Vector Calculus B

7/30/2019 Vector Calculus B

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AER 504: Aerodynamics

Points in three dimensional Cartesian space are

defined by three coordinates defined in terms of

lengths along the Cartesian axisx, y, z. A location

of a point is defined in terms of distance from theorigin, P(x,y,z). The distance from the origin to a

point P is given by,

Crash Course in Vector Calculus

Vector Calculus is used extensively in advanced fluid dynamics and Aerodynamics. This

summary gives a brief overview of some of the topics in this field of mathematics that appliesdirectly to Aerodynamics. The summary begins with a review of partial differentiation and its

applications along with vector notation and basic vector operations. Topics related to vector

differentiation and multiple integration and applications are then presented. All figures are taken

from reference 1.

Review: Vector Notation

= || = 2 + 2 + 2A vector in three dimensional space is defined byscalars applied to unit vectors which are alignedwith the positive coordinate axis. A unit vector is

one with a magnitude or length of 1. A vector

defining point P is written as

(,, ) = + + The scalarsx, y, z are the lengths from the origin

to the point P. The bold variables i, j, k refer tothe reference vectors, shown at left, aligned with

the Cartesian axis,x, y, z respectively. These

reference vectors are unit vectors. All boldvariables commonly refer to vectors. The vector

P has a length or magnitude defined by OP

above and points in the direction defined by aline joining the origin O and point P. The vector

P can be converted to a unit vector,

= || = + + 2 + 2 + 2

The Cartesian components of the unit vector form ofP can be define as the direction Cosines of

the vector,


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= 2 + 2 + 2 , = 2 + 2 + 2 , = 2 + 2 + 2The angle x

refers to the angle between the line OP and the positive Cartesianx axis.

= + + Review: Vector Addition, Subtraction

If vectorsA andB are defined asA = ax i + ayj + azk andB = bx i + byj + bzk, then when thevectors are added and subtracted, the operation is carried out on each individual component,

+ = ( + ) + + + ( + )

= (

)

+

+ (

)

Geometrically, the addition and subtraction

ofA andB creates entirely new directions.Multiple vectors can be added or subtracted

in an easy fashion. Note that the vectorA-B

will have the same magnitude as vectorB-A,but will be in completely the opposite

direction.

Vector Differentiation:

Vector differentiation is conducted in a similar fashion to the differentiation of single variable

functions. By example, consider a point particle moving in time, where time is defined by the

variable t. At every point in time the location of the particle is defined by the Cartesiancoordinates,x, y, z. These coordinates can be viewed as functions of a single parameter, t,

= (), = (), = ()This defines the position of the particle at any time. The

position vectorr, can be used to locate the particle,

= () = () + () + ()The velocity of the particle can be found in a manner similar tosingle variable calculus, by taking the limit of the difference

between positions at time t+tand t.

= = = ( + ) ()


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This yields,

= = + + By considering that the derivative of the position vector was formed by taking the limit of thedifference ofrat two points in time, r, the derivative vector v, is tangent to the path traced out

by the particle. In essence the velocity vector points in the direction that the particle is moving atone instance in time, t. The magnitude ofv, gives the speed of the particle.

= || = dxdt2 + dy

dt2 + dz

dt2

Scalar Products:The scalar product of two vectorsa andb is defined as:

= ||||were is the angle betweena andb. In effect, this product

represents the magnitude ofamultiplied by the projection

ofb ontoa. Calculation of the scalar product for the twovectors is done via,

= + + Since cos() =cos(-), the roles ofa andb can be reversed, a b = b a.

Vector Products:

The vector product ofa andb is defined as:

| | = ||||were is the angle betweena andb. In effect, this product

represents the area of a parallelogram formed from the vectors

aandb as shown. The product yields a vector in a direction

normal to bothaandb. Calculation of the vector product forthe two vectors is done via,

= = + ( )+


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The derivative of the cross product is,

( ) = ddt + ddt The Gradient of a Scalar Function:It is assumed that the reader is familiar with partial derivatives (eg. f/x) which represent rates

of change of some function, f, in the directionx whenfis a function of more than one variable,

f=f(x,y,z,.). The differential, df, is defined as an estimate of the change in a function of more

than one variable,

= + + +The gradient of a scalar functionf(x,y,z) is denoted as f, which is a vector, and is formed:

= + + Assume now that the function represents a surface in 3D space. If the function fis now written in

the form:f(x,y) = zo, wherezo is a constant, then this form provides the level curves of the

surface. As an example, if the function represents the surface of a mountain, then the form f(x,y)

= zo gives the curve that is the cross-section of the mountain at a constant height ofzo. In otherwords, this form gives the contour of the mountain at a specific height. By definition, the value

offalong a level curve will be fixed. Thus the differential, df, will be zero along these level

curves. A vector tangent to a level curve is,

= + or = + Then, on a level curve, = + = = |||| cos = 0

Since neither f nordr are identically zero, then the angle between these two vectors must

always be 90 degrees; they must be perpendicular. Sincedr is always tangent to the curve, then

the direction of the gradient fmust always be at right angles to it, or always pointing in thedirection of the largest change inf.

f

f

f

fconstant

dr


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Sample Calculation of a Scalar Field Gradient

The gradient of a scalar field appears frequently in the analysis of fluid systems. The drivingforce behind all fluid motion is ultimately the change in pressure inside a moving fluid. An

example of the calculation of a gradient, and some interpretation is given below.

If the pressure in a fluid is given by,

20)2()2(),(22+= ybxayxP

Find,

a) the gradient of Pb) the location of the largest pressurec) the gradient at (2,0), (1,1), and (0,2)

d) the equation of a contour of constant pressure, P = 5.a) The gradient is

jybixakz

Pj

y

Pi

x

PP

)2(2)2(2 =

+

+

=

b) The location of the largest pressure is at the local maximum,0)2(20)2(2 == ybxa

Which is at the point x = 2, y = 2.

c) The gradient of pressure at the points (2,0), (1,1), and (0,2),jbijbiaP

40)20(2)22(2)0,2( +==

jbiajbiaP

22)21(2)21(2)1,1( +==

jiajbiaP

04)22(2)20(2)2,0( +==

All three of these vectors point upslope approximately towards the point of highest

gradient at (2,2). It can be shown (math texts) that the gradient of a scalar field provides

the local magnitude and direction of the largest increase of the scalar. It should be

emphasized that since the scalar field is continuous, the gradient is also; meaning its

magnitude and direction will vary continuously.


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d) The equation of a contour of constant pressure is essentially the equation of a curvecorresponding to a value of P equal to some fixed value. In this case we wish P = 5.

20)2()2(522+= ybxa

22)2()2(15 += ybxa

Ifa = b, then we would obtain the equation of a circle; meaning the level contours of

pressure would be circular curves centered at the point (2,2). Ifa is not equal to b, then the

curves would be elliptical. The vectors computed in part c) would all be normal to the local

contours of constant pressure show below.


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The Directional Derivative:

Given that the gradient of a scalar points in

the direction of maximum increase of the

scalar function, we can use this property dodetermine the rate of change of the scalar

in any direction, s. An estimate of thechange in a functionf, with a change in anarbitrary direction s is given by,

=( + , + ) (,)2 + 2 By the mean value theorem,

= +and since, = cos() = sin ()Taking the limit as s goes to 0,

= lim0 cos() + sin() = cos() + sin()The components cos() and sin() can be interpreted as the components of a unit vectorpointing in the desired s direction,

= cos(

)

+ sin(

)

Therefore the derivative of any scalar function in an arbitrary direction s, can be computed from,

= = cos() + sin()wherenis a unit vector pointing in the direction s.

Line Integrals:

A number of applications within Aerodynamics require the calculation of the total influence of acomponent of a vector field along a given path. For example the work done by an arbitrary force

vectorF along a path Cgoing from point Poto P1. To calculate the work first recall that a vectortangent to the curve Cis,

= + + Therefore, the work done by the force moving from point Poto P1 is


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1 If the force vector is given in the form:

= 1(,, )+ 2(,, ) + 3(,, )The work calculation reduces to, = +1 2 +

This form may not be easily used since the force components F1, F2, F3 must be evaluated along

the given curve in terms of the appropriate variable. For example, the integration ofF1 with

respect tox will require the variablesy, z in F1 to be substituted for relations only inx. An easier

approach would be to write the force vector, the curve, anddr

in terms of a single parametert.Then,

1 = 1

If the position vector is, () = ()+ ()+ ()Then,

1

= + 2 + 3 1

Example: IfF =xyi + (y

2+ 1)j, find the work done in moving from Po(0,0,0) to P1(1,1,0) when,

i. the curve is a straight lineii. the curve is along the x axis to (1,0,0), then alongx = 1 to P1

iii. the curve isy2 =xi. The curve is the liney = x. The most convenient method is to change variables to a single

parameter. In this case the easiest choice: x = t, y = t. Therefore, t= 0 referes to point(0,0) and t= 1 refers to point (1,1).

= 2 + (2 + 1) = + (2 + (2 + 1))1 (+) = (22 + 1) = 5 3

1

0


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ii. Here, the curve is made of two straight line components. Referring to point (1,0) aspoint R, the first line is from (0,0) toR, then fromR to (1,1). The integral is broken into

1

=

+

1

On the first segment,y = 0,x is the parameter and rangesfrom 0 to 1, andF = j and dr = dxj.

= = 01

0

On the next segment,x = 1, and dr = dyj,F =yi+(y2+1)j,

withy going from 0 to 1.

1

= ( + (2 + 1)) 1

0

= (2 + 1) = 4 310

iii. On this segment,y2= x, and the parameterization isx = t

2, y = t, were tranges from 0 to 1. On the line,

= 2+ = (2+)

=

+ (

2 + 1)

1 = ( + (2 + 1)) (2+) = (24 + 2 + 1)1

0

= 26 1510

Double Integrals:

Double integrals operate in a similar way to their 1D counterparts where the area under a curve

y = f(x), was easily computed by integration along thex coordinate. In a double integral, the

integrand,z = f(x,y), when integrated over an areaR in thex-y plane, gives the volume under thesurface z = f(x,y). To be more precise, consider the figure below. The surface represented by,

z = f(x,y), sits above an areaR in thex-y plane with vertical walls fromR toz . In the 1D case,

the area under the curve was found by taking the limit as the spacings x where reduced to aninfinitesimally small value. In the double integral case, the area of regionR, is found by taking

the limit as small areas A = xy, are reduced in size as the limit is taken. In this case, both x

and y shrink simultaneously as the limit is taken.

If the areaR is divided along lines ofx = xiwhere i = 1,2,3,.m, andy =yj wherej =

1,2,3, n, an approximation to the volume under the curve over the element Aij = xiyj is

, = ( , )


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The exact volume is found when the limit ofthe above expression is computed as m and n

go to infinity. The variables and are the

values ofx andy on the range,xi-1 < i


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Example of Double Integration

Find the volume common to two cylinders, x2 + y2 = a2 and x2 + z2 = a2

The first cylinder is centred on thez axis while the other is centred on they axis. The volume thatrepresents the intersection of the two cylinders is desired. The volume is symmetric with respect

to all Cartesian coordinate planes,x-y, x-z, y-z. The volume shown in the figure when multipliedby 8 will give the total volume. In this problem, the function that represents the top of thevolume will be chosen as,

= 2 2The projected regionR will lie in thex-y plane. So, the total volume is found from,

= 82 2 Considering the regionR as shown, the top bound of the region is = 2 2. The lowerbound isy = 0. Therefore,

= 8 2 2=22=0 ==0 = 8 2 2

0

22 ==0

= 8 (2 2)=

=0 = 8 2 3

3 0 =16

3

3


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Triple Integrals:

Triple integrals are performed in a manner similar to double integrals. The only difference is that

an extra layer of integration is added making the result applicable over a volume and not an area.

The concept of using a curve as the boundsfor the first integration in a double integral,

is extended such that in a triple integral,the first bounds are surfaces. As anexample, consider the calculation to find

the total mass inside a volume where the

density of the material varies in space,

(x,y,z). Following the indexing developedunder double integrals, the volume is

divided along lines ofx = xiwhere i =1,2,3,.m,y =yj wherej = 1,2,3, n, and

z =zkwhere k = 1,2,3, p. Here,

xi=xi-1-xi , yj=yj-1-yj, zk=zk-1-zk.

An approximation to the mass in thevolume is,

, , = ( , , ) The variable kdenotes a value ofz such that zk-1 < k


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Then the coordinates of the centre of mass of the volume are,

= = = Example calculation, triple integral,Find the volume enclosed by the two surfaces: z = x

2+ 3y

2 and z = 8 x

2 y

2.

The two surfaces intersect on the elliptic cylinder,

2 + 32 = 8 2 2or 2 + 22 = 4The volume projects onto a region in thex-y plane bounded byx2+2y2=4. In this projected

region, the limits of the ellipse are 2 inx, and 21/2 iny. Take the first integration to be inz, the

second iny, then inx.

= ==+ =(42)/2

=(42)/2 =2=2 = (8 22 42)=(42)/2=(42)/2 =2=2

=

2(8

2

2)

4 2

2

8

3

4 2

2

3

2

=2

=

2

= 423 (4 2)3 2=2=2 = 82

Surface Integrals:

Some fields of study may require the evaluation of a function over a 3D surface. For example the

force on the surface of a wing due to surface pressure can be evaluate provided the pressure isknownp =p(x,y,z), and the wing surface shape Sis defined by z =f(x,y). The pressure on the

wing surface is thenp =p(x,y,f(x,y)), which must be integrated over S. Using the nomenclature of

the double integral, this surface force calculation is then,

(,, ) = m,n( , , )Sij

=1=1

Here, ij is the value offwhen evaluated at the points i, j, or ij =f(i, j). The parameter Sij

is an element of surface area on the surface S. The integration can be completed if the surface


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element Sij can be equated to a projected element

on thex-y plane Aij, or some other plane. Whenthis is done, we can use the double integral as

previously defined to do the analysis. First,

consider a surface in 3D space represented by the

equation z =f(x,y). The differential of this surfacecan be written,

= + = +or

+ = 0This equation can be viewed as the product of the

vector

= + with a vector on the surface

= + + or = 0Since the vector dr is always along the surface, then

the only way the equation above is true is if the

vectorn is normal to the surface. Consider a smallelement on the surface dS, which has a projection

onto thex-y plane called dA. On the element dS,

the angle betweenn and the vertical vectorkis ,

which is given by

cos = |||| = 1 +2 +21 2

This relation gives the angle of inclination between the surface dS and its corresponding

projection dA, so that dA = cos() dS.

= 1 +2 +21 2 = 1 +2 +21 2 So long as the surface Sis simple, this relation allows the surface integral to be evaluated as wehad done for the double integral. A simple surface is one where a vertical line cuts through the

surface at only one point. The surface integral forp above can be written,

Sij

xi

yj

Y

X

Z

Aij

(i,j)x=xi

x=xi-1

y=yjy=yj-1

z=f(x,y)


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(,, ) = ,,(,)1 +2 +21 2 In this equation the projected region onto thex-y plane is denoted asR. If the projection was ontothez-x plane instead (the surface defines asy = f(x,z),

(,, ) = (,(, ), )(1 +2 +2)1 2 Note that if the value of the integrand is 1, the surface integral gives the area of the surface.

Example Surface Integral

Find the area of the plane ax + by + cz +d =0, inside the cylinder x2

+ y2

= r2.

Using a surface integral with an integrand of 1, the surface is easily projected onto the x-y plane.Re-writing,

= (,) = ( + + )/and = = = = 1 +

2 + 21

2

= 1 + 2

+ 2

1

2

Since the remaining integral is just the area of the projected region R, which is a circle of radius

r, = 2 [2 + 2 + 2]1 2 The Curl of a Vector Field,

The curl of a vector field V= ui + vj + wk, is defined as, curl V= V,

= x y z = wy vz + uz wx+ vx uy

If the vector Vis the velocity of the fluid, the curl is twice the local rotation rate of the fluid.


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The Divergence of the Vector Field,

The divergence of a vector field V, is div V= V,

=

=

+

+

The divergence of a velocity field can be shown to be an expression of mass conservation.

The Divergence theorem,The divergence theorem, which is also known as Gauss theorem, states that the divergence of a

vector when integrated over a closed volume (V) is equal to the scalar product of the vector withthe unit normal of the volume surface integrated over the volume surface (S),

dV

=

Gauss theorem in the plane is an extension of the divergence theorem applied to a 2D surface A

with a perimeter l,

= The divergence theorem is used to simplify the control volume approach to conservation relation

in fluid dynamics.

The Gradient theorem,The gradient theorem, like the divergence theorem, allows simplification of the basic

conservation relations of fluid dynamics.

dV = However, in this case the theorem can be used to show that the pressure force action on a surfaceis equal to the gradient of the pressure integrated over the enclosed volume.

References

1. Greenspan, H.P., Benney, D.J., Calculus: An Introduction to Applied Mathematics

, McGraw-

Hill Co., New York, 1973.

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