29 1 Line Ints Vecs

8/13/2019 29 1 Line Ints Vecs

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Contentsontents Integral VectorCalculus

29.1 Line Integrals Involving Vectors 2

29.2 Surface and Volume Integrals 34

29.3 Integral Vector Theorems 54

Learning

In this Workbook you will learn how to integrate functions involving vectors. You will learn

how to evaluate line integrals i.e. where a scalar or a vector is summed along a line or

contour. You will be able to evaluate surface and volume integrals where a function

involving vectors is summed over a surface or volume. You will learn about some theorems

relating to line, surface or volume integrals i.e Stokes' theorem, Gauss' theorem and

Green's theorem.

outcomes
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Line Integrals

Involving Vectors

29.1Introduction

28 considered the differentiation of scalar and vector fields. Here we consider how to integratesuch fields along a contour of integration. Firstly, integrals along a line will be considered in a general(non-vector) context. Subsequently, line integrals involving vectors will be considered. These canintegrate either to a scalar or to a vector depending on the form of integral used. Of particular

interest are the integrals of conservative vector fields.

Prerequisites

Before starting this Section you should. . .

have a thorough understanding of the basictechniques of integration

be familiar with the operators div, grad andcurl

Learning OutcomesOn completion you should be able to. . .

find the divergence, gradient or curl of avector or scalar field

2 HELM (2005):Workbook 29: Integral Vector Calculus


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1. Line integrals28 was concerned with evaluating an integral overall points within a rectangle or other shape

(or over a cuboid or other volume). In a related manner, an integral can take place over a line orcurve running through a two-dimensional (or three-dimensional) shape.

Line integrals in two dimensions

A line integral in two dimensions may be written asC

F(x, y)dw

There are three main features determining this integral:

F(x, y) This is the function to be integrated e.g. F(x, y) =x2 + 4y2.

C This is the curve along which integration takes place. e.g.y=x2 orx= sin y

or x=t 1; y=t2. The last case is wherexandy are expressed in termsof a parametert.

dw This states the variable of the integration. Three main cases aredx,dy andds.

Here s is arc length and so indicates position along the curveC.

ds may be written asds=

(dx)2 + (dy)2 or ds=

1 +

dy

dx

2dx.

A fourth case is whenF(x, y)dw has the form:F1dx+F2dy. This is a combination of the cases

dxanddy.

The technique with a line integral is to express all quantities in an integral in terms of a singlevariable. Often, if the integral is with respect to x or y, the curve C and thefunction F may be expressed in terms of the relevant variable. If the integral is carried outwith respect tods, normally everything is expressed in terms ofx. Ifx andy are given in terms ofa parametert, normally everything is expressed in terms oft.

HELM (2005):Section 29.1: Line Integrals Involving Vectors

3


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Example 1

Find

c

x (1 + 4y) dx whereC is the curvey =x2, starting fromx= 0, y = 0

and ending at x= 1, y= 1.

Solution

As this integral concerns only points alongCand the integration is carried out with respect tox,y may be replaced byx2. The limits on x will be 0 to 1. So the integral becomes

C

x(1 + 4y)dx =

1x=0

x

1 + 4x2dx=

1x=0

x+ 4x3dx

= x2

2 +x4

1

0

= 12

+ 1 (0) =3

2

Example 2

Find

c

x (1 + 4y) dy where C is the curve y = x2, starting from

x= 0, y = 0 and ending at x= 1, y= 1. This is the same as Example 1 otherthandxbeing replaced bydy.

Solution

As this integral concerns only points alongCand the integration is carried out with respect toy,everything may be expressed in terms ofy, i.e. x may be replaced byy1/2. The limits on y willbe 0 to 1. So the integral becomes

C

x(1 + 4y)dy =

1y=0

y1/2 (1 + 4y) dx=

1y=0

y1/2 + 4y3/2dx

=

23y3/2 +8

5x5/210

=

23

+85

(0) =3415



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Example 3

Find

c

x (1 + 4y) ds whereC is the curvey=x2, starting fromx= 0, y = 0

and ending at x = 1, y = 1. Once again, this is the same as the previous

two examples other than the integration being carried out with respect to s, thecoordinate along the curveC.

Solution

As this integral is with respect tox, all parts of the integral can be expressed in terms ofx, Along

y =x2, ds=

1 +

dy

dx

2dx=

1 + (2x)2dx=

1 + 4x2dx

So, the integral isc

x (1 + 4y) ds=

1

x=0

x

1 + 4x2

1 + 4x2 dx=

1

x=0

x

1 + 4x23/2

dx

This can be evaluated using the transformationU = 1 + 4x2 so dU = 8xdx i.e. dx =dU

8 .

When x= 0, U= 1 and when x= 1, U= 5.The integral therefore equals 1

x=0

x

1 + 4x23/2

dx = 1

8

5U=1

U3/2dU

= 1

82

5

U5/251

= 1

20

55/2 1 2.745

Example 4

Find

C

xy dxwhere, onC,xandyare given byx= 3t2,y=t31fortstartingatt= 0 and progressing tot= 1.

Solution

Everything can be expressed in terms oft, the parameter. Herex= 3t2 sodx= 6t dt. The limitsont aret= 0 andt= 1. The integral becomes

C

xy dx =

1t=0

3t2 (t3 1) 6t dt= 1t=0

(18t6 18t3)dt

= 18

7t7 18

4t4

1

0

=18

7 9

2 0 = 27

14


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Key Point 1

A line integral is normally evaluated by expressing all variables in terms of one variable.

In general C

f(x, y)ds =C

f(x, y)dy=C

f(x, y)dx

TaskaskForF(x, y) = 2x+y2, find CF(x, y)dx, CF(x, y)dy and CF(x, y)dswhereC is the liney= 2x from (0, 0) to (1, 2).

Express each integral as a simple integral with respect to a single variable and hence evaluate eachintegral:

Your solution

Answer 1x=0

(2x+ 4x2)dx, 7

3,

2y=0

(y+y2)dy, 14

3 ,

1x=0

(2x+ 4x2)

5dx, 7

3

5

TaskaskFindCF(x, y)dx,CF(x, y)dy andCF(x, y)ds whereF(x, y) = 1 andC

is the curvey= 12x2 1

4ln x from(1, 1

2) to (2, 2 1

4ln 2).

Your solution



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Answer 21

1dx= 1,

21/4ln21/2

1dy=3

21

4ln 2,

21 (x+

1

4x)dx=

3

2+

1

4ln 2.

TaskaskFind

C

F(x, y)dx,

C

F(x, y)dyand

C

F(x, y)dswhereF(x, y) = sin2xand

C is the curvey= sin x from(0, 0) to (

2, 1).

Your solution

Answer /2

0

sin2x dx= 1, /2

0

2sin x cos2 x dx= 23 /2

0

sin2x

(1 + cos2 x dx Using the substitutionu= 1 + cos2 xgives 23

(2

2 1).

2. Line integrals of scalar products

Integrals of the formC

Fdr, referred to at the end of the previous sub-section, occur in applicationssuch as the following.

r

vT

T

B

A

v

S(current position)

Figure 1


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Consider a cyclist riding along the road fromAtoB (Figure 2). Suppose it is necessary to find thetotal work the cyclist has to do in overcoming the wind.

The work done is proportional to the component of the wind speed in the direction travelled i.e.proportional tov r

The total work done overAB is approximatelyall r v r, where the summation is carried out over

all small elements making upAB.

In the limitr 0, the total work done overAB is limr0

allv r = BA

v r.This is an example of the integral along a specific line of the scalar product of a vector field anda vector describing the contour. The term scalar line integral is often used for integrals of thisform but the role of the vectorv should not be forgotten. The vectordr may be considered to bedx i+dy j+dz k.

Multiplying out the scalar product, in three dimensions, the scalar line integral of the vector F

along contourC is given byC

F dr and equals C{Fxdx+Fydy +Fzdz} in three dimensions

(

C

{Fxdx+Fydy} in two dimensions.)If the contourChas its start and end points in the same positions i.e. it represents a closed contour,

the symbol

C

rather than

C

is used, i.e.

C

F dr .As before, to evaluate the line integral, express the path and the functionFin terms of eitherx,yandz, or in terms of a parametert. Note that in examplest often represents time.

Example 5

Find

C

{2xy dx 5x dy} whereCis the curvey=x3 withxvarying fromx= 0tox= 1.

Solution

It is possible to split this integral into two different integrals and express the first term as a functionofx and the second term as a function ofy. However, it is also possible to express everything interms ofx. Note that onC,y=x3 sody = 3x2 dxand the integral becomes

C

{2xy dx 5x dy} = 1x=0

2x x3 dx+ 5x3x2 dx

=

10

(2x4 15x3)dx

=

2

5x5 15

4x410

=2

515

4 0 = 67

20



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Key Point 2

An integral of the form C

F

dr may be expressed as C{

Fxdx+F

ydy+F

zdz

}. Knowing the

expression for the pathC, every term in the integral can be further expressed in terms of one ofthe variablesx,y orzor in terms of a parametertand hence integrated.

If an integral is two-dimensional there are no terms involvingz.

The integral

C

F dr evaluates to a scalar.

Example 6Three paths from (0, 0) to (1, 2) are defined by

(a) C1: y = 2x(b) C2: y = 2x

2

(c) C3: y = 0 from(0, 0) to (1, 0) andx= 1 from (1, 0) to (1, 2)

Sketch each path and find

F dr, whereF =y2

i+xyj , along each path.

Solution

(a)

F dr=

y2dx+xydy

. Alongy= 2x,dy

dx= 2 sody= 2dx. Then

C1

F dr = 1x=0

(2x)2 dx+x (2x) (2dx)

= 10

4x2 + 4x2dx= 10

8x2dx=

83x210

=83

y = 2x

C1

A(1, 2)

x

y

1

2

Figure 2(a)


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Solution (contd.)

(b)

F dr=

y2dx+xydy

. Alongy= 2x2,dy

dx= 4xsody= 4xdx. Then

C2

F dr =

1

x=0

2x22

dx+x

2x2

(4xdx)

=

10

12x4dx=

12

5x510

=12

5

y= 2x2

A(1, 2)

C2

x

y

1

2

Figure 2(b)

(c) As the contourC3, has two distinct parts with different equations, it is necessary to break thefull contourOA into the two parts, namelyOB andBA whereB is the point (1, 0). Hence

C3

F dr= BO

F dr+ AB

F dr

AlongOB,y= 0 sody = 0. Then BO

F dr= 1x=0

02dx+x 0 0= 1

0

0dx= 0

AlongAB,x= 1 sodx= 0. Then BA

F dr= 2y=0

y2 0 + 1 y dy= 2

0

ydy =

1

2y220

= 2.

Hence

C3

F dr= 0 + 2 = 2

x

y

1

2

y= 0

x= 1

C3

A(1, 2)

Figure 2(c)



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Key Point 3

In general the value of the line integral depends on the path of integration as well as the end points.

Example 7

Find

OA

F dr, whereF =y2i+xyj (as in Example 6) and the path fromAtoOis the straight line from (1, 2) to (0, 0), that is the reverse ofC1 in Example6(a).

DeduceC

F dr, the integral around the closed pathCformed by the parabolay = 2x2 from (0, 0) to (1, 2) and the liney= 2x from (1, 2) to (0, 0).

Solution

Reversing the path swaps the limits of integration, this results in a change of sign for the value ofthe integral.

OA

F

dr=

A

O

F

dr=

8

3

The integral along the parabola (calculated in (iii) above) evaluates to 12

5 , then

C

F dr=C2

F dr+C4

F dr= 125 8

3= 4

15 0.267

Example 8Consider the vector field

F =y2z3i+ 2xyz3j+ 3xy2z2k

LetC1 andC2 be the curves fromO= (0, 0, 0) toA= (1, 1, 1), given by

C1 : x=t, y =t, z=t (0 t 1)C2 : x=t, y =t

2, z=t2 (0 t 1)

(a) Evaluate the scalar integral of the vector field along each path.

(b) Find the value ofC

F dr whereC is the closed path alongC1 fromO toAand back alongC2 toO.


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Solution

(a) The pathC1 is given in terms of the parametertbyx=t,y=tandz=t. Hence

dxdt

=dydt

=dzdt

= 1 anddrdt

=dxdt

i+dydt

j+dzdt

k =i+j+k

Now by substituting forx=y=z=t inFwe have

F =t5i+ 2t5j+ 3t5k

HenceFdrdt

=t5 + 2t5 + 3t5 = 6t5. The values oft= 0 andt= 1 correspond to the

start and end point ofC1 and so these are the required limits of integration. Now

C1

F

dr=

1

0

F

dr

dt

dt= 1

0

6t5dt= t61

0

= 1

For the pathC2 the parameterisation isx=t2,y=tandz=t2 so

dr

dt = 2ti +j+ 2tk.

Substitutingx=t2,y=tandz=t2 inFwe have

F =t8i+ 2t9j+ 3t8k andFdrdt

= 2t9 + 2t9 + 6t9 = 10t9

C2

F dr= 10

10t9dt=

t1010

= 1

(b) For the closed pathCC

F dr=C1

F dr C2

F dr= 1 1 = 0

Summary

Vector Field Path Line IntegralF C

1 1

F C2 1F closed 0

Note that the line integral ofF is 1 for both paths. This would hold for any path from (0, 0, 0) to(1, 1, 1). The fieldF is an example of a conservative vector field; these are discussed in detail inthe next subsection.

In

C

Fdr, the vector fieldF may be the divergence of a scalar field or the curl of a vectorfield.



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TaskaskConsider the vector field

G=xi+ (4x y)jLetC1 andC2 be the curves fromO= (0, 0, 0) toA= (1, 1, 1), given by

C1 : x=t, y=t, z=t (0 t 1)C2 : x=t

2, y=t, z=t2 (0 t 1)

(a) Evaluate the scalar integral

C

G dr of each vector field along eachpath.

(b) Find the value of

C

G dr whereC is the closed path alongC1 fromO toAand back alongC2 toO.

Your solution


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Answer

(a) The pathC1 is given in terms of the parametert byx=t,y=tandz=t. Hence

dx

dt =

dy

dt =

dz

dt = 1 and

dr

dt =

dx

dti+

dy

dtj+

dz

dtk=i+j+k

Substituting forx=y=z=t inG we have

G=ti+ 3tj andG drdt

=t+ 3t= 4t

The limits of integration aret= 0 andt= 1, thenC1

G dr= 10

G drdt

dt=

10

4tdt=

2t210

= 2

For the pathC2 the parameterisation isx=t2

,y=tandz=t2

so

dr

dt = 2ti +j+ 2tk.

Substitutingx=t2,y=tandz=t2 inG we have

G=t2i+

4t2 tj andG drdt

= 2t3 + 4t2 t

C2

G dr= 10

2t3 + 4t2 t dt= 1

2t4 +

4

3t3 1

2t210

=4

3

(b) For the closed pathC C

Gdr=

C1

Gdr

C2 G dr= 2 4

3=

2

3

Example 9

Find

C

(x2y) dr whereC is the contoury= 2x x2 from (0, 0) to (2, 0).

Solution

Note that(x2y) = 2xyi+x2j so the integral isC

2xy dx+x2 dy

.

Ony= 2x x2,dy= (2 2x)dxso the integral becomesC

2xy dx+x2 dy

=

2x=0

2x(2x x2)dx+x2(2 2x)dx

= 2

0

(6x2 4x3)dx= 2x3 x4

2

0

= 0



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TaskaskEvaluate

C

F dr, whereF = (x y)i+ (x+y)j along each of the followingpaths

(a)C1 : from (1, 1) to (2, 4) along the straight liney= 3x 2:(b)C2 : from (1, 1) to (2, 4) along the parabolay=x

2:

(c)C3 : along the straight linex= 1 from (1, 1) to (1, 4) then along thestraight liney= 4 from(1, 4) to (2, 4).

Your solution

Answer

(a)

21

(10x 4)dx= 11,

(b) 2

1

(x+x2 + 2x3)dx=34

3 ,

(c)

41

(1 +y)dy+

21

(x 4)dx= 8

TaskaskFor the functionFand paths in the last Task, deduce

F dr for the closed

paths

(a)C1 followed by the reverse ofC2.

(b)C2 followed by the reverse ofC3.

(c)C3 followed by the reverse ofC1.

Your solution


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Answer

(a)13

, (b) 10

3, (c)3.

Exercises

1. Consider

C

F dr, whereF = 3x2y2i + (2x3y 1)j. Find the value of the line integral alongeach of the paths from (0, 0) to (1, 4).

(a)y= 4x (b)y= 4x2 (c)y= 4x1/2 (d)y= 4x3

2. Consider the vector fieldF = 2xi + (xz 2)j+ xyk and the two curves between (0, 0, 0) and(1,1, 2) defined byC1 :x =t

2,y=

t,z= 2t for0

t

1.

C2 :x =t 1,y= 1 t,z= 2t 2 for1 t 2.

(a) Find

C1

F dr,C2

F dr

(b) Find

C

F dr whereC is the closed path from (0, 0, 0) to (1, 1, 2) alongC1 and backto (0, 0, 0) alongC2.

3. Consider the vector fieldG=x2zi + y2zj + 13

(x3 + y3)k and the two curves between (0, 0, 0)and (1,

1, 2) defined by

C1 :x =t2,y= t,z= 2t for0 t 1.C2 :x =t 1,y= 1 t,z= 2t 2 for1 t 2.

(a) Find

C1

G dr,C2

G dr

(b) Find

C

G dr whereC is the closed path from (0, 0, 0) to (1, 1, 2) alongC1 and backto (0, 0, 0) alongC2.

4. FindCF (dxi+dyj) alongy= 2x from (0, 0) to (2, 4) for

(a)F = (x2y)(b)F = (1

2x2y2k)

Answers

1. All are 12.

2. 2, 53

, 0.

3. 0, 13

, 0.

4. 16, 16.



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3. Conservative vectorfieldsFor certain of the line integrals in the previous section, the integral depended only on the vector fieldFand the start and end points of the contour but not on the actual path of the contour betweenthe start and end points. However, for other line integrals, the result depended on the actual details

of the path of the contour.

Vector fields are classified according to whether the line integrals are path dependent or path in-dependent. Those vector fields for which all line integrals between all pairs of points are pathindependent are called conservative vector fields.

There are five properties of conservative vector field (P1 to P5). Since it is impossible to check thevalue of every line integral over every path, it is possible to use these five properties (and in particularproperty P3 below to determine whether a vector field is conservative. They are also used to simplifycalculations with conservative vector fields.

P1 The line integral BA F dr depends only on the end pointsA andB and is independent of

the actual path taken.

P2 The line integral around any closed curve is zero. That is

C

F dr= 0 for allC.

P3 The curl of a conservative vector fieldF is zero i.e. F = 0.P4 For any conservative vector fieldF, it is possible to find a scalar field such that =F.

Then,

C

F dr=(B) (A) whereAandB are the start and end points of contourC.

P5 All gradient fields are conservative. That is,F = is a conservative vector field for anyscalar field.

Example 10The following vector fields were considered in the Examples of the previous sub-section.

1.F1=y2i + xyj 2.F2= 2xi + 2yj 3.F3=y

2z3i + 2xyz3j+ 3xy2z2k

4.F4=xi+ (4x

y)j

Determine which of these vector fields are conservative e.g. by referring to theanswers given in the solution. For those that are conservative find a scalar fieldsuch thatF = and use property P4 to verify the line integrals found.

Solution

1. Two different values were obtained for line integrals over the paths C1 andC2. Hence,by P1,F1 is not conservative. [It is also possible to reach this conclusion by finding that

F= yk= 0.]


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Solution

2. Both line integrals from(0, 0) to (4, 2) had the same value i.e. 20 and for the closed path theline integral was0. This alone does not mean thatF2is conservative as there could be other untriedpaths giving different values. So by using P3

F2=

i j k

x

y

z

2x 2y 0

=i(0 0) j(0 0) +k(0 0) = 0

As F2= 0, P3 gives thatF2 is a conservative vector field.Now, find asuch thatF2= . Then

xi+

yj = 2xi+ 2yj.

Thus

x= 2x =x2 +f(y)

y = 2y =y2 +g(x)

=x2 +y2(+ constant)

Using P4:

(4,2)(0,0)

F2 dr= (4,2)(0,0)

() dr=(4, 2) (0, 0) = (42 + 22) (02 + 02) = 20.

3. The fact that line integrals by two different paths between the same start and end points is

consistent withF3 being a conservative field. So too is the fact that the integral around a closedpath is zero. However, neither fact can be used to conclude thatF3 is a conservative field. Thiscan be done by showing that F3= 0.

Now,

i j k

x

y

z

y2z3 2xyz3 3xy2z2

= (6xyz2 6xyz2)i (3y2z2 3y2z2)j+ (2yz3 2yz3)k= 0.

As F3= 0, P3 gives thatF3 is a conservative field.

To findthat satisfies=F3, it is necessary to satisfy

x=y2z3 =xy2z3 +f(y, z)

y = 2xyz3 =xy2z3 +g(x, z)

z = 3xy2z2 =xy2z3 +h(x, y)

=xy2z3

Using P4: (1,1,1)

(0,0,0)

F3 dr=(1, 1, 1) (0, 0, 0) = 1 0 = 1.



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Solution

4. As the integral alongC1 is2 and the integral alongC2 (same start and end points but differentintermediate points) is 4

3,F4 isnota conservative field.

Note that

F4= 4k= 0 so this is an independent conclusion thatF4 isnotconservative.

Engineering Example 1

Work done moving a charge in an electric field

Introduction

If a charge,Q, is moved through an electric field,E, fromAtoB, then the workrequired is given by the line integral

WAB = Q BA

E dl

Problem in words

Compare the work done in moving a charge through an electric field via two different paths.

Mathematical statement of problem

For a fieldEgiven by

E = Q

40r2r

= Q

40(x2 +y2 +z2) xi+yj +zk

x2 +y2 +z2

=Q(xi+yj +zk)

40(x2 +y2 +z2)3

2

withQ = 108C (the field given by a single charge), find the work done in bringing a charge ofq= 1010C from the pointA = (10, 10, 0) to the pointB = (1, 1, 0) (where the dimensions are inmetres)

(a) by the straight liney=x,z= 0

(b) by the straight line pair viaC= (10, 1, 0)

Mathematical analysis

(a) HereQ/(40) = 90 so

E=90[xi+yj ]

(x2 +y2)3

2


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asz= 0 over the region of interest. The work done

WAB =

Q

B

A

E

dl

= 1010 BA

90

(x2 +y2)3

2

[xi+yj ] [dxi+dyj ]

Using y=x, dy=dx

WAB = 1010 1

x=10

90

(2x2)3

2

{x dx+x dx}

= 1010 110

90

(2

2)x3 2x dx

= 90 1010

2

110

x2 dx

= 9 109

2

x1110

= 9 109

2

x1110

= 9 10

92

[1 0.1]= 5.73 109 J

O

A

B Cx

y

ab

b

Figure 3

The path comprises two straight lines fromA= (10, 10, 0)toB = (1, 1, 0)viaC= (10, 1, 0)(Figure3).



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The first part of the path isAtoC wherex= 10,dx= 0 andy goes from 10 to 1.

WAC = Q CA

E dl

= 10

10 1y=10

90

(100 +y2)3

2[xi+yj ] [0i+dyj ]

= 1010 110

90y dy

(100 +y2)3

2

= 1010 101u=200

45du

u3

2

substituting u= 100 +y2, du= 2y dy

= 45 1010 101200

u3

2 du

=

45

1010

2u1

2 101

200

= 45 1010

2101

2200

= 2.59 1010J

The second part isC toB, wherey= 1,dy = 0 andxgoes from 10 to 1.

WCB = 1010 1x=10

90

(x2 + 1)3

2

[xi+yj ] [dxi+ 0j]

= 1010 110

90x dx

(x2 + 1)3

2

= 1010

2

u=101

45du

u3

2

substituting u=x2 + 1, du= 2x dx

= 45 1010 2101

u3

2 du

= 45 10102u 122101

= 45 1010

22

2101

= 5.468 109J

The sum of the two componentsWACandWCB is5.73

109J. Therefore the work done over the

two routes is identical.

Interpretation

In fact, the work done is independent of the route taken as the electric fieldE is a conservativefield.


21


22/33

Example 11

1. Show thatI =

(2,1)(0,0)

(2xy+ 1)dx+ (x2 2y)dy

is independent of the

path taken.

2. FindIusing property P1.

3. FindIusing property P4.

4. FindI=

C

(2xy+ 1)dx+ (x2 2y)dy whereC is

(a) the circlex2 +y2 = 1

(b) the square with vertices (0, 0),(1, 0), (1, 1),(0, 1).

Solution

1. The integralI=

(2,1)(0,0)

(2xy+ 1)dx+ (x2 2y)dymay be re-written

C

F dr whereF = (2xy+ 1)i+ (x2 2y)j.

Now F=

i j k

x

y

z

2xy+ 1 x2 2y 0

= 0i+ 0j+ 0k= 0

As F= 0,Fis a conservative field andIis independent of the path taken between (0, 0)and (2, 1).

2. AsI is independent of the path taken from (0, 0) to (2, 1), it can be evaluated along anysuch path. One possibility is the straight liney = 1

2x. On this line,dy = 1

2dx. The integral

Ibecomes

I =

(2,1)(0,0)

(2xy+ 1)dx+ (x2 2y)dy

=

2x=0

(2x 1

2x+ 1)dx+ (x2 4x) 1

2dx

=

20

(3

2x2 1

2x+ 1)dx

= 1

2x3 1

4x2 +x

2

0

= 4 1 + 2 0 = 5



23/33

Solution (contd.)

3. IfF = then

x

= 2xy+ 1 =x2y+x+f(y)

y =x2 2y =x2y y2 +g(x)

=xy2z3.

These are consistent if=x2y+x y2 (plus a constant which may be set equal to zero).SoI=(2, 1) (0, 0) = (4 + 2 1) 0 = 5

4. AsF is a conservative field, all integrals around a closed contour are zero.

Exercises

1. Determine whether the following vector fields are conservative

(a)F = (x y)i+ (x+y)j(b)F = 3x2y2i+ (2x3y 1)j(c)F = 2xi+ (xz 2)j+xyk(d)F =x2zi +y2zj + 1

3(x3 +y3)k

2. Consider the integral

C

FdrwithF = 3x2y2i+ (2x3y1)j. Noting thatFis a conservative

vector field, find a scalar fieldso that =F. Hence evaluate the integralC

F dr whereCis an integral with start-point (0, 0) and end point (1, 4).

3. For the following conservative vector fieldsF, find a scalar field such that =F andhence evaluate theI=

C

F dr for the contoursC indicated.

(a)F = (4x3y

2x)i+ (x4

2y)j; any path from (0, 0) to (2, 1).

(b)F = (ex +y3)i+ (3xy2)j; circlex2 +y2 = 1.

(c)F = (y2 + sin z)i+ 2xyj +x cos zk ; any path from (1, 1, 0) to (2, 0, ).

(d)F = 1

xi+ 4y3z2j+ 2y4zk; any path from (1, 1, 1) to (1, 2, 3).

Answers

1. (a) No, (b) Yes, (c) No, (d) Yes

2.x3y2 y, 12

3. (a)x4y x2 y2,11; (b)ex +xy3,0; (c)xy2 +x sin z,1; (d) ln x+y4z2,143


23


24/33

4. Vector line integrals

It is also possible to form the integrals

C

f(x,y,z)drand

C

F(x,y,z)dr. Each of these integralsevaluates to a vector.

Remembering thatdr=dx i+dy j+dz k, an integral of the formC

f(x,y,z)drbecomesC

f(x,y,z)dx i +

C

f(x,y,z)dy j +

C

f(x,y,z)dz k. The first term can be evaluated by

expressingyandzin terms ofx. Similarly the second and third terms can be evaluated by expressingall terms as functions ofy andzrespectively. Alternatively, all variables can be expressed in termsof a parametert. If an integral is two-dimensional, the term inzwill be absent.

Example 12

Evaluate the integralC

xy2drwhereCrepresents the contoury=x2 from(0, 0)

to (1, 1).

Solution

This is a two-dimensional integral so the term inzwill be absent.

I =

C

xy2dr

=C

xy2(dxi+dyj)

=

C

xy2dx i+

C

xy2 dy j

=

1x=0

x(x2)2dx i+

1y=0

y1/2y2 dy j

=

10

x5dx i+

10

y5/2 dy j

= 16x6

1

0

i+ 27x7/2

1

0

j

= 1

6i+

2

7j



25/33

Example 13

FindI=

C

xdr for the contourCgiven parametrically byx = cos t,y = sin t,

z =t

starting att = 0 and going tot = 2, i.e. the contour starts at

(1, 0, ) and finishes at (1, 0, ).

Solution

The integral becomes

C

x(dxi+dyj +dzk).

Now,x= cos t,y= sin t,z=t sodx= sin t dt,dy= cos t dtanddz=dt. So

I = 20

cos t(

sin t dti+ cos t dtj+dtk)

= 20

cos t sin t dt i+

20

cos2 t dt j+

20

cos t dt k

= 12

20

sin2t dt i+1

2

20

(1 + cos 2t)dt j+

sin t20

k

= 1

4

cos2t20

i + 1

2

t+

1

2sin 2t

20

j + 0k

= 0i+ j=j

Integrals of the form

C

F dr can be evaluated as follows. The vector fieldF =F1i+F2j+F3kanddr=dx i+dy j+dz k so

F dr=

i j k

F1 F2 F3

dx dy dz

= (F2dz F3dy)i+ (F3dx F1dz)j+ (F1dy F2dx)k

= (F3j F2k)dx+ (F1k F3i)dy+ (F2i F1j)dzThere are a maximum of six terms involved in one such integral; the exact details may dictate whichform to use.


25


26/33

Example 14

Evaluate the integral

C

(x2i+ 3xyj) dr whereCrepresents the curvey= 2x2from (0, 0) to (1, 2).

Solution

Note that thezcomponent ofF anddzare both zero.

SoF dr=

i j k

x2 3xy 0

dx dy 0

= (x2dy 3xydx)k

andC

(x2i+ 3xyj) dr= C

(x2dy 3xydx)kNow, onC,y= 2x2 sody= 4xdxand

C

(x2i+ 3xyj) dr =C

{x2dy 3xydx}k

=

1x=0

x2 4xdx 3x 2x2dx k

=

10

2x3dxk

=

12x410

k

= 12k



27/33


Force on a loop from a magnetic field

Introduction

A current in a magnetic field is subject to a force given by

F =J B =I dl Bwhere the current can be regarded as having magnitudeI and flowing (positive charge) in thedirection given by the vectordl. The force is known as the Lorentz force and is responsible for theworkings of an electric motor. If current flows around a loop, the total force on the loop is given bythe integral ofFaround the loop, i.e.

F =

(I dl B) = I

(B dl)

where the closed path of the integral represents one circuit of the loop.

x

y

z

Figure 4

Problem in words

A current of 1 A flows around a circuit in the shape of the unit circle in the Oxy plane. A field of 1G in the positivez-direction is present. Find the total force on the circuit loop.


Choose an origin at the centre of the circuit and use polar coordinates to describe the position of

any point on the circuit and the length of a small element.

Calculate the line integral around the circuit representing the force using the given values of currentand magnetic field.


The circuit is described by

x= cos y = sin z= 0

with

dl = sin d i+ cos d jB = B k


27


28/33

sinceB is constant. Therefore, the force on the circuit is given by

F = IB

k dl=

k dl (sinceI=1A andB=1G)

where

k dl =

i j k

0 0 1

sin d cos d 0

= cos i sin j d

So

F =

2

=0

cos i

sin j d

=

sin i cos j2=0

= (0 0)i (1 1)j= 0

Hence there is no net force on the loop.

Interpretation

At any given point of the circle, the force on the point opposite is of the same magnitude but oppositedirection, and so cancels, leaving a zero net force.



29/33


Magnetic field from a line current

Introduction

A chargeQ, moving at a steady velocityv produces a magnetic field given by

dB =Q04r2

(v r)

where0 is the permeability of free space (4107H m1),r is the distance from the point ofinterest,P, to the line current and r is a unit vector alongr.If, instead of a single charge, a current is used, then it is necessary to integrate over all charges inthe current. So, the total magnetic field due to the current is given by

B =

DC

dB =

DC

0I

4

dl rr2

whereIdl is the continuous form ofQv and the current extends fromC toD. Note that the fieldis perpendicular to both the current and the line from the current toP.

Problem in words

Find the magnetic field strength (or magnetic flux density), measured in tesla (T), due to a currentIdirected vertically downwards, starting atz=cand ending atz= d. What is the field 1 m fromthe current whenc= 5 m,d= 10 m andI= 1 A?


z

h

rdl

h2 +z2

P

Figure 5

Here

Idl= Ikdzi.e.dl = k dz(pointing downwards). Imagine (without loss of generality) a pointPa distancehfrom the line current and a distancezbelow a typical line element of the current. The increment offield is given by

dB = 0I

4(h2 +z2)dl r

where

h2 +z2 is the distance ofP from the typical element. Sincedl =k dz and r is a unitvector, the magnitude of the vector product is

|dl r| = sin dz= h dzh2 +z2


29


30/33

and is in a direction which (by the right-hand-rule) points OUT of the page to the right of theline and IN to it on the left. Knowing the direction of the field, now calculate the magnitude: theincrement of field is given by

dB = 0I

4(h2

+z2

)

h dz

h2 +z2=

0I

4

h(h2 +z2)3/2 dz

so that the total field at a point is

B =

cz=d

0I

4h(h2 +z2)3/2 dz


This integral can be evaluated by means of the substitutionz=htan uwhere

z=h tan u dz=hsec2 u du

z=c u= tan1

(c/h) =ucz= d u= tan1(d/h) =udSubstituting into the total field integral gives

B = 0I

4

ucud

h(h2 sec2 u)3/2h sec2 u du

= 0I

4

ucud

cos u du

h

= 0I

4hsin uuc

ud

= 0I

4h

c/h1 + (c/h)2

+ d/h1 + (d/h)2

as sin(tan1 y) =

y1 +y2

= 0I

4h

c(h2 +c2)

+ d(h2 +d2)

andB =B where is a unit vector in a circumferential direction around the line current. Now ifI= 1 A,c= 5 m,d= 10 m andh= 1 m the magnetic field becomes

B = 10

7 526 + 10

101= 1.98 107T= 1.98 milli-gauss.Interpretation

Note that ifc andd then

B = 0I

4h

c(h2 +c2)

+ d(h2 +d2)

0I

4h[2] =

0I

2h

i.e. the field lines are circles around the line current and the magnetic field strength is inverselyproportional to the distance of the point of interestP from the current.

A scalar or vector involved in a vector line integral may itself be a vector derivative as this nextExample illustrates.



31/33

Example 15

Find the vector line integral

C

(F)drwhereF is the vectorx2i+ 2xyj + 2xzkandC is the curvey =x2,z =x3 fromx = 0 tox = 1 i.e. from (0, 0, 0) to

(1, 1, 1).

Solution

AsF =x2i+ 2xyj + 2xzk, F= 2x+ 2x+ 2x= 6x.The integral

C

( F)dr =C

6x(dxi+dyj +dzk)

=C

6x dx i+C

6x dy j+C

6x dz k

The first term isC

6x dx i=

1x=0

6x dx i=

3x210

i= 3i

In the second term, asy =x2 onC,dy may be replaced by 2x dxsoC

6x dy j =

1x=0

6x 2x dx j= 10

12x2 dx j =

4x310

j = 4j

In the third term, asz=x3 onC,dzmay be replaced by 3x2 dxsoC

6x dz k=

1x=0

6x 3x2 dx k= 10

18x3 dx k=

9

2x410

k=9

2k

On summing,

C

( F)dr= 3i+ 4j+ 92k.


31


32/33

TaskaskFind the vector line integral

C

fdr wheref=x2 andC is

(a) the curvey=x1/2 from (0, 0) to (9, 3).

(b) the liney=x/3 from (0, 0) to (9, 3).

Your solution

Answer(a)

90

(x2i+1

2x3/2j)dx= 243i+

243

5 j, (b)

90

(x2i+1

3x2j)dx= 243i+

243

2 j.

TaskaskEvaluate the vector line integral

C

F dr whenCrepresents the contoury =4 4x,z= 2 2x from(0, 4, 2) to (1, 0, 0) andF is the vector field (x z)j.

Your solution

Answer

10 {(4 6x)i+ (2 3x)k} =i+

1

2k



33/33

Exercises

1. Evaluate the vector line integral

C

( F)dr in the case whereF =xi+xyj +xy2k andC is the contour described byx = 2t,y =t2,z= 1 t fort starting att = 0 and going tot= 1.

2. WhenC is the contoury =x3,z = 0, from (0, 0, 0) to (1, 1, 0), evaluate the vector lineintegrals

(a)

C

{(xy)} dr

(b)

C

(x2i+y2k) drAnswers

1. 4i+73j 2k,2. (a) 0, (b)k

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