to move a unit mass in a force field F orG, is usually depending on the chosen path �(t) between the points A and B. Thework is then evaluated by a line integral

W =tB∫

tA

(F(�)

)�Tdt,

W =tB∫

tA

(G(�)

)�Tdt,

(3.1)

where

• F(�) or G(�) is the force field evaluated along the curve,• T = � is the tangent vector of the curve,• and tA and tB are the parameters of the curve to reach the points A and B, respec-tively.

In case of a conservative vector field, a scalar potential (field) � can be derived bycomparing the gradient in this frame – including the norm of the ‘frame vectors’ –with the vector field:

Fx i + Fy j + Fz k!= ∂�

∂xi + ∂�

∂yj + ∂�

∂zk, (3.2)

Gα hα + Gβ hβ + Gγ hγ!= 1

‖hα‖∂�

∂αhα + 1

‖hβ‖∂�

∂βhβ + 1

‖hγ ‖∂�

∂γhγ . (3.3)


57


58 3 Work, Line Integral and Potential

This leads to a sequence of integrations and differentiations where the result� is thepotential up to an unknown integration constant. The workW is hereby independentof the path and calculated by the potential differences between start and end point:W = �(B) − �(A).

Exercises

Line Integral within a Vector Field

27. A unit mass is moved in the two-dimensional vector field F =(

11+y2 , −x

)�

along the curve �(t) = (e2t − et , et

)�.

(a) Determine the work/energy for moving a unit mass along the curvebetween t ∈ [0, ln 4].

(b) Verify – without using the curl operator – that the work is path-dependentand so the vector field cannot be conservative.

28. Calculate thework formoving a unitmass along the curve�(t) =(

3t2

1+t3 ,3t

1+t3

)�

for t ∈ [0,∞[ in the force field F = (−y, x)�.

29. Calculate the work for moving a unit mass in the non-conservative vector field

G(r, ϑ, λ) = 1

r2hr − cos λ sin ϑ hϑ + sin 2ϑ sin λ hλ

(of exercise 19) along the intersection curve of a circular cylinder

Z :{x ∈ R

3 :(y − 12

)2+ x2= 14

}and the surface+ =

{x ∈ R

3 :‖x‖ = 1, z ≥ 0}.

30. Calculate the energy W for moving a unit mass in the vector fieldF = (

x, y, y ln(x2 + y2))�

along the curve �(t) which is defined by the inter-

sectionof theplaneE : 1 − x2 = z and the cylinderZ :

{x ∈ R

3 : x2 + y2 = R2}.

Evaluate the energy for the two cases R = 1 and R = 2.

Potential and Gradient

31. Calculate the energy W for moving a unit mass along the curve�(t) = (

t2, cos t2, t2)�

with t ∈ [0,√π]in the radial-dependent vector field

G(r, ϑ, λ) = (r + 1)(3r − 1)

r3 + r2 − r + 100hr in spherical coordinates.

Exercises 59

32. Determine the parameters {a, b, c, d} in such a way that the vector field

F = 1

(x + y + z)3

⎛⎝ x + y − 3zax + by + czdx + 3y − z

⎞⎠

is conservative and derive also the corresponding potential.33. The vector field

G(r, ϑ, λ) = eλ+ϑ hr +(eλ+ϑ + 2

r

)hϑ + eλ+ϑ + 1

r

sin ϑhλ

is given in spherical coordinates. Find the corresponding potential.34. A vector field is given in cylindrical coordinates by

Gη(ρ, φ, z) = − sin φ tanh ρ hρ +(ln | cosh z| + η ln | cosh ρ|

)cosφ

ρhφ

+ sin φ tanh z hz

with η ∈ R.

(a) Determine the parameter η in such a way, that the vector field is conser-vative and find the corresponding potential �(ρ, φ, z).

(b) Find the work for moving a unit mass along the curve � = 2 hρ + 2 hz

with φ ≡ t and fix the parameter to η = 0 and η = −1, respectively.

35. Parabolic coordinates are given by the relationship


2

with γ ∈ [0, 2π ], α, β ∈ R+.

(a) Determine the ‘frame vectors’ hqi and the gradient in this system.(b) Assume the vector field

G(α, β, γ ) = 1

α2hα − 1

αβhβ + 1

αβhγ

to be conservative and calculate the corresponding potential �.

36. A modified spherical coordinate system is defined by

x = √2

α(sin β − cosβ

) 1

cosh γ

y = √2

α(cosβ + sin β

) 1

cosh γ


z = √2

α+1tanh γ

with α, γ ∈ R and β ∈ [−π, π).

(a) Calculate the normalized and simplified ‘frame vectors’ { hα, hβ, hγ }.(b) Derive the gradient in this system.(c) Assume the vector field

G(α, β, γ ) = cosh γ√2

α+1cosβ

hβ + sinh γ√2

α+1 hγ

to be conservative and determine the corresponding potential �.

Solutions

3.27. Work in a Non-Conservative Vector Field

F =(

11+ y2 , −x

)�

a. Work for the Curve �(t) =(e2t − et , et

)�

We evaluate the vector field along the curve and multiply with the tangent vector

F(�) = ( 11+e2t , −e2t + et

)�,

F�T = ( 11+e2t , −e2t + et

) (2e2t − et

et

)= 2e2t − et

1 + e2t− e3t + e2t .

The work of moving a unit mass is found by the line integral

W =∫

F�Tdt =ln 4∫

0

2e2t − et

1 + e2t− e3t + e2tdt =

=[−1

3e3t + 1

2e2t + ln

(1 + e2t

)− arctan et

]ln 40

= −27

2+ ln

17

2− arctan 4 + π

4.

Solutions 61

b. Non-Conservative Vector Field

In case of a conservative vector field, the following statements are equivalent:

1. Each conservative vector field has a corresponding potential.2. The field is curl-free for every point in space.3. The work for moving a unit mass between two points is independent of the

chosen path.4. For every closed curve, the work for moving a unit mass will vanish.

The given vector field is non-conservative if we can prove one of these conditionswrong. According to the question we should not use the curl, but we can calculatethe work along different paths (Fig. 3.1).An independent path is the straight line between the start point A = (

0, 1)�

and the

end point B = (12, 4

)�:

G :(xy

)=(01

)+ t

(123

), t ∈ [0, 1]

As the work

WG =1∫

0

(1

1+(1+3t)2

−12t

)� (123

)dt =

[12

3arctan(1 + 3t) − 18t2

]10

= 4 arctan 4 − π − 18 = W

differs from the previous results, this field is non-conservative.

Fig. 3.1 Force field F =(

11+y2

,−x)�

and two different paths between A and B (exercise 27)


3.28. Work Along the Curve �(t) =(

3t2

1+t3, 3t

1+t3

)�in the

Force Field F = (− y, x)�

We calculate the tangent vector

T =⎛⎝

(1+t3)6t−3t2·3t2(1+t3)2

(1+t3)3−3t ·3t2(1+t3)2

⎞⎠ =

⎛⎝

6t−3t4

(1+t3)2

3−6t3

(1+t3)2

⎞⎠

and evaluate the vector field along the curve

F = (−y, x)� =

(− 3t

1+t3 ,3t2

1+t3

)�,

F�T = −3t (6t − 3t4) + 3t2(3 − 6t3)

(1 + t3)3= −18t2 + 9t5 + 9t2 − 18t5

(1 + t3)3= −9t2

(1 + t3)2.

The work is determined by the line integral

W =∞∫

0

F�Tdt =∞∫

0

−9t2

(1 + t3)2dt =

[3

1 + t3

]∞

0

= −3.

• The curve �(t) =(

3t2

1+t3 ,3t

1+t3

)�is a parametric representation of the

Folium of Descartes.• This particular parametrization has a discontinuity in the origin 0, which ishidden in the double point of Fig. 3.2.

Fig. 3.2 Folium of Descartes in the vector field F = (−y, x)� (exercise 28)

Solutions 63

3.29. Work in a Non-Conservative Vector Fieldin Cylindrical Coordinates

To determine the curve, we use polar coordinates in the plane z = 0:(y − 1

2

)2

+ x2 =(

ρ sin φ − 1

2

)2

+ ρ2 cos2 φ = 1

4

ρ2(sin2 φ + cos2 φ) − ρ sin φ = 0

ρ = sin φ.

Hence, the projection of the cylinder onto the plane z = 0 is represented by ρ = sin φ

with φ ∈ [0, π ]. For the z-component we consider, that the curve is lying on the unitsphere:

z =√1 − ρ2 =

√1 − sin2 φ = | cosφ|.

Similar to the arc length we have to consider the modulus here. The effect of themodulus is shown by the black and the gray curve in Fig. 3.3b.We split the curve into two parts

� = sin φ hρ + | cosφ| hz =

⎧⎪⎨⎪⎩

�+ = sin φ hρ + cosφ hz for φ ∈ [0, π2

]

�− = sin φ hρ − cosφ hz for φ ∈ [π2 , π

]

depending on the sign of the cosine-term.

Fig. 3.3 Visualization of Vivani’s figure by the intersection of a (semi)-sphere and a cylinder(exercise 29)


In cylindrical coordinates, the tangent vectors are

T = ρ(1) hρ + φ(ρ) hφ + z(1) hz ⇒

⎧⎪⎨⎪⎩T+ = cosφ hρ + sin φ hφ − hz sin φ

T− = cosφ hρ + sin φ hφ + hz sin φ.

We insert the curve into the vector field – expressed in cylindrical coordinates (cf.exercise 20) and consider

√ρ2 + z2 = r = 1 and λ = φ:

G(�) =(

ρ√ρ2 + z2

3 − cos λzρ

ρ2 + z2

)hρ +

(z√

ρ2 + z23+ cos λ

ρ2

ρ2 + z2

)hz +

+(2

ρz√ρ2 + z2

2

)sinλ hλ =

= sin φ (1 − cosφ| cosφ|) hρ +(| cosφ| + cosφ sin2 φ

)hz + 2 sin2 φ| cosφ| hφ.

To avoid mistakes with the modulus, we solve the two intervals separately. The firstinterval is limited by φ ∈ [0, π

2

]with the integrand:

[G�T ]+ =

= cosφ sin φ (1−cosφ cosφ)−sin φ(cosφ+cosφ sin2 φ

)+sin φ2 sin2 φ cosφ == cosφ sin φ − cos3 φ sin φ − cosφ sin φ − cosφ sin3 φ + 2 sin3 φ cosφ == − cos3 φ sin φ + sin3 φ cosφ = cosφ sin φ(− cos2 φ + sin2 φ) == − sin 2φ

2cos 2φ = − sin 4φ

4

and the work

W+ =π/2∫

0

[G�T ]+dφ =

π/2∫

0

− sin 4φ

4dφ = −1

4

[−cos 4φ

4

]π/2

0

= 0.

Analogous, we calculate for the second interval φ ∈ [π2 , π

]

[G�T ]− = cosφ sin φ (1 + cosφ cosφ) + sin φ

(− cosφ + cosφ sin2 φ)−

− sin φ2 sin2 φ cosφ == cosφ

(sin φ + sin φ cos2 φ − sin φ + sin3 φ − 2 sin3 φ

)=

= cos3 φ sin φ − sin3 φ cosφ

Solutions 65

and

W− =π∫

π/2

[G�T ]−dφ =

π∫

π/2

cos3 φ sin φ − sin3 φ cosφdφ =

=[−cos4 φ

4− 3

sin4 φ

4

]π

π/2

= 0.

The total work is then the sum of its components:

W = W+ + W− =∮

G�Tdφ = 0.

• Although the vector field is non-conservative (cf. exercise19), the energyvalues in this closed curve and also in the two parts are zero. However,other closed curves with non-vanishing work must exist as well.

• The intersection of a sphere and a cylinder – in the given geometri-cal relationship – leads to the so-called Viviani’s curve, also known asViviani’s window or Viviani’s figure. The curve in this exercise consistin the upper part of this figure.

• Depending on the context, also the partial surface on the sphere might becalled Viviani’s window (cf. Fig.3.3).

3.30. Work in a Non-Conservative Field

The intersection of a plane and a circular cylinder leads to a conic section as well. Ina geometrical parametrization, we would prove via eigen values that the intersectioncurve is an ellipse and determine dimension and rotation (Fig. 3.4).In the algebraic method, we parametrize the cylinder with x = R cos t andy = R sin t and insert into the equation of the plane z = 1 − R cos t

2 to derive thetangent vector:

T = (−R sin φ, R cosφ,R sin φ

2

)�.

Now we evaluate the vector field along curve and multiply with the tangent vector

F = (x, y, y ln(x2 + y2)

)� = (R cosφ, R sin φ, R sin φ ln R2

)�,

F�T = −R2 sin φ cosφ + R2 sin φ cosφ + R2

2sin2 φ ln R2 = R2

2sin2 φ ln R2.


Fig. 3.4 Intersection of the cylinder x2 + y2 = 1 and the plane z = 1 − x2 (exercise 30)

We obtain the energy via the line integral

W =∫

F�Tdφ =2π∫

0

R2

2sin2 φ ln R2dφ = πR2 ln R.

The work along this closed curve is depending on the radius R of the cylinder Z.By inserting R = 1, we find the work W = 0, while for the radius R = 2 the workis equal to W = π4 ln 2.

3.31. Radial-Dependent Vector Field

• Every radial-dependent vector field of the form G = f (r)hr in sphericalcoordinates is conservative with the potential �(r) = ∫

f (r)dr .

We recognize that the denominator is the derivative of the nominator:

�(r) =∫

(r + 1)(3r − 1)

r3 + r2 − r + 100dr =

∫3r2 − r + 3r − 1

r3 + r2 − r + 100dr =

= ln∣∣∣r3 + r2 − r + 100

∣∣∣.

Solutions 67

The work of moving a unit mass is equal to the potential difference between startand end point of the curve, unless the curve is passing a singularity of the field. Thegiven field has no singularity as the nominator has only complex or negative roots,while the spherical radius is positive.The parameter t = 0 is equivalent to the point A = �(0) = (

0, 1, 0)�

with the

radius r = 1.Analogous,we obtain the point�(√

π) = (π, −1, π

)�with the radius

r =√2π2 + 1 for the value t = √

π .For computing the energy, we insert the 2 radii into the potential difference:

W = ln

∣∣∣∣∣(2π2 + 1)

√2π2 + 1 + (2π2 + 1) − √

2π2 + 1 + 100

100 + 1

∣∣∣∣∣ =

= ln

∣∣∣∣∣(2π2)

√2π2 + 1 + 2π2 + 101

101

∣∣∣∣∣ .

3.32. Finding a Potential in Cartesian Coordinates

To determine the coefficients, we derive the curl of the vector field:

∇ × F =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

∂dx + 3y − z

(x + y + z)3∂y −

∂(ax + by + cz)

(x + y + z)3∂z

∂x + y − 3z

(x + y + z)3∂z −

∂(dx + 3y − z)

(x + y + z)3∂x

∂(ax + by + cz)

(x + y + z)3∂x −

∂(x + y − 3z)

(x + y + z)3∂y

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=

=

⎛⎜⎜⎜⎝

(x + y + z)3(3) − (dx + 3y − z)3(x + y + z)2 − (x + y + z)3(c) + (ax + by + cz)3(x + y + z)2

(x + y + z)3(−3) − (x + y − 3z)3(x + y + z)2 − (x + y + z)3(d) + (dx + 3y − z)3(x + y + z)2

(x + y + z)3(a) − (ax + by + cz)3(x + y + z)2 − (x + y + z)3(1) + (x + y − 3z)3(x + y + z)2

⎞⎟⎟⎟⎠

(x + y + z)6=

= 1

(x + y + z)4

⎛⎜⎜⎜⎝

(3a − 3d − c + 3)x + (3b − 3c − c + 3)y + (3c + 3 − c + 3)z

(−3 − 3 − d + 3d)x + (−3 − 3 − d + 3c)y + (−3 + 9 − d − 3)z

(a − 3a − 1 + 3)x + (a − 3b − 1 + 3)y + (a − 3c − 1−9)z

⎞⎟⎟⎟⎠ .

The condition of a curl-free field leads to a set of linear equations for the unknowns.We solve the linear equations by the values {a = 1, b = 1, c = −3, d = 3} andobtain a conservative vector field

F = 1

(x + y + z)3

⎛⎝ x + y − 3z

x + y − 3z3x + 3y − z

⎞⎠ .


a. Potential

First we integrate the gradient component ∂�∂x via partial fraction decomposition

� =∫

x + y − 3z

(x + y + z)3dx =

∫A

(x + y + z)+ B

(x + y + z)2+ C

(x + y + z)3dx =

=∫

1

(x + y + z)2+ −4z

(x + y + z)3dx =

= −1

x + y + z+ 2z

1

(x + y + z)2+ c1(y, z) = z − x − y

(x + y + z)2+ c1(y, z)

with a ‘constant’ depending on (y, z). Then we differentiate our result w.r.t. thecoordinate y and compare with the gradient’s component:

∂�

∂y= 1

(x + y + z)2+ −4z

(x + y + z)3+ ∂c1

∂y= x + y + z − 4z

(x + y + z)3+ ∂c1

∂y!= x + y − 3z

(x + y + z)3

⇒ ∂c1∂y

= 0 ⇒ c1 = c12(z).

We repeat the previous step for the z-component:

∂�

∂z= 1

(x + y + z)2+ 2(x + y + z) − 2z · 2

(x + y + z)3+ ∂c12(z)

∂z!= 3x + 3y − z

(x + y + z)3

⇒ ∂c12∂z

= 0.

By combining we find the potential

� = z − x − y

(x + y + z)2+ const.

corresponding to the vector field

F = 1

(x + y + z)3

⎛⎝ x + y − 3z

x + y − 3z3x + 3y − z

⎞⎠ .

3.33. Finding a Potential in Spherical Coordinates

For the potential, we compare the gradient (on the left) and the given vector field (onthe right):

∂�

∂rhr + 1

r

∂�

∂ϑhϑ + 1

r sin ϑ

∂�

∂λhλ

!= eλ+ϑ hr +(eλ+ϑ + 2

r

)hϑ + eλ+ϑ + 1

r

sin ϑhλ

Solutions 69

First we integrate the gradient component ∂�∂r

∂�

∂r!= eλ+ϑ

⇒ � =∫

eλ+ϑdr = reλ+ϑ + c1(λ, ϑ)

with a ‘constant’ depending on (λ, ϑ). Then we differentiate our result w.r.t. thecoordinate ϑ and compare with the gradient’s component:

1

r

∂�

∂ϑ= eλ+ϑ + 1

r

∂c1(λ, ϑ)

∂ϑ

!= eλ+ϑ + 2

r

⇒ ∂c1(λ, ϑ)

∂ϑ= 2 ⇒ c1(λ, ϑ) =

∫2dϑ = 2ϑ + c12(λ)

⇒ � = reλ+ϑ + 2ϑ + c12(λ).

We repeat the previous step for the λ-component:

1

r sin ϑ

∂�

∂λ= 1

r sin ϑ

(reλ+ϑ + ∂c12

∂λ

)!= eλ+ϑ + 1

r

sin ϑ

c12 =∫

1dλ = λ.


� = reλ+ϑ + 2ϑ + λ + const.


G = eλ+ϑ hr +(eλ+ϑ + 2

r

)hϑ + eλ+ϑ + 1

r

sin ϑhλ.

• The order of integration and differentiation can be chosen freely. It is rec-ommended to start with the simplest integral.

• In this particular example, all three integrations are necessary to find thepotential. In many cases, the potential is derived already after the first orsecond integration step.


3.34. Finding a Potential in Cylindrical Coordinates

a. Conservative Vector Field and its Potential

To find a conservative vector field, we investigate the curl:

∇ × G =⎡⎣ 1

ρ

∂{sin φ tanh z

}

∂φ−

∂{(ln | cosh z| + η ln | cosh ρ|) cosφ

ρ

}

∂z

⎤⎦ hρ + 0 hφ+

+ 1

ρ

⎡⎣∂{ρ(ln | cosh z| + η ln | cosh ρ|) cosφ

ρ

}

∂ρ−

∂{

− sin φ tanh ρ}

∂φ

⎤⎦ hz =

=[1

ρcosφ tanh z − tanh z

cosφ

ρ

]hρ + 1

ρ[η cosφ tanh ρ + cosφ tanh ρ] hz =

= cosφ

ρ

(η tanh ρ + tanh ρ

)hz .

To obtain a curl-free vector field with a potential �, we have to set η = −1.For the potential, we compare the gradient (on the left) and the given vector field (onthe right):

∂�

∂ρhρ + 1

ρ

∂�

∂φhφ + ∂�

∂zhz

!= − sin φ tanh ρ hρ + ln

∣∣∣∣ cosh zcosh ρ

∣∣∣∣ cosφ

ρhφ + sin φ tanh z hz .

First we integrate the gradient component ∂�∂ρ

∂�

∂ρ

!= − sin φ tanh ρ

⇒ � =∫

(− sin φ) tanh ρdρ = − sin φ ln cosh ρ + c1(φ, z)

with a ‘constant’ depending on (φ, z). Then we differentiate our result w.r.t. thecoordinate z and compare with the gradient’s component:

∂�

∂z= ∂c1(φ, z)

∂z!= sin φ tanh z

c1(φ, z) =∫

sin φ tanh zdz = sin φ ln cosh z + c12(φ)

⇒ � = sin φ ln


∣∣∣∣+ c12(φ).

Solutions 71

We repeat the previous step for the φ-component:

1

ρ

∂�

∂φ= 1

ρ

(cosφ ln


∣∣∣∣+ ∂c12∂φ

)!= 1

ρ

(cosφ ln


∣∣∣∣)

⇒ ∂c12∂φ

= 0.


� = sin φ ln


∣∣∣∣+ const.,

b. Work

The curve � = 2 hρ + 2 hz is part of a circle in the plane E : z = 2 with radiusρ = 2.

case 1: η = −1In the case η = −1, we find a conservative vector field with the corresponding poten-tial� = sin φ ln |1| + const. and obtain the workWη=−1 = 0 for all parts of the path.

case 1: η = 0We evaluate the non-conservative field along the curve (with ρ = 2 and z = 2)

Gη=0(ρ, φ, z) = − sin φ tanh 2 hρ + (ln cosh 2)cosφ

2hφ + sin φ tanh 2 hz

and insert it into the line integral with the tangent vector T = 2 hφ :

W =∫

G�η=0(2 hφ)dφ =

∫ln(cosh 2)

cosφ

22dφ = ln(cosh 2)

[sin φ

]φB

φA

.

For all intervals φB = φA + kπ, k ∈ N the energy will also vanish!

• The work vanishes for the vector fields Gη=0 and Gη=−1 for all closed circles� = 2 hρ + 2 hz .

• In case of a conservative vector field with η = −1, the work is always zero, when

start and end point of the curve are lying on the coneC :{x ∈ R

3 : x2 + y2 = z2}

due to the factor ln∣∣∣ cosh zcosh ρ

∣∣∣ in the potential.

3.35. Finding a Potential in Parabolic Coordinates

a. ‘Frame Vectors’ and Gradient of Parabolic Coordinates

The ‘frame vectors’ have been derived in exercise 14 and we can directly write downthe gradient


∇� = 1√β2 + α2

∂�

∂αhα + 1√

β2 + α2

∂�

∂βhβ + 1

βα

∂�

∂γhγ .

b. Potential


1√β2 + α2

∂�

∂αhα + 1√

β2 + α2

∂�

∂βhβ + 1

βα

∂�

∂γhγ

!= 1

α2hα − 1

αβhβ + 1

αβhγ

First we integrate the gradient component ∂�∂γ

1

βα

∂�

∂γ

!= 1

βα

⇒ � =∫

1dγ = γ + c1(α, β)

with a ‘constant’ depending on (α, β). Then we differentiate our result w.r.t. thecoordinate α and compare with the gradient’s component:

1√β2 + α2

∂�

∂α= 1√

β2 + α2

∂{γ + c1(α, β)}∂α

!= 1

α2

∫0 + ∂c1(α, β)

∂αdα =

∫ √β2 + α2

α2dα

⇒ c1(α, β) = −√

α2 + β2

α+ arsinh

α

β+ c12(β).

We repeat the previous step for the β-component:

∂�

∂β= − 1

α

2β

2√

α2 + β2+ (−α)

β2

√1 +

(αβ

)2 + ∂c12(β)

∂β= −

√α2 + β2

αβ+ ∂c12

∂β

1√α2 + β2

∂�

∂β= 1√

α2 + β2

[−√

α2 + β2

αβ+ ∂c12

∂β

]!= − 1

αβ

⇒ c12 = 0.

Solutions 73


� = γ −√

α2 + β2

α+ arsinh

α

β+ const.


G = 1

α2hα − 1

αβhβ + 1

αβhγ

in the parabolic coordinate system.

3.36. Finding a Potential in Modified Spherical Coordinates

a. ‘Frame Vectors’ and Gradient

The ‘frame vectors’ have been derived in exercise 16.We insert them into the formulafor the gradient:

∇� = 1√2

α+1ln

√2

∂�

∂αhα + cosh γ√

2α+1

∂�

∂βhβ + cosh γ√

2α+1

∂�

∂γhγ .

b. Potential


1√2α+1

ln√2

∂�

∂αhα + cosh γ√

2α+1

∂�

∂βhβ + cosh γ√

2α+1

∂�

∂γhγ

!= cosh γ√2α+1

cosβhβ + sinh γ√

2α+1 hγ .

The corresponding potential has no terms of variable α, hence, only two integrationsare necessary.First we integrate the gradient component ∂�

∂γ

cosh γ√2

α+1

∂�

∂γ

!= sinh γ√2

α+1

⇒ ∂�

∂γ= tanh γ

⇒ � = ln | cosh γ | + C(α, β)

with a ‘constant’ depending on (α, β). Then we differentiate our result w.r.t. thecoordinate β and compare with the gradient’s component:


cosh γ√2α+1

∂�

∂β= cosh γ

√2α+1

∂C(α, β)

∂β

!= cosh γ√2α+1

cosβ

⇒ ∂C(α, β)

∂β= 1

cosβ

⇒ C(α, β) =∫

1

cosβdβ = ln

∣∣∣∣tan(

π

4+ β

2

)∣∣∣∣+ const.


� = ln | cosh γ | + ln

∣∣∣∣tan(

π

4+ β

2

)∣∣∣∣+ const.

• In geosciences, the integral

∫1

cosβdβ = ln

∣∣∣∣tan(

π

4+ β

2

)∣∣∣∣is sometimes called Mercator integral.The solution canbeused to figureout isometric coordinates on the sphere or ellipsoid.

• The integral can be solved in many different ways, e.g. the substitution ofWeierstraß, trigonometric identities, integration by parts or simply byexpanding. The given solution is preferred for the isometric coordinates.

Chapter 4Integral Theorems of Vector Analysis

The integral theorems of vector analysis build a relation between differentiation andintegration and reduce often the ‘dimension of integration’. The theorems introducedin this chapter are used to determine

• the area and geometrical center of planar figures (theorem of Green),• the flux through a surface or volume (theorem of Gauß),• or the circulation within a (curved) surface (theorem of Stokes).

Green’s Theorem in the Plane

The integral theorem of Green in the plane enables the evaluation of a (differ-entiated) vector field F = (F1, F2

)�within a closed and bounded region R by the

line integral along its boundary ∂B:∫ ∫

R

(∂ F2

∂x− ∂ F1

∂y

)dxdy =

∮

∂B

F1dx + F2dy. (4.1)

In particular, the theorem is applied for calculating the enclosed area of the region:

A =∫ ∫

R

dxdy =∮

∂B

xdy = −∮

∂B

ydx = 1

2

∮

∂B

xdy − ydx . (4.2)

In case of a parametric representation, we obtain in the general case

∮

∂B

F1dx + F2dy =∮

∂B

F�Tdt (4.3)


75


76 4 Integral Theorems of Vector Analysis

and for the area

A = 1

2

∮

∂B

x y − yxdt. (4.4)

Flux Through Surface and Volume

The flux F of a vector field through a surface S(u, w) is calculated by the surfaceintegrals

F =∫ ∫

F�N du dw,

F =∫ ∫

G�N du dw,

(4.5)

where N = N(u, w) denotes the normal vector of the surface. The vector fields areevaluated on the surface, which might be highlighted by a notation like F(S(u, w))

or G(u, w).The integral theorem of Gauss relates the surface integral to a volume integral.It can be applied for a closed volume V with the corresponding surface S in a vectorfield in the form

F =∫∫∫

Vdiv FdV =

∫ ∫

SF�N du dw,

F =∫∫∫

Vdiv GdV =

∫ ∫

SG�N du dw,

(4.6)

where dV is the volume element.In Cartesian coordinates, the volume element is given by dV = dxdydz (up to theorder of the variables). In general, we find for a volume V = V (u, w, ξ) the elementdV = |J|dξdudw with the Jacobian determinant

|J| = det

(∂V

∂ξ,∂V

∂u,∂V

∂w

).

If we recall the interpretation of the ‘divergence’ as the measurement of productionor annihilation of energy or material, then the integral theorem of Gauß states thatthe total balance of production or annihilation in the volume equals the outflux orinflux through the boundary surface.In many cases, the volume integral is easier to solve and can be performed in onestep, while the solution by surface integrals consist in several partitions.

Circulation Within a Surface

The integral theorem of Stokes – in this form also called Kelvin-Stokes’theorem – is used to calculate the (infinitesimal) circulation of a vector field within abounded surface. The theorem builds a relation between integration over the surfaceS and the line integral along its (curved) boundary ∂S:

4 Integral Theorems of Vector Analysis 77

∮

∂S

F�Tdt =⎧⎨

⎩

∫∫S(∇ × F)�N du dw

∫∫S(∇ × G)�N du dw

(4.7)

The boundary curve must have a positive orientation – i.e. anti-clockwise enclosingthe area – and the path must be closed, but not necessarily planar. If the integrationof the boundary is performed in several line segments, then the orientation must beconsidered.

Why Curvilinear Coordinates?

The concept of curvilinear coordinates plays different roles in the integral theorems:

• The introduction of polar coordinates simplifies the theorem of Green to the the-orem of Leibniz.

• The vector field might be given due to its symmetry in a curvilinear coordinatesystem. Hence, it might be natural, to calculate curl or divergence in these coor-dinates.

• The parametrization of surface or volume is often performed in an adequate curvi-linear coordinate system (cf. exercise 47, 48, 51).

• In the case of partial coordinate surfaces – and a corresponding vector field – thecalculation is simplified and often very compact (cf. exercise 52, 53, 58, 61).

Acurvilinear coordinate system is in particularmeaningful andhelpful, if the problemcontains the corresponding symmetry. All problems related to rotational figures canobviously be solved in cylindrical coordinates. For problems without symmetry,another parametrization might be more adequate, which is not based on orthogonalcurvilinear coordinates. The volume or surfaces of the mathematical cylinder inexercise 50 can be seen as an example here.

A Helpful Integral

In case of rotation symmetries, the integrals often contain products of (multiple)sine or cosine terms. For a shortcut, one can derive – via integration by parts – therecursion formula

∫sinn(x + φ)dx =

[− sinn−1(x + φ) cos(x + φ)

n

]+ n − 1

n

∫sinn−2(x + φ)dx (4.8)

for n > 1 with the initial values∫

sin0(x + φ)dx = x,

∫sin1(x + φ)dx = − cos(x + φ),

∫sin2(x + φ)dx = − sin(x + φ) cos(x + φ)

2+ 1

2x = − sin 2(x + φ)

4+ 1

2x .


Exercises

Area Calculation by the Theorem of Green

37. A closed curve is given by

� = (cos3 t, sin3 t)�

t ∈ [0, 2π ].

Determine the arc length of the boundary and the enclosed area.38. Calculate the area enclosed by the curve x3 + y3 − 3xy = 0 in the first quadrant.

39. Given the implicit curve L ={x ∈ R

2 : 2(x2 − y2) = (x2 + y2)2, x ≥ 0}.

Determine the enclosed area

(a) in a rational parametrization with y = x · t ,(b) and by introducing polar coordinates.

40. Calculate the area enclosed by the curve ρ(φ) = cos(nφ) in polar coordinatesfor n ∈ N.

41. Calculate the geometrical center of the domain

D ={x ∈ R

2 : 3x2 + y3 − y2 = 0, y ≥ 0}

(a) by the standard surface integrals

x = 1

A

∫ ∫xdxdy, y = 1

A

∫ ∫ydxdy,

(b) and by application of Green’s theorem.

42. Determine the geometrical center of the domain

A+ ={x ∈ R

2 : x2/3 + y2/3 ≤ 1, y ≥ x}.

43. Figure out the area which is enclosed by the implicit curve

(x2 + y2)2 − 2(x2 + y2) − 1 = 0.

44. A pedal curve P�,E(t) is a planar figure which is defined by the orthogonalintersection of

• the tangent vectors of the curve �(t)• and the family of straight lines passing the pedal point E.

(a) Derive a parametrization of all pedal curves of the parabola x = 0.5y2 withan arbitrary pedal point.

Exercises 79

(b) Evaluate the area of the closed loop for the pedal point E = (− 32 , 0

)�.

Flux Through Volume and Surface

45. Determine the surface integral for the vector field

F = 1√

x2 + y2

⎛

⎝x − yz · arccos zy + xz · arccos z

ln |x2 + y2|

⎞

⎠

through the ‘mathematical cylindrical’ C with x = 2φ sin φ, y = 2φ cosφ forφ ∈ [0, 4π ] and 0 ≤ z ≤ 1.

46. Evaluate surface and volume integral of the theorem of Gauß for the flux ofthe vector field F = −x i + y j + 6z k through the torus T . The surface of thetorus is given by

x = (4 + cos u) sinw

y = (4 + cos u) cosw u, w ∈ [0, 2π ]z = sin u.

47. Evaluate the surface integral of the vector field

G(r, ϑ, λ) = ln(1 +

√1 − cos2 ϑ

)hr + tan4 λhλ + arctan ϑ hϑ

given in spherical coordinates through the upper semi-sphere

+ ={x ∈ R

3 : x2 + y2 + z2 = 4, z ≥ 0}.

48. Evaluate thefluxof the vector fieldG(ρ, φ, z) = 2ρ hρ + z hφ + 83 z2 hz through

the common volumeV of the cylinderZ ={x ∈ R

3 : (y − 12

)2 + x2 ≤ 14

}and

the semisphere + ={x ∈ R

3 : x2 + y2 + z2 ≤ 1, z ≥ 0}.

For parametrization, consider the plane z = 0 and express the non-centeredcircle in polar coordinates.

49. Given a regular tetraeder T with the corner points {A, B,C, D}, where• all triangles are equilateral with the sides = √

3,• the triangle �ABC lies in the plane z = 0,• the point A has the coordinates

(1, 0, 0

)�,

• and the point D is on the positive z-axis.

(a) Determine the coordinates of {B,C, D}.(b) Move the whole tetraeder – without rotations – along the x−axis, so that

in the shifted tetraeder Ts the x-component of Cs is zero, while the y- andz-coordinates remain unchanged.


(c) Determine the flux through the tetraeder Ts for the vector field

F = (−y, x,√3e−z

)�.

In case of surface integrals, the 4 faces can be solved together by anadequate parametrization of the triangles. Consider the orientation of thenormal vectors.

50. Move the area enclosed by the boundary curve � = (cos3 t, sin3 t, 0)�

without

rotations along the vector v = (1, 1, 1)� to create a solid mathematical cylinderMC with the height zmax = 2. Determine the flux of the vector field

F =(2y2 − ez2 , y 1

(z+2) ln |z+2| , sin(ex2−y) − x

)�

through the volume MC .51. Given the vector field G(ρ, φ, z) = 25

4 ρ3 hρ + 253 z3 hz in cylindrical coordi-

nates, the plane E : z = x7 + 24

35√2and the cone C =

{x ∈ R

3 : x2 + y2 = z2}.

(a) Determine the curl and the divergence of the vector field G and usethe result to figure out the circulation within the ellipse defined by theintersection of the plane E and the cone C.

(b) Calculate the flux through the smaller Dandelin sphere D1, which istangential to the cone and the plane.The radius can be found in the plane y = 0 by the incircle formula R = 2A

uwith the area A and the perimeter u of the corresponding triangle.

52. Calculate theflux through the areaOp ={x ∈ R

3: z = 13

√9 − x2 − y2, |y| ≤ 1

}

with the vector field

G(α, β, γ ) = sinh α hβ + cosβ hγ

given in oblate spheroidal coordinates

x = √8 cosh α cosβ cos γ

y = √8 cosh α cosβ sin γ

z = √8 sinh α sin β.

Consider, that the surface is a partial coordinate surface of the system.

53. Given the vector field G = (α2 + β2) cos γ hα + 1(α2+β2)

hβ in parabolic coor-dinates


2.

Exercises 81

Evaluate the flux through the parabolic surface

P :{x ∈ R

3 : x2 + y2 − 1 = 2z,−0.5 ≤ z ≤ 4}

(a) in parabolic coordinates,(b) and verify the result via cylindrical coordinates.

54. Evaluate the flux of the vector field F = (2x + y) i − (4y − x) j + (2z −16) k through the surface C, which is defined by the rotation of the curve� = (3 + cos(z cos 3

√z), 0, z

)�for 0 ≤ z ≤ 6 around the z-axis with the angle

φ ∈ [0, 2π ]. Consider the theorem of Gauß.

Circulation Within a Surface

55. Verify Stokes’ theorem by evaluating line integrals and surface integral for thevector field F = ((1 − 2z)ex , xy, xz2

)�and the surface

S ={x ∈ R

3 : y = 4 − z2, 0 ≤ x ≤ 1, y ≥ 0}.

56. Verify Stokes’ theorem by evaluating line and surface integrals in cylindricalcoordinates for the vector field G(ρ, φ, z) = −ρ cosφ hρ + ρz hz and the sur-face

V+ ={x ∈ R

3 : ‖x‖ = 1, z ≥ 0,

(y − 1

2

)2

+ x2 ≤ 1

4

}.

57. Calculate the circulation of the vector field G = ρ3 cos2 φ hφ + z4 hz throughthe surface z = xy

(a) in the domain x2 + y2 ≤ 1,(b) and in the domain |x | + |y| ≤ 1.

58. Given the vector field G = cos(λ + ϑ)(hr + hϑ + hλ

). Determine the circu-

lation within the spherical triangle on the unit sphere :{x ∈ R

3 : ‖x‖ = 1}

with the corner points A = (1, 0, 0)�, B = (0, 1, 0)�, C = (0, 0, 1)�. Whatis the result for the planar triangle with the same corner points?

59. Evaluate the circulationof thevectorfield F =(2z, x2, ln

∣∣∣(x − 2)2 + 4(y + 1)2

∣∣∣)�

within the parabolic partial area

P :{x ∈ R

3 : x2 + y2 − z = 0, (x − 2)2 + 4(y + 1)2 ≤ 1}.

60. Given an arbitrary planar surface SE in the plane E : y + 4x − 2z = 0. Find anon-conservative vector field F so that the circulation for all figures SE is zero.The solution is not unique.


61. Evaluate the circulation of the vector fieldG(α, β, γ ) = α2 cos γ hα + hβ givenin cardioid coordinates

x = αβ

(α2 + β2)2cos γ

y = αβ

(α2 + β2)2sin γ

z = α2 − β2

2(α2 + β2)2

through the partial coordinate surfaces with γ ∈ [0, π ] and α = const. andβ = const., respectively.

Solutions

4.37. Enclosed Area of � = (cos3 t, sin3 t

)

We calculate the tangent vector T and simplify the inner product by trigonometricidentities

T = (−3 cos2 t sin t, 3 sin2 t cos t)�

,

T�T = (3 sin2 t cos t)2 + (3 cos2 t (− sin t))2 = 9

(1

2sin 2t

)2

.

Fig. 4.1 Astroid curve with

� = (cos3 t, sin3 t)�

and itsenclosed area (exercise 37)

Solutions 83

For the arc length, we must consider the modulus

s =2π∫

0

√T�Tdt = 3

2

2π∫

0

| sin 2t |dt =

= 3

2· 12

([− cos 2t

]π/2

0−[

− cos 2t]π

π/2+[

− cos 2t]3π/2

π−[

− cos 2t]2π

3π/2

)= 6.

We re-write the integrand of Green’s theorem

xdy − ydx = 3 cos3 t sin2 t cos t − 3 sin3 t cos2 t (− sin t)dt =

= 3 cos2 t sin2 t (cos2 t + sin2 t) = 3

(1

2sin(2t)

)2

dt =

= 3

4sin2 2tdt.

Hence, we obtain the enclosed area by

A = 1

2

∮xdy − ydx = 3

8

2π∫

0

sin2(2t)dt = 3

8

[t

2− sin 2t cos 2t

4

]2π

0

= 3π

8.

• The planar figure � = (cos3 t, sin3 t)�

is known as astroid, but also ascubocycloid, or paracycle.

• In a geometrical definition, the astroid is created by tracing a marked pointon a circle, which is rolling inside a fixed circle. Therefore, the curves belongsto the family of hypocycloids.

• The algebraic equation of the curve is x2/3 + y2/3 = 1. Therefore, the curvebelongs to the family of superellipses.

• Due to the symmetry, it is also possible to integrate only in the first quadrant(Fig.4.1).

4.38. Area Enclosed by x3 + y3 − 3x y = 0

Obviously, the origin 0 is a point of the curve. Hence, we try to parametrize by afamily of straight lines, e.g. with y = x · t


Fig. 4.2 Folium ofDescartes withx3 + y3 − 3xy = 0(exercise 38)

x3 + x3t3 − 3x2t = 0,

x(1 + t3) = 3t,

⇒ x = 3t

1 + t3.

For this kind of parametrization, we can avoid computing y and y

xdy − ydx =(

x(x t + x) − xt x)dt = x2dt.

The curve is closed in the origin, if we consider the interval t ∈ [0,∞[.Hence, we calculate the area by Green’s theorem:

A = 1

2

∞∫

0

9t2

(1 + t3)2dt

u=t3== 1

2

∞∫

0

3

(1 + u)2du = 3

2

[−(1 + u)−1]∞0 = 3

2.

• The figure x3 + y2 − 3xy = 0 – in particular its closed part – is known asFolium of Descartes and the relevant part of the curve is visualized inFig.4.2.

• In exercise 28, the x- and y-components are exchanged. Due to its symmetry,the result is the same curve (after the parametrization x = y · t).

• The parametrization by y = xt or x = yt might work for curves which con-tain the origin 0. In the area calculation by Green’s theorem, the calculationof {x, x} or {y, y} can be avoided in this case.

Solutions 85

4.39. Area Enclosed by the Curve 2(x2 − y2) = (x2 + y2)2

Rational Parametrization

The point 0 fulfills the equation of the curve. Hence, we try again a parametrizationby y = x · t :

2x2(1 − t2) = x4(1 + t2)2

⇒ x2 = 21 − t2

(1 + t2)2.

For the integrand, we consider the simplification

(yx − yx)dt = x2dt,

which avoids the calculation of y and y.The curve is closed in the domain x ≥ 0 for the interval t ∈ [−1, 1]. Therefore, wecalculate the area by

A = 1

2

1∫

−1

21 − t2

(1 + t2)2dt =

1∫

−1

2

(1 + t2)2− 1

1 + t2dt =

[t

t2 + 1+ arctan t − arctan t

]1

−1= 1.

Polar Representation

Firstwe re-write the integrand ofGreen’s theorem in polar coordinates for an arbitrarycurve:

x = ρ(φ) cosφ

y = ρ(φ) sin φ

y′x − x ′ydφ =(ρ′ sin φ + ρ cosφ

)ρ cosφ −

(ρ′ cosφ − ρ sin φ

)ρ sin φdφ = ρ2dφ

Then we insert polar coordinates into the equation

2ρ2(cos2 φ − sin2 φ) = ρ4

⇒ ρ2 = 2 cos 2φ

and recognize the representation of the lemniscate of Bernoulli (exercise 5b)(Fig. 4.3).


Fig. 4.3 Lemniscate ofBernoulli with2(x2 − y2) = (x2 + y2)2

(exercise 39)

The curve is lying in the domain x ≥ 0 for the intervalφ ∈ [−π4 , π

4

]with the enclosed

area

A = 1

2

π/4∫

−π/4

2 cos 2φdφ = 1

2

[sin 2φ

]π/4

−π/4= 1.

Due to symmetry reasons, the complete lemniscate encloses the doubled areaAlemniscate = 2.

• In polar coordinates, the theorem of Green reduces to the sector formulaof Leibniz:

A = 1

2

φ1∫

φ0

ρ2dφ.

This formula holds for closed curves which are either passing the originor going around the origin. In particular it holds for ‘sectors’, which aredefined by one curved line ρ(φ) and two straight lines intersecting in theorigin (cf. Fig.4.4).

Solutions 87

Fig. 4.4 Example of a sector with two straight lines and a curved line

4.40. Area Enclosed by ρ = cos nφ

First we have to find the interval of φ, so that the curve

� = (cos nφ cosφ, cos nφ sin φ)�

is closed. For each choice of n ∈ N the closed figure will have a blossom-like shapewith different widths of the petals (cf. Fig. 4.5). We assume φ0 = 0 as the startingpoint with the coordinates�(0) = (1, 0)�. The curve will be closed when this pointis reached again, which is equivalent to φ = kπ with k ∈ N.

• For an even number n, we find x = cos(n · kπ) cos(kπ) = + cos(kπ)which leadsto an upper limit φe = 2π .

• For an odd number n, we find x = cos(n · kπ) cos(kπ) = − cos(kπ) which leadsto an upper limit φe = π .

According to the sector formula of Leibniz, the area is calculated by

A = 1

2

φe∫

0

ρ2dφ = 1

2

φe∫

0

(cos nφ)2dφ = 1

2

φe∫

0

(sin(nφ + 0.5π)

)2dφ =

= 1

2

[φ

2− sin(nφ + 0.5π) cos(nφ + 0.5π)

2n

]φe

0

= φe

4={

π2 for n evenπ4 for n odd.


Fig. 4.5 Area enclosed by ρ = cos nφ (exercise 40)

4.41. Geometrical Center of the Loop: 3x2 + y3 − y2 = 0

Different parametrizations of this curve have been discussed already in exercise 6.The question here considers only the area of the upper part with y ≥ 0 (cf. Fig. 4.6).

a. ‘Standard Integrals’

In case of a double integral, we use the parametrization in x(y):

x = ± 1√3

√y2 − y3.

The enclosed area is given by

A =1∫

0

1√3

√y2−y3∫

− 1√3

√y2−y3

dxdy =1∫

0

2√3

√y2 − y3dy = 2√

3

1∫

0

y√1 − ydy =

y=sin2 t== 2√3

π/2∫

0

sin2 t√1 − sin2 t · 2 sin t cos tdt = 4√

3

π/2∫

0

sin3 t − sin5 tdt =

= 4√3

[sin4 t cos t

5+ 1

5

(− sin2 t cos t

3+ 2

3(− cos t)

)]π/2

0

= 8

15√3.

Due to symmetry considerations, the x-coordinate of the geometrical center is zero:

x = 1

A

∫ ∫xdA = 15

√3

8

1∫

0

1√3

√y2−y3∫

− 1√3

√y2−y3

xdxdy = 15√3

8

1∫

0

[x2

2

] 1√3

√y2−y3

− 1√3

√y2−y3

dy = 0.

Solutions 89

Fig. 4.6 Geometrical centerof the loop3x2 + y3 − y2 = 0(exercise 41)

For the y-component, we use the integral recursion (4.8) and consider for the first

term[sinn t cos t

]π/2

0= 0. This leads to the coordinate:

y = 1

A

1∫

0

1√3

√y2−y3∫

− 1√3

√y2−y3

ydxdy = 15√3

8

1∫

0

2√3

y√

y2 − y3dy =

y=sin2 t== 15

4

π/2∫

0

sin4 t√1 − sin2 t2 sin t cos tdt = 15

2

π/2∫

0

sin5 t − sin7 tdt =

= 15

2

⎛

⎝[sin6 t cos t

7

]π/2

0

+ 1

7

π/2∫

0

sin5 tdt

⎞

⎠ = 15

2· 17

⎛

⎝4

5

π/2∫

0

sin3 tdt

⎞

⎠ =

= 15

2· 17

· 45

· 23

[− cos t

]π/2

0= 4

7.

b. Theorem of Green

For this question we have to recall the fact that the formula

A =∫ ∫

1dxdy = 1

2

∮xdy − ydx =

∮xdy = −

∮ydx

is only the most popular form of Green’s theorem. The original formula

∫ ∫ (∂ F2

∂x− ∂ F1

∂y

)dxdy =

∮F1dx + F2dy

enables other applications then the calculation of enclosed areas.


When we compare the right hand side with the geometrical center

x = 1

A

∫ ∫xdxdy

we conclude F1 = 0 and F2 = 0.5x2. By inserting into the line integral we obtain

x = 1

A

∮x2

2dy = 1

2A

∮x2dy.

We use the parametrization x = yt with the integrand

xdy − ydx = −y2dt.

In addition, we know y = 1 − 3t2 and so we get the area

A=−1

2

−1/√3∫

1/√3

(1 − 3t2)2dt = −1

2

[

t − 6t3

3+ 9

t5

5

]−1/√3

1/√3

= 1√3

[1 − 2

3+ 9

5· 19

]= 1√

3· 8

15.

For the geometrical center we calculate

x = 1

2A

∮x2dt = 15

√3

16

−1/√3∫

1/√3

(−6)(t3 − 6t5 + 9t7)dt =

= 15√3

16(−6)

[t4

4− 6

6t6 + 9t8

8

]−1/√3

1/√3

= 0,

In an analogous way, we conclude F1 = − y2

2 and F2 = 0 for the y-component:

y = 1

2A

∮y2dx = 15

√3

16

−1/√3∫

1/√3

(1 − 3t2)2(1 − 9t2)dt =

= 15√3

16

−1/√3∫

1/√3

1 − 6t2 + 9t4 − 9t2 + 54t4 − 81t6dt =

= 15√3

16

[t − 5t3 + 63

5t5 − 81

7t7]−1/

√3

1/√3

= 4

7.

Solutions 91

Fig. 4.7 Geometrical centerof a partial astroid figure(exercise 42)

4.42. Geometrical Center

For the parametrization, we identify the domain as part of the astroid of exercise 37.When we remove the parts below the line x = y, we obtain the interval t ∈ [π

4 , 5π4

]

for the boundary part ∂ B1. In addition we consider the straight line y = x in theinterval y ∈ [−23/2, 23/2

]as boundary ∂ B2:

x = 1

2A

⎛

⎜⎝

5π/4∫

π/4

(cos3 t)23 sin2 t cos tdt +2−3/2∫

−2−3/2

y2dy

⎞

⎟⎠ =

= 1

2A

⎛

⎜⎝3

5π/4∫

π/4

(1 − sin2 t)3 sin2 t cos tdt +[

y3

3

]2−3/2

−2−3/2

⎞

⎟⎠ =

= 1

2A

⎛

⎜⎝3

5π/4∫

π/4

sin2 t cos t − 3 sin4 t cos t + 3 sin6 t cos t − sin8 t cos tdt + 1

24√2

⎞

⎟⎠ =

= 1

2A

⎛

⎝

[

3sin3 t

3− 9

sin5 t

5+ 9

sin7 t

7− 3

sin9 t

9

]5π/4

π/4

+ 1

24√2

⎞

⎠ =

= 8

3π

(− 319

840√2

+ 1

24√2

)= −142

√2

315π≈ −0.203...

In a similar way, we could calculate the y-coordinate of the geometric center:

y = − 1

2A

∮y2dx = − 1

2A

⎛

⎜⎝

5π/4∫

π/4

(sin3 t)23 cos2 t (− sin t)dt +2−3/2∫

−2−3/2

x2dx

⎞

⎟⎠= · · · = 142

√2

315π,


but we can also argue with the symmetry:The complete astroid is symmetric w.r.t. the axes x = 0, y = 0 and x = y andx = −y. After removing the part below the line x = y, the remaining figure is stillsymmetric w.r.t. to the line y = −x . Therefore, we get the relation y = −x for thegeometrical center (Fig. 4.7).

4.43. Area Enclosed by (x2 + y2)2 − 2(x2 + y2) − 1 = 0

The formula contains only terms like x2 + y2 =: ρ2. Therefore, we introduce polarcoordinates

ρ4 − 2ρ2 − 1 = 0.

By adding zero we obtain

ρ4 − 2ρ2 + (1 − 1) − 1 = 0

(ρ2 − 1)2 = 2.

In other words, we figure out that the curve consists of concentric circles with thetwo radii

(ρ2)2 − 2(ρ2) − 1 = 0

ρ21/2 = 2 ± √

4 + 4

2= 1 ± √

2.

The first solution describes a circle with the radius ρ1 =√1 + √

2 and the enclosed

area A = ρ21π =

(1 + √

2)

π .

The second solution might be an artifact as the squared radius ρ22 = 1 − √

2 shouldnot be negative. In previous examples, it was possible to interpret a negative radius

as the reflected curve. We insert the radius ρ2 =√

|1 − √2| =

√√2 − 1 into the

equation and obtain

(√2 − 1

)2 − 2(√

2 − 1)

− 1 = 2 − 2√2 + 1 − 2

√2 + 2 − 1 = 0.

Hence, the second solution turns out to be an algebraic artifact and the only real

solution is the circle with the radius ρ =√1 + √

2.

Solutions 93

• In this question, we could avoid the theorem of Green due to known prop-erties of the circle.

• The existence of (x2 + y2)-terms is often an indication for polar/cylindricalcoordinates.

• The formal result of a negative radius can be a real solution (exercise 4) oran algebraic artifact.

4.44. Pedal Curve of a Parabola

a. Parametrization

The parabola x = 0.5y2 can be represented by

�(t) = (0.5t2, t)�

with the tangent vector T = (t, 1)�. The tangent line through an arbitrary point ofthe parabola is given by

G :(0.5t2

t

)+ p

(t1

), p ∈ R.

In a similar way, we set up the straight line through the pedal point E(x0, y0) withunknown coordinates

H :(

x0y0

)+ q

(−1t

), q ∈ R,

where the orthogonality is already ensured by(t, 1) (−1, t

)� = 0.We intersect the two lines G and H

0.5t2 + pt = x0 − q

t + p = y0 + qt

and eliminate p and q:

0.5t2 − t2 = x0 − q − y0t − qt2

q = (0.5)t2 + x0 − y0t

(1 + t2).


Fig. 4.8 Parabola x = 0.5y2

and some of its pedal curves(exercise 44)

We obtain the pedal curve of the parabola by inserting q into the equation ofH :

x(t) = x0 − (0.5)t2 + x0 − y0t

(1 + t2)= (x0 − 0.5)t2 + y0t

1 + t2,

y(t) = y0 + (0.5)t2 + x0 − y0t

(1 + t2)t = y0 + 0.5t3 + x0t

1 + t2.

Three different pedal curves are shown in Fig. 4.8. The curve with the solid blackline is the one where we want to calculate the enclosed area.

b. Area

We insert the values y0 = 0 and x0 = − 32 into the general expression:

x(t) = (−1.5 − 0.5)t2

1 + t2= −2t2

1 + t2,

y(t) = 0.5t3 − 1.5t

1 + t2= 1

2

t3 − 3t

1 + t2.

For the area we use the theorem of Green and find after some simplifications theintegrand

x y − yx = t2

(1 + t2)2(t2 + 3).

By polynomial long division, we split into the ‘integer’ and ‘fractional’ part andapply partial fraction decomposition

(t4 + 3t2)/(t4 + 2t2 + 1) = 1 + t2 − 1

(1 + t2)2= 1 + 1

1 + t2+ −2

(1 + t2)2.

Solutions 95

Due to symmetry, the curve is closing somewhere on the x-axis, which is equivalentto the condition t (t2 − 3) = 0, i.e. t = ±√

3 or t = 0. Now we consider, that thepath of integration must be anti-clockwise enclosing the area. Hence, the integrationinterval starts at t = √

3:

A = 1

2

−√3∫

√3

x y − yx dt = 1

2

−√3∫

√3

1 + 1

1 + t2+ −2

(1 + t2)2dt =

= 1

2

[t + arctan t − 2

(arctan t

2+ t

2t2 + 2

)]−√3

√3

= 3√3

4.

4.45. Mathematical Cylinder

First we parametrize the cylindrical surface CE (φ, z) = (2φ sin φ, 2φ cosφ, z)�

inFig. 4.9 with the normal vector

N = ∂CE (φ, z)

∂φ× ∂CE (φ, z)

∂z=⎛

⎝2 sin φ + 2φ cosφ

2 cosφ − 2φ sin φ

0

⎞

⎠×⎛

⎝001

⎞

⎠ =⎛

⎝2 cosφ − 2φ sin φ

−2 sin φ − 2φ cosφ

0

⎞

⎠ .

Then we evaluate the vector field F on the surface and introduce the abbreviationαz = arccos z

F = 1√

(2φ sin φ)2 + (2φ cosφ)2

⎛

⎝(2φ sin φ) − (2φ cosφ)zαz

(2φ cosφ) + (2φ sin φ)zαz

ln |(2φ sin φ)2 + (2φ cosφ)2|

⎞

⎠ =

= 1

2φ

⎛

⎜⎜⎝

2φ(sin φ − cosφzαz

)

2φ(cosφ + sin φzαz

)

ln |4φ2|

⎞

⎟⎟⎠

Fig. 4.9 Mathematical cylinder (exercise 45)


and multiply with the normal vector

F�N = 1

2φ

⎛

⎝2φ(sin φ − cosφzαz)

2φ(cosφ + sin φzαz)

ln |4φ2|

⎞

⎠

�⎛⎝

2 cosφ − 2φ sin φ

−2 sin φ − 2φ cosφ

0

⎞

⎠ =

= −2φ − 2z arccos z.

The flux is then calculated by the surface integral

F =∫ ∫

F�Ndφdz =1∫

0

4π∫

0

−2φ − 2z arccos zdφdz =

=1∫

0

[−2

φ2

2− 2φz arccos z

]4π

0

dz =1∫

0

−16π2 − 8π z arccos zdz =

= [−16π2z]10 − 8π

1∫

0

z arccos zdz =

= −16π2 − 8π

⎛

⎝[

z2

2arccos z

]1

0

−1∫

0

z2

2

−1√1 − z2

dz

⎞

⎠ =

= −16π2 − 8π

⎛

⎝0 + 1

2

1∫

0

z21√

1 − z2dz

⎞

⎠ =

z=cos t== −16π2 + 4π

0∫

π/2

cos2 t√1 − cos2 t

sin tdt =

= −16π2 + 4π

([1

2cos t sin t

]0

π/2

+ 1

2

∫ 0

π/2dt

)= −17π2.

• A mathematical cylinder is defined by a planar curve � and its translation�v in an arbitrary direction v outside the plane. The straight lines between� and �v form the surface of the cylinder. The curve � is not necessarilyclosed. If the curve is closed, then also the whole body is sometimes calleda cylinder in the mathematical sense.

Solutions 97

4.46. Flux of the Vector Field F = (−x, y, 6z)� Througha Torus

a. Surface Integral

First we calculate the normal vector of the torus surface:

N = ∂T∂u

× ∂T∂w

=⎛

⎝− sin u sinw

− sin u cosw

cos u

⎞

⎠×⎛

⎝(4 + cos u) cosw

−(4 + cos u) sinw

0

⎞

⎠ =

= (4 + cos u)

⎛

⎝cos u sinw

cosw cos usin u

⎞

⎠ .

Then we evaluate the vector field at the location of the surface and multiply with thenormal vector

F =⎛

⎝−(4 + cos u) sinw

(4 + cos u) cosw

6 sin u

⎞

⎠ ,

F�N =⎛

⎝−(4 + cos u) sinw

(4 + cos u) cosw

6 sin u

⎞

⎠ (4 + cos u)

⎛

⎝cos u sinw

cosw cos usin u

⎞

⎠ =

= (4 + cos u)2(− sin2 w cos u + cos2 w cos u) + 24 sin2 u + 6 cos u sin2 u == (4 + cos u)2 cos u(cos 2w) + 24 sin2 u + 6 cos u sin2 u.

We obtain the flux by the surface integral

F =∫ ∫

F�Ndudw =

=2π∫

0

2π∫

0

(4 + cos u)2 cos u cos 2w + 24 sin2 u + 6 cos u sin2 udwdu =

= 0 + 48π

2π∫

0

sin2 udu + 12π

2π∫

0

sin2 u cos udu = 48π2.

b. Volume Integral

The divergence div F = 6 is independent of the position, which makes the problemequivalent to determining the volume of the torus in Fig. 4.10. It is possible, to workwith cylindrical coordinates, but we want to practice the volume integral with theJacobian determinant here as well.For a volume integral, a third parameter ξ ∈ [0, 1] has to be introduced:


Fig. 4.10 Torus with R = 4 and r = 1 (exercise 46)

x = (4 + ξ cos u) sinw,

y = (4 + ξ cos u) cosw,

z = ξ sin u.

The volume element dV = |J| dw dξ dw requires the Jacobian determinant

|J| = det

⎛

⎝−ξ sin u sinw (4 + ξ cos u) cosw cos u sinw

−ξ sin u cosw −(4 + ξ cos u) sinw cos u cosw

ξ cos u 0 sin u

⎞

⎠ =

= ξ(4 + ξ cos u) det

⎛

⎝− sin u sinw cosw cos u sinw

− sin u cosw − sinw cos u cosw

cos u 0 sin u

⎞

⎠ = ξ(4 + ξ cos u).

The flux is then calculated by the volume integral

F =∫∫∫

Vdiv FdV =

2π∫

0

1∫

0

2π∫

0

6ξ(4 + ξ cos u)dwdξdu = 12π

2π∫

0

1∫

0

4ξ + ξ2 cos udξdu =

= 12π

2π∫

0

[

4ξ2

2+ ξ3

3cos u

]1

ξ=0

du = 12π

2π∫

0

2 + 1

3cos udu = 48π2.

4.47. Flux Through a Semisphere in Spherical Coordinates

• The normal vector N of a sphere around the origin is proportional to theradial vector hr with N = r2 sin ϑ hr .

Solutions 99

• In case of surface integral over a (partial) spherical surface around the ori-gin, the field components hλ and hϑ cannot contribute due to orthogonalityof the frame vectors.

We multiply the vector field with the normal vector of the sphere

G�N =[ln(1 +

√1 − cos2 ϑ

)hr + tan4 λ hλ + arctan ϑ hϑ

]�[4 sin ϑ hr

]=

= ln(1 +

√1 − cos2 ϑ

)4 sin ϑ.

The flux is calculated by the surface integral

F =∫∫

G�Ndλdϑ =π/2∫

0

2π∫

0

ln(1 +

√1 − cos2 ϑ

)4 sin ϑdλdϑ = 8π

π/2∫

0

ln (1 + sin ϑ) sin ϑdϑ =

= 8π

[− cosϑ ln(1 + sin ϑ)

]π/2

0

− 8π

π/2∫

0

− cosϑcosϑ

1 + sin ϑdϑ =

= 8π

π/2∫

0

cosϑcosϑ

1 + sin ϑ

1 − sin ϑ

1 − sin ϑdϑ = 8π

[ϑ + cosϑ

]π/2

0

= 4π2 − 8π.

4.48. Flux of the Vector FieldG(ρ, φ, z) = 2ρ hρ + z hφ + 8

3 z2 hz Through

Vivani’s Figure

The divergence of the field G(ρ, φ, z) = 2ρ hρ + z hφ + 83 z2 hz is

div G = 1

ρ

∂{ρ · 2ρ

}

∂ρ+ 1

ρ

∂{0}

∂φ+

∂{83 z2}

∂z= 4 + 16

3z,

which is depending only on the z-coordinate. Hence, a volume integral within theupper Vivani’s figure (cf. Fig. 4.11) might be easier to evaluate. We express theshifted cylinder in polar coordinates. In the xy-plane we find

(y − 1

2

)2

+ x2 =(

ρ sin φ − 1

2

)2

+ ρ2 cos2 φ = 1

4,

ρ2(sin2 φ + cos2 φ) − ρ sin φ = 0,

⇒ ρ = sin φ.


Fig. 4.11 Vivani’s figure(upper part) (exercise 48)

Therefore, the circle is represented in polar coordinates by ρ = sin φ withφ ∈ [0, π ].The maximum z-component of the volume integral is obviously z = √1 − ρ2:

F =∫∫∫

Vdiv GdV =

π∫

0

sin φ∫

0

√1−ρ2∫

0

(4 + 16

3z

)ρdzdρdφ =

=π∫

0

sin φ∫

0

[4z + 8

3z2]√1−ρ2

0

ρdρdφ =

=π∫

0

sin φ∫

0

(4ρ√1 − ρ2 + 8

3ρ(1 − ρ2)

)dρdφ =

=π∫

0

[−4 · 1

3

√1 − ρ2

3 + 8

3· ρ2

2− 8

3· ρ4

4

]sin φ

0

dφ =

=π∫

0

(−4

3cos3 φ + 4

3+ 4

sin2 φ

3− 2 sin4 φ

3

)dφ = 7

4π.

• In case of volume integrals there are two fundamental ‘options’:

– Adapt the volume element dV by the Jacobian determinant to the bodyand obtain simple integration limits (e.g. exercise 46b).

– Use standard volume elements of cylindrical, spherical or Cartesian coor-dinates and adapt the integration limits (e.g. exercise 48 and 51).

Solutions 101

4.49. Flux Through a Tetraeder

a. Determine the Corner Coordinates

Based on the given conditions,we have to determine the corner points of the tetraeder.

For D we use the known distance to A and obtain D = (0, 0, √2)�.

For B,C we rotate the location A around the z-axis by the angle φ = ± 2π3 :

B = (cos 2π3 , sin 2π

3 , 0)� =

(− 1

2 ,√32 , 0

)�,

C = (cos (− 2π3

), sin

(− 2π3

), 0)� =

(− 1

2 , −√32 , 0

)�.

b. Translation of Coordinates

A translation by the vector v = ( 12 , 0, 0)�

leads to the new coordinates:

A′ =(32 , 0, 0

)�, B′ =

(0,

√32 , 0

)�, C ′ =

(0, −

√32 , 0

)�, D′ =

(12 , 0,

√2)�

.

c. Flux Through the Tetraeder

If we want to solve the problem by surface integrals, then we have to calculate theflux through four faces of the tetraeder. Therefore, we consider the parametrizationof an arbitrary triangle �(P2, P1, P3) in space

S(u, w) = P1 + u · (P2 − P1) + w · (P3 − P1)

with 0 ≤ u ≤ 1 − w and 0 ≤ w ≤ 1 and also the direction of the normal vectors.This will lead – after simplifications – to the double integral

FT =1∫

0

1−w∫

0

3

2e−√

2+√2u+√

2w+3

2e−√

2u + 3

2e−√

2w − 9

2dudw = · · · = 9

4

(e−√

2 − 2 + √2)

.

Alternative Solution by the Principle of Cavalieri

The calculation via four faces and the double integral is time consuming. A smartermethod is a ‘one-dimensional volume integral’. The divergence

div F = −√3e−z

is only dependent on z. Therefore, we can apply the principle of Cavalieri for theintegration

∫∫∫

Vf (z)dV =

∫f (z)A(z)dz


where A(z) is the area while cutting the volume at height z.With the intercept theorem or some drawings, we find

A(z) = 1

2h = 1

2

H − z

H·

H − z

H

√3

2= 1

2

(√3

√2 − z√2

)2 √3

2= 3

√3

8(2 − 2

√2z + z2).

The flux is calculated by a one-dimensional integration over z:

FT =∫

f (z)A(z)dz =

=√2∫

0

−√3e−z

(3√3

8(2 − 2

√2z + z2)

)

dz = −9

8

√2∫

0

2e−z − 2√2ze−z + z2e−zdz =

= −9

8

{[−2e−z]

√2

0 + 2√2[ze−z + e−z]

√2

0 +[−z2e−z − 2ze−z − 2e−z

]√2

0

}

=

= 9

4

(e−

√2 − 2 + √

2)

.

4.50. Mathematical Cylinder

A closer inspection of the divergence

div F = 1

(z + 2) ln |z + 2|shows that the principle of Cavalieri could be applied again and we don’t need acomplete parametrization of body visualized in Fig. 4.12. The area was determinedalready in exercise 37with A(z) = 3π

8 . Themaximumheight is given by the questionand so we obtain the flux

Fig. 4.12 Mathematicalcylinder where the bottomsurface is defined by theastroid of exercise 37(exercise 50)

Solutions 103

F =∫∫∫

Vdiv FdV = 3π

8

2∫

0

1

(z + 2) ln |z + 2|dz =

= 3π

8

[ln (ln |z + 2|)

]2

0= 3π

8ln

ln 4

ln 2= 3π

8ln 2.

4.51. Dandelin Sphere: Flux and Circulation

a. Divergence, Curl and Circulation

We insert the vector field into the formula of the divergence

div G = 1

ρ

∂{ρ 25

4 ρ3}

∂ρ+ 1

ρ

∂{0}

∂φ+

∂{253 z3}

∂z= 25ρ2 + 25z2

and the curl

∇ × G =⎛

⎝ 1

ρ

∂{253 z3}

∂φ−

∂{0}

∂z

⎞

⎠ hρ +(

∂{254 ρ3

}

∂z− ∂

{253 z3}

∂ρ

)

hφ

+⎛

⎝ 1

ρ

∂{ρ · 0

}

∂ρ− 1

ρ

∂{254 ρ3

}

∂φ

⎞

⎠ hz = 0 hz .

The curl of the vector field is the null vector. Hence, the circulation within the ellipse– or any other surface – is also zero.

b. Flux Through Dandelin Sphere

This question has the remarkable property, that we can integrate the flux withoutknowing the exact parametrization like radius or center coordinates.

Cylinder Coordinates with Unknown Radius

Due to symmetry, the center of the sphere is somewhere on the z-axis with thecoordinates Z = (0, 0, z0

)�. We assume a radius of R, introduce the abbreviation

σ = √R2 − ρ2 and solve the problem in cylindrical coordinates:


FD1 =R∫

0

z0+σ∫

z0−σ

2π∫

0

25(ρ2 + z2)ρdφdzdρ = 50π

R∫

0

[

ρ3z + ρz3

3

]z0+σ

z0−σ

dρ =

= 100π

3

R∫

0

σρ(3ρ2+ (3z20 + σ 2)

)dρ = 100π

3

R∫

0

ρ

√R2−ρ2

(3ρ2+ 3z20 + R2−ρ2

)dρ =

= 100π

3(3z20 + R2)

[−1

3(R2 − ρ2)3/2

]R

0+ 100π

3

[−2

15(R2 − ρ2)3/2(3ρ2 + 2R2)

]R

0=

= 20

3π R3

(5z20 + 3R2

).

Geometry

Now, we figure out the geometry of the problem (cf. Figs. 4.13 and 4.14). In the

plane y = 0, we draw a triangle with C ′ = (0, 0)�, A′ =(

45√2, 4

5√2

)�and B′ =

( −35√2, 3

5√2

)�representing the cone and the plane.

The lattices have the dimensions b = |C ′A′| = 45 , a = |C ′B′| = 3

5 and c = |A′B′| = 1.Hence we calculate the perimeter u = 12

5 and the area A = 625 , which leads to the

radius R = 2Au = 1

5 . For the z-component of the center we obtain

z0 = √2R =

√25 .

After returning to the 3D space, we determine the center Z =(0, 0,

√25

)�and the

radius R = 15 . Hence, we insert the geometry to the flux integral and obtain

Fig. 4.13 2D representation of cone, plane and Dandelin sphere (exercise 51)

Solutions 105

Fig. 4.14 Cone, plane and one Dandelin sphere (exercise 51)

FD1 =∫∫∫

Vdiv GdV = 20

3π R3(5z20 + 3R2) = 20

3π

1

125

(5 · 2 + 3

25

)= 52

1875π.

c. Spherical Coordinates

As an alternative, we could solve the integration in spherical coordinates. We want tointegrate a sphere centered around theorigin0 for simplicity. Therefore,we ‘translate’the divergence of the vector field in the opposite direction

div G = 25ρ2 + 25(z + z0)2 = 25r2 sin2 ϑ + 25

(

r cosϑ +√2

5

)2

= 25r2 + 10√2r cosϑ + 2.

We determine the flux by integration:

FD1 =∫∫∫

Vdiv V dV =

1/5∫

0

π∫

0

2π∫

0

(25r2 + 10

√2r cosϑ + 2

)· r2 sin ϑdλdϑdr =

= 2π

1/5∫

0

π∫

0

(25r2 sin ϑ + 10

√2r cosϑ sin ϑ + 2 sin ϑ

)· r2dϑdr =

= 2π

1/5∫

0

[(25r2(− cosϑ) + 10

√2r

1

2sin2 ϑ + 2(− cosϑ)

)· r2]π

0dr =

= 2π

1/5∫

0

25r4(2) + 0 + 2(2)r2dr = 4

54π + 8

3 · 53 π = 52

1875π.


4.52. Oblate Spheroidal Coordinates

The surface Op ={x ∈ R

3 : z = 13

√9 − x2 − y2, |y| ≤ 1

}is a partial area of an

ellipsoid (cf. Fig. 4.15). We re-formulate

9z2 = 9 − x2 − y2

x2 + y2

9+ z2 = 1

and find the semi-major axis a = 3 and the semi-minor axis b = 1.The oblate spheroidal coordinates represent also ellipsoid surfaces for all constantpairs {p, α}. Our calculation will be very simple, if we find a value of α so that thesurface OP is a coordinate surface.We solve

√8 cosh α = 3√8 sinh α = 1

Fig. 4.15 Ellipsoidal partial area Op ={x ∈ R

3 : z = 13

√9 − x2 − y2, |y| ≤ 1

}(exercise 52)

Solutions 107

and obtain α = artanh 13 for the correct ratio between the axis. In the remaining step,

we prove that p = √8 is the correct scaling:

√8 cosh α

!= 3√8

1√1 − tanh2 α

= √8

1√

89

!= 3 ��.

Hence, we conclude that in the oblate spheroidal coordinate system, the parameterp = √

8 and the coordinate α = artanh 13 represent the ellipsoidal surface with the

equation

x2 + y2

9+ z2 = 1.

The normal vector of the ellipsoid (with α = const.) points in the same direction ashα . Therefore, the product with the vector field G(α, β, γ ) = sinh α hβ + cosβ hγ

will vanish:

G�N = 0,

FOp =∫ ∫

G�Ndβdγ = 0.

The given vector field provides a zero flux through every partial area on the ellipsoid

O ={x ∈ R

3 : x2+y2

9 + z2 = 1}.

• A coordinate surface is given by a fixed value qi = const. (for i = {1, 2, 3})in the system

x = x(q1, q2, q3)

y = y(q1, q2, q3)

z = z(q1, q2, q3).

• For every orthogonal coordinate system, the normal vector of a coordinatesurface (qi = const.) is parallel to the corresponding frame vector hqi .Assuming qi = β, the normal vector is found by the non-normalized framevectors:

N = hα × hγ = ±‖hα‖ · ‖hγ ‖ · hβ.

The positive and negative signs are introduced here, to consider the right-

handed orientation of the ‘frame vectors’{hα, hβ, hγ

}.


4.53. Flux Through a Paraboloid

a. Parabolic Coordinates

The surface

P(α, γ ) =(α cos γ, α sin γ, α2−1

2

)�

fulfills the equation x2 + y2 − 1 = α2 − 1 = 2z. This is equivalent to the coordinatesurface β ≡ 1 with u = α and w = γ in parabolic coordinates. In other words, theparaboloid in Fig. 4.16 is a coordinate surface of the applied coordinate system.We use the ‘frame vectors’ of exercise 14 and obtain the normal vector

N = hα × hγ = −‖hα‖ · ‖hγ ‖ · hβ = −αβ√

α2 + β2 hβ,

where the minus sign is a consequence of the right-handed coordinate system.We evaluate the vector field on the surface and multiply with the normal vector

G = (α2 + 1) cos γ hα + 1

(α2 + 1)hβ,

G�N =(

(α2 + 1) cos γ hα + 1

(α2 + 1)hβ

)� (−α√

α2 + 1 hβ

)= − α√

α2 + 1.

The flux of the vector field is calculated by the surface integral

F =∫ ∫

G�Ndαdβ =2π∫

0

3∫

0

−α√

α2 + 1dαdγ = −2π

[√1 + α2

]3

0= −2π

(√10 − 1

).

Fig. 4.16 Paraboloid:x2 + y2 − 1 = 2z(exercise 53)

Solutions 109

b. Cylindrical Coordinates

For the solution in cylindrical coordinates, we have to re-write the vector field similarto exercise 20. We identify φ ≡ γ and solve for the ‘frame vectors’:

⎛

⎜⎝

hα

hβ

hγ

⎞

⎟⎠ =

⎛

⎜⎜⎝

1√β2+α2

β cos γ 1√β2+α2

β sin γ 1√β2+α2

α

1√β2+α2

α cos γ 1√β2+α2

α sin γ − 1√β2+α2

β

− sin γ cos γ 0

⎞

⎟⎟⎠

⎛

⎜⎝

ijk

⎞

⎟⎠ ,

⎛

⎜⎝

ijk

⎞

⎟⎠ =

⎛

⎝cos γ − sin γ 0sin γ cos γ 00 0 1

⎞

⎠

⎛

⎜⎝

hρ

hγ

hz

⎞

⎟⎠ =

⎛

⎜⎝

hρ cos γ − sin γ hγ

hρ sin γ + cos γ hγ

hz

⎞

⎟⎠ .

By multiplication, we obtain

hα =[β cos γ

(hρ cos γ − sin γ hγ

)+ β sin γ

(hρ sin γ + cos γ hγ

)+ α hz

]

√β2 + α2

=

= β hρ + α hz√β2 + α2

,

hβ =[α cos γ

(hρ cos γ − sin γ hγ

)+ α sin γ

(hρ sin γ + cos γ hγ

)− β hz

]

√β2 + α2

=

= α hρ − β hz√α2 + β2

,

with α = ρ, β = 1 and z = 0.5(α2 − 1). This choice has the consequence, that weevaluate the vector field already on the given paraboloid

G = (ρ2 + 1) cosφ

(hρ + ρ hz√1 + ρ2

)

+ 1

(ρ2 + 1)

(ρ hρ − 1 hz√

ρ2 + 1

)

,

while the general expression in cylindrical coordinates might be more demanding.The normal vector in cylindrical coordinates is

N =⎛

⎝cos γ

sin γ

α

⎞

⎠×⎛

⎝−α sin γ

α cos γ

0

⎞

⎠ =⎛

⎝−α2 cos γ

−α2 sin γ

α

⎞

⎠ = −ρ2 hρ + ρ hz

We evaluate the vector field on the surface and multiply with the normal vector:

G�N =(

(ρ2 + 1) cosφ

(hρ + ρ hz√1 + ρ2

)

+ 1

(ρ2 + 1)

(ρ hρ − 1 hz√

ρ2 + 1

))�(−ρ2 hρ + ρ hz) =

=√

ρ2 + 1 cosφ(−ρ2 + ρ2) + 1√

ρ2 + 13 (−ρ3 − ρ) = − ρ

√ρ2 + 1

.


Hence, we obtain the again flux

F =2π∫

0

3∫

0

−ρ√

ρ2 + 1dρdφ = −2π

[√1 + ρ2

]3

0= −2π

(√10 − 1

).

• This exercise demonstrates the benefit of the ‘optimal’ coordinate system.For known ‘frame vectors’ and coordinate surfaces, the calculation of thenormal vector is performed in one line only.

4.54. ‘Theorem of Gauß’ for Surface Integrals

The surface C is defined by a rotation of a one-dimensional function, which enables aparametrization in cylindrical coordinates with a variable radius ρ(z). Nevertheless,the integrals might not be handy.To apply the theorem of Gauß, we calculate the divergence div F = 2 − 4 + 2 = 0.Therefore, every closed volume will have a total flux of zero through its total surface.The rotational surface C is not closed, but open on both ends (Fig. 4.17). We addthe missing partial areas and calculate the flux through them. For simplicity wechoose circles in the plane z = 0 and z = 6. The normal vector of these circles isN = ρ(± k) with the radius ρ depending on the height.

Fig. 4.17 Surface definedby the rotation ofx = z cos 3

√z around the

z-axis (exercise 54)

Solutions 111

For the ‘bottom’ of the surface C we find ρ = 3 and the flux

F�N =[(2ρ cosφ + ρ sin φ) i − (4ρ sin φ − ρ cosφ) j + (2 · 0 − 16) k

]�[−ρ k]

F1 =2π∫

0

3∫

0

16ρdρdφ = 144π.

In a similar way we calculate for the ‘top’ of C

F�N =[(2ρ cosφ + ρ sin φ) i − (4ρ sin φ − ρ cosφ) j + (2 · 6 − 16) k

]�[ρ k]

F2 =2π∫

0

R∫

0

−4ρdρdφ = −4R2π = −4(3 + cos(6 cos 3

√6))2

π.

The flux through the rotational surface C is

FC = −(F1 + F2) = −144π + 4(3 + cos(6 cos 3

√6))2

π.

4.55. Verification of Stokes’ Theorem for the Parabolic CylinderS = {x ∈ R

3 : y = 4 − z2, 0 ≤ x ≤ 1, y ≥ 0}

a. Circulation via the Surface Integral

We introduce the obvious parametrization S = (x, 4 − z2, z)�

with x ∈ [0, 1] andz ∈ [−2, 2] for the surface, which is visualized in Fig. 4.18. Then we calculate thenormal vector

N =⎛

⎝0

−2z1

⎞

⎠×⎛

⎝100

⎞

⎠ =⎛

⎝012z

⎞

⎠ .

The curl of the vector field is

∇ × G =⎛

⎝

∂∂x∂∂y∂∂z

⎞

⎠×⎛

⎝(1 − 2z)ex

xyxz2

⎞

⎠ =⎛

⎝0

−2ex − z2

y

⎞

⎠ .


Fig. 4.18 Parabolic cylinder S = {x ∈ R3 : y = 4 − z2, 0 ≤ x ≤ 1, y ≥ 0} (exercise 55)

Hence, we obtain the circulation

� =∫ ∫

(∇ × G)�Ndxdz =2∫

−2

1∫

0

⎛

⎝0

−2ex − z2

4 − z2

⎞

⎠

�⎛

⎝012z

⎞

⎠ dxdz =

=2∫

−2

1∫

0

−2ex − z2 + 8z − 2z3dxdz =2∫

−2

−2e + 2 − z2 + 8z − 2z3dz =

=[−2ze + 2z − z3

3+ 4z2 − 1

2z4dz

]2

−2

= −8e + 8

3

• By interchanging the parameters – here x and y – the opposite directionof the normal vector is used. Without further condition (e.g. ‘pointing awayfrom ...’), also the negative answers are correct in Stokes’ theorem.

b. Circulation via the Line Integral

For the line integral, we split the boundary into four parts.We start with the boundary �1 : (t, 0, 2)� with t ∈ [0, 1] and obtain

ω1 =1∫

0

[G�T ]1dt =1∫

0

⎛

⎝(1 − 2 · 2)et

t · 04t

⎞

⎠

�⎛

⎝100

⎞

⎠ dt =1∫

0

(−3)etdt = −3(e − 1).

Solutions 113

Thenext path is thengivenby�2 : (1, 4 − t2, −t)�

and t ∈ [−2, 2]with the integral

ω2 =2∫

−2

[G�T ]2dt =2∫

−2

⎛

⎝(1 − 2(−t))e1

1(4 − t2)1(−t)2

⎞

⎠

�⎛

⎝0

−2t−1

⎞

⎠ dt =2∫

−2

−8t + 2t3 − t2dt =

=[−8t2

2+ 2

t4

4− t3

3

]2

−2

= −16

3.

Considering the orientation, we continue with the curve �3 : (1 − t, 0, −2)�

andwith t ∈ [0, 1]

ω3 =1∫

0

[G�T ]3dt =1∫

0

⎛

⎝(1 − 2(−2))e1−t

0(1 − t)(1 − t)(−2)3

⎞

⎠

�⎛⎝

−100

⎞

⎠ dt =1∫

0

−(5)e1−tdt = −5(e − 1).

For the last boundary, we find the curve �4 : (0, 4 − t2, t)�

with t ∈ [−2, 2]:

ω4 =2∫

−2

[G�T ]4dt =2∫

−2

⎛

⎝(1 − 2t)e0

0(4 − t2)0t2

⎞

⎠

�⎛

⎝0

−2t1

⎞

⎠ dt = 0.

The total circulation is the sum

� =∮

G�Tdt =∑∫

[G�T ]idt = ω1 + ω2 + ω3 + ω4 = −8e + 8

3.

• In case of line integrals, it might be necessary to split the path and to considerthe orientation. The path should enclose the area anti-clockwise.

4.56. Circulation of the FieldG(ρ, φ, z) = −ρ cos φ hρ + ρ z hz in CylindricalCoordinates

The boundary of the surface is again part of Vivani’s curve of exercise 29. We havealready determined its tangent vectors:

T+ = cosφ · (1) hρ + 1 · (sin φ) hφ − hz sin φ, φ ∈ [0, π/2[,T− = cosφ · (1) hρ + 1 · (sin φ) hφ + hz sin φ, φ ∈ [π/2, π [.


Fig. 4.19 Vivani’s surface(upper part) (exercise 56)

We evaluate the vector field along the two curves (with ρ = sin φ and z = | cosφ|)and multiply with the tangent vectors

[G�T ]+ =(− sin φ cosφ hρ + sin φ cosφ hz

)(cosφ hρ + sin φ hφ − sin φ hz) =

= − cos2 φ sin φ − sin2 φ cosφ,

[G�T ]− =(− sin φ cosφ hρ − sin φ cosφ hz

)(cosφ hρ + sin φ hφ + sin φ hz) =

= − cos2 φ sin φ − sin2 φ cosφ.

The modulus of the z-component has no effect on the integrand and we can integratein one step:

� =∮

GTdφ =π∫

0

− cos2 φ sin φ − sin2 φ cosφdφ =[cos3 φ

3− sin3 φ

3

]π

0

=−2

3.

a. Surface Integral

To parametrize the surface in Fig. 4.19, we have to vary the radius in the intervalρ ∈ [0, sin φ]

V(ρ, φ) =⎛

⎝ρ cosφ

ρ sin φ√1 − ρ2

⎞

⎠ φ ∈ [0, π ].

The corresponding normal vector can be expressed in cylindrical coordinates

Solutions 115

N =⎛

⎜⎝

cosφ

sin φ

− ρ√1−ρ2

⎞

⎟⎠×

⎛

⎝−ρ sin φ

ρ cosφ

0

⎞

⎠ = ρ

⎛

⎜⎝

ρ√1−ρ2

cosφ

ρ√1−ρ2

sin φ

1

⎞

⎟⎠ = ρ2

√1 − ρ2

hρ + ρ hz .

We determine the curl of the vector field

∇×G=(1

ρ

∂{ρz}∂φ

− ∂{0}∂z

)hρ +

(∂{−ρ cosφ}

∂z− ∂{ρz}

∂ρ

)hφ +

(1

ρ

∂{ρ0}∂ρ

− 1

ρ

∂{−ρ cosφ}∂φ

)hz =

= 0 hρ + (−z) hφ + (− sin φ) hz

and insert into the surface integral

� =∫∫

(∇ × G)�Ndρdφ =π∫

0

sin φ∫

0

ρ (− sin φ) dρdφ =π∫

0

sin2 φ

2(− sin φ) dφ = − 1

2

π∫

0

sin3 φdφ = − 2

3.

4.57. Circulation Within the Hyperbolic Surface z = x yfor the Vector Field G = ρ3 cos2 φ hφ + z4 hz

a. In the Domain x2 + y2 ≤ 1

In the domain x2 + y2 ≤ 1 the surface is ‘circular pattern’ in space (Fig. 4.20, left).Hence, we find – in cylindrical coordinates – the boundary

� = (cos t, sin t, cos t sin t)� = hρ + cos t sin t hz

Fig. 4.20 Hyperbolic partial areas of z = xy in the domain x2 + y2 ≤ 1 and |x | + |y| ≤ 1(exercise 57)


with the tangent vector

T = (− sin t, cos t, − sin2 t + cos2 t) = hφ + cos 2t hz .

We evaluate the vector field along the curve and multiply with the tangent vector

G = cos2 t hφ + (cos t sin t)4 hz,

G�T = cos2 t + (cos t sin t)4 cos 2t = cos2 t + 1

16sin4 2t cos 2t.

The circulation is calculated by the line integral

� =2π∫

0

cos2 t + 1

16sin4 2t cos 2tdt = π +

[1

16 · 10 sin5 2t

]2π

0

= π.

b. In the Domain |x | + |y| ≤ 1

This figure is a pattern within a square domain (Fig. 4.20, right). It is possible, butnot recommended, to solve the question in cylindrical coordinates. The radius canbe found by ρ = 1

| cosφ|+| sin φ| according to exercise 3. We prefer to re-write intoCartesian coordinates. For the vector field, we obtain

G = ρ3 cos2 φ hφ + z4 hz =⎛

⎝−yx2

x3

z4

⎞

⎠ =: F

with the corresponding curl

∇ × F =⎛

⎝

∂∂x∂∂y∂∂z

⎞

⎠×⎛

⎝−yx2

x3

z4

⎞

⎠ =⎛

⎝004x2

⎞

⎠ .

The normal vector of the surface H(x, y) = (x, y, xy)�

is easy to find

N = ∂H∂x

× ∂H∂y

=⎛

⎝10y

⎞

⎠×⎛

⎝01x

⎞

⎠ =⎛

⎝−y−x1

⎞

⎠ ,

Solutions 117

which leads to the circulation

� =∫ ∫

(∇ × F)�NdA =1∫

−1

1−|x |∫

|x |−1

4x2dydx =1∫

−1

4x2(1 − |x | − |x | + 1)dx =

=0∫

−1

4x2(2 + 2x)dx +1∫

0

4x2(2 − 2x)dx = 4

3.

4.58. Circulation of G = cos(λ + ϑ)( hr + hϑ + hλ) Withina Spherical Triangle

We choose the surface integral with the normal vector N = sin ϑ hr of the unitsphere. Due to later multiplication with the normal vector, we have to derive onlythe hr -component of the curl:

∇ × G∣∣∣hr

= 1

1 sin ϑ

(∂{Gλ sin ϑ}

∂ϑ− ∂{Gϑ }

∂λ

)hr =

= 1

sin ϑ

(− sin(λ + ϑ) sin ϑ + cos(λ + ϑ) cosϑ − sin(λ + ϑ)

)hr =

= 1

sin ϑ

(cos(λ + 2ϑ) − sin(λ + ϑ)

)hr =

(∇ × G)�N = cos(λ + 2ϑ) − sin(λ + ϑ).

We evaluate the circulation within the spherical triangle by the surface integral

� =∫ ∫

(∇ × G)�Ndϑdλ =π/2∫

0

π/2∫

0

cos(λ + 2ϑ) − sin(λ + ϑ)dϑdλ =

=π/2∫

0

[sin(λ + 2ϑ)

2+ cos(λ + ϑ)

]π/2

0

dλ =π/2∫

0

−2 sin λ − cos λdλ = −3.

The vector field depends on the angles but not on the distance to the origin. Theplanar triangle is described by the same angles as the spherical one. Therefore, thecirculation within the planar triangle must be the same ��ABC = −3.


4.59. Circulation within a Parabolic Partial Area

The domain (x − 2)2 + 4(y + 1)2 = 1 describes an ellipse in the plane z = 0 withthe center (2,−1). The semi-major axis a = 1 and the semi-minor axis b = 1

2 areoriented parallel to the coordinate axes (cf. Fig. 4.21). Its projection on the paraboloidz = x2 + y2 has the boundary

x = 2 + cos t

y = −1 + 1

2sin t

z = 4 + 4 cos t + cos2 t + 1 − sin t + 1

4sin2 t.

We evaluate the vector field along the curve and multiply with its tangent vector

F =⎛

⎝2(5 + 4 cos t + cos2 t − sin t + 1

4 sin2 t)

(2 + cos t)2

0

⎞

⎠ ,

F�T =⎛

⎝2(5 + 4 cos t + cos2 t − sin t + 1

4 sin2 t)

(2 + cos t)2

0

⎞

⎠

�⎛⎝

− sin t12 cos t

−4 sin t − cos t − 32 sin t cos t

⎞

⎠ =

= −21

2sin t − 8 sin t cos t − 3

2sin t cos2 t + 2 + 3

2cos t − 1

2cos t sin2 t.

Hence, we obtain the circulation

Fig. 4.21 Partial areas onparaboloid z = x2 + y2

(exercise 59)

Solutions 119

� =∮

F�Tdt =

=2π∫

0

−21

2sin t − 8 sin t cos t − 3

2sin t cos2 t + 2 + 3

2cos t − 1

2cos t sin2 tdt =

=[−4 sin2 t + 3

6cos3 t + 2t − 1

6sin3 t

]2π

0

= 4π

within the parabolic partial area.

4.60. No Circulation for Planar Figures

Based on the Stokes’ theorem we conclude that the circulation will vanish when theproduct of normal vector and curl is zero for every figure. The normal vector of theplane y + 4x − 2z = 0 is N = (4, 1, −2

)�. Hence, we have to determine a vector

field whose curl is orthogonal to N .This problem is under-determined and we can use different methods to find an arbi-trary solution. The methods and the results will differ in complexity. To keep thecalculation simple, we assume a vector field with linear components

F =⎛

⎝a11x + a12y + a13za21x + a22y + a23za31x + a32y + a33z

⎞

⎠

and calculate the corresponding curl

∇ × F =⎛

⎝a32 − a23

a13 − a31

a21 − a12

⎞

⎠ .

Due to the curl operator, every linear or non-linear function g1(x) is eliminated forthe first component, and in a similar way also functions of g2(y) and g3(z) for thesecond and third components, respectively. For the other variables wemake an ansatzof a (linear) vector field

F =⎛

⎝g1(x) + a12y + a13za21x + g2(y) + a23za31x + a32y + g3(z)

⎞

⎠ =⎛

⎝g1(x)

g2(y)

g3(z)

⎞

⎠+⎛

⎝a12y + a13za21x + a23za31x + a32y

⎞

⎠ .

The inner product


(∇ × F)�N =⎛

⎝a32 − a23a13 − a31a21 − a12

⎞

⎠

�⎛⎝

41

−2

⎞

⎠ = 4(a32 − a23) + (a13 − a31) − 2(a21 − a12)!= 0

still contains six degrees of freedom.We normalize by one value aik and set some others to zero to derive a possiblesolution:

• For a21 = 1 and a12 = a23 = a13 = 0 we obtain a31 = 4a32 − 2 and the field

F1 = (0, x, (4a32 − 2)x + a32y)�

• For a21 = 1 and a23 = a31 = a12 = 0 we obtain a13 = 2 − 4a32

F2 = ((2 − 4a32)z, x, a32y)�

.

• Further solutions F j are found by cyclic permutation or setting other componentsto zero.

After deriving a set of independent vectors F j , also the linear combinations solvethe problem:

F =⎛

⎝g1(x)

g2(y)

g3(z)

⎞

⎠+ ξ1

⎛

⎝(2 − 4a32)z

xa32y

⎞

⎠+ ξ2

⎛

⎝0x

(4a32 − 2)x + a32y

⎞

⎠ . . . =

= g +∑

j=1

ξ j F j .

We want to point out once more that many solutions exist for a question in thisgeneral form without further conditions.

4.61. Circulation of the Field G = α2 cos γ hα + hβ inCardioid Coordinates

We use the ‘frame vectors’ of exercise 15 to calculate the curl of the vector field bythe formal determinant

∇ × G = 1

hαhβhγ

det

⎛

⎝hα hα hβ hβ hγ hγ

∂∂α

∂∂β

∂∂γ

hαGα hβGβ hγ Gγ

⎞

⎠ =

= (α2 + β2)5

αβdet

⎛

⎜⎜⎝

1√α2+β2

3 hα1√

α2+β23 hβ

αβ

(α2+β2)2hγ

∂∂α

∂∂β

∂∂γ

1√α2+β2

3 α2 cos γ 1√

α2+β23 0

⎞

⎟⎟⎠ =

Solutions 121

= (α2 + β2)5

αβ

(0 hα + α2

α2 + β2(− sin γ ) hβ+

+ αβ

(α2 + β2)2

−3

2(α2 + β2)−5/2(2α + 2βα2 cos γ ) hγ

)=

= − (α2 + β2)4α

βsin γ hβ − 3α

√α2 + β2(1 + αβ) cos γ hγ .

Coordinate Surface of α = const.In this coordinate system, the coordinate surfaces for constant α or β form rotationalcardioids. Three examples are shown in Fig. 4.22. Hence, we find the normal vectorvia:

Nα = ±hβ × hγ = ±‖hβ‖ hβ × ‖hγ ‖ hγ = −‖hβ‖ · ‖hγ ‖ · hα =

= − αβ

(α2 + β2)2

√1

(α2 + β2)3hα.

The inner product

(∇ × G)�Nα = 0

Fig. 4.22 Partial cardioid surface with γ ∈ [0, π ], β ∈ [0,∞[ and α = const (exercise 61)


is zero and also the circulation:

�α =∫ ∫

(∇ × G)�Nαdβdγ = 0.

Coordinate Surface of β = const.

In an analogous way, we calculate

Nβ = −hα × hγ = −‖hα‖ hα × ‖hγ ‖ hγ = −‖hα‖ · ‖hγ ‖ · hβ =

= − αβ

(α2 + β2)2

√1

(α2 + β2)3hβ

with the inner product

(∇ × G)�Nβ = αβ

(α2 + β2)2

√1

(α2 + β2)33α√

α2 + β2(1 + αβ) cos γ =

= 3α2β(1 + αβ)

(α2 + β2)3cos γ

and the circulation:

�β =∞∫

0

π∫

0

3α2β(1 + αβ)

(α2 + β2)3cos γ dγ dα =

[− sin γ

]π

0

∞∫

0

3α2β(1 + αβ)

(α2 + β2)3dα = 0.

Although the vector field is non-conservative, we found two families of surfaces withvanishing circulation:

• All partial areas on the surface α = const. have zero circulation, because of theorthogonality between the surface’s normal vector and the curl of the vector field.

• On the surface β = const., the circulation vanishes due to symmetry of sin γ forthe interval γ ∈ [0, π ].

Index

AAngle between two curves, 29Astroid, 83, 91

CCardioid, 15, 36Cardioid coordinates, 10, 36, 82Cartesian base vectors, 2Cassini ovals, 20Confocal paraboloids, 34Conic section, 65Conservative vector field, 42, 61Coordinate surface, 107

cardioid coordinates: surface, 36oblate spheroidal coordinates: surface,32

parabolic coordinates: surface, 34Curl-free, 42Curl in cylindrical coordinates, 42Curl in spherical coordinates, 42Curl via formal determinant, 51, 120Curvature, 21, 25Curve lying on a sphere, 25, 27

loxodrome, 29Curve lying on a surface, 21Curvilinear coordinates

arc length, 5curl, 42divergence, 54gradient, 40

DDandelin sphere, 105

EEigen values, 65Ellipse, 65, 118Ellipsoid, 106Elliptic integral, 20Epicycloids, 15

FFolium of Descartes, 62, 84Frame vectors, 3

GGeometrical center, 88

HHelmholtz decomposition, 42Hypocycloids, 83

IIntegral theorem of Gauss, 76Integral theorem of Green, 75Integral theorem of Stokes, 76Intersection of cylinder and (semi)sphere, 63

JJacobian determinant, 76, 100

LLemniscate of Bernoulli, 20, 85

© Springer Nature Switzerland AG 2019M. Antoni, Calculus with Curvilinear Coordinates,https://doi.org/10.1007/978-3-030-00416-3

123

124 Index

MMathematical cylinder, 95, 96, 102Mercator integral, 74Meridian, 35Modified spherical coordinates, 11, 59Multiple points, 13

NNegative radius, 15, 93Non-conservative vector field, 61Non-regular point, 13Normal form of a quadric, 24

OOblate spheroidal coordinates, 10, 32, 80One-sheeted hyperboloid, 32Orthogonality of frame vectors, 52

PParabolic coordinates, 10, 34, 44, 59, 80Paraboloid, 108, 118Parametric representation, 18Parametrization w.r.t. arc length, 21, 22Pedal curve, 94Principle of Cavalieri, 101, 102

QQuadric form, 23

RRadial-dependent vector field, 66Rational parametrization, 18Rotational ellipsoid, 32

SSector formula of Leibniz, 86, 87Spherical loxodrome, 30Square in polar coordinates, 14Substitution of Weierstraß, 30Superellipses, 83

TTangent vector, 2Tetraeder, 101Torus, 24, 97

VVillarceau circles, 24Viviani’s curve, 65, 113

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