UNIT_7.PDF Engg Math

8/12/2019 UNIT_7.PDF Engg Math

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Line and Surface

IntegralsUNIT 7 LINE AND SURFACE INTEGRALS

Structure

7.1 Introduction

Objectives

7.2

Integration of a Vector7.2.1 Line Integral

7.2.2 Types of Line Integral

7.2.3 Evaluation of Line Integral

7.2.4 Path Independence Conservative Fields

7.3 Double Integrals

7.3.1 Properties of Double Integrals

7.3.2 Evaluation of Double Integrals

7.3.3 Applications of Double Integrals

7.3.4

Change of Variables in Double Integrals

7.4

Transformation of Double Integrals into Line Integrals Greens Theorem

7.5

Surface Integrals

7.6

Transformation of Surface Integrals into Line Integrals Stokes Theorem

7.7

Triple Integral

7.7.1 Definition

7.7.2 Properties of Triple Integrals

7.7.3

Volume

7.7.4 Evaluation of Triple Integrals

7.7.5 Physical Applications in Three Dimensions

7.8

Transformation of Volume Integrals into Surface Integrals

7.8.1 Gauss Divergence Theorem

7.8.2 Consequences and Applications of Divergence Theorem

7.8.3

Integral Definitions of Gradient, Divergence Theorem

7.8.4 Physical Interpretation of Divergence and Curl

7.8.5 Modelling of Heat Flow

7.8.6 Greens Theorem and Greens Formula

7.8.7 A Basic Property of Solutions of Laplaces Equation

7.9 Solenoidal and Irrorational Vector Fields Revisited

7.10 Summary

7.11

Answers to SAQs

7.1 INTRODUCTION

In Unit 6, you have learnt about the differentiation of vectors and arrived at the important

concepts of directional derivative, gradient, divergence and curl, in this unit, we shall talk

about integration of vector functions.

In this unit, we shall begin our discussion in section 7.2 with the definition of a line

integral and see that it is a natural generalization of a definite integral.

We shall discuss the concepts involving double integrals and their evaluation, (iii) plane

region may be transformed in double integrals in Section 7.3. Double integrals over a

plane region may be transformed into the line integrals over the boundary of the regionand conversely. This is achieved through the Greens theorem in the plane and will form

the subject of our discussion in Section 7.4. Section 7.5 is devoted to the discussion of

surface integrals.


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Engineering Mathematics The transformation of surface integrals into line integrals and conversely is done through

Stokes Theorem, which will form our subject matter for Section 7.6. In this section, we

shall also take up some applications of the Stokes theorem and give another physical

interpretation of Curl.

You have studied about line integrals, double integrals and surface integrals. In the

process you have learnt to transform double integrals and surface integral into line

integrals. You had learnt that line integrals are the generalisation of a single integral anda surface integral is a sort of generalization of a double integral. In this unit, we shall give

another generalization of double integral called triple integrals or volume integrals.

We shall first of all define triple integrals in Section 7.2, wherein we shall also give the

properties and evaluation of such integrals. This section will be closed with some

physical applications of triple integrals.

We had made use of integral transform theorems-Greens Theorem and Stokes Theorem

in the last unit. In this unit, we shall discuss another important integral transform

theorem, known as Gauss Divergence Theorem, which helps in the transformation of

volume integral to surface integral and conversely. Divergence Theorem has many

important consequences and various applications, some of which have been discussed in

Section 7.3.

We had earlier discussed solenoidal vector fields and irrotational vector fields in Unit 6.

With our knowledge of vector calculus, we have revisited these concepts in Section 7.4,

where we have now given integral form conditions for vector fields to be solenoidal and

irrotational. The summary of the results discussed in this unit is presented at the end of

this unit.

Objectives

After going through this unit, you should be able to

integrate a vector with respect to a scalar and solve problems based on it,

define a line integral, state various types of it and evaluate it, compute double integrals and state the conditions under which the order

to integration can be changed,

explain the need of change of variables in double integrals andevaluation of double integrals by change of variables,

outline method/conditions of transformation of double integral into lineintegral and conversely,

solve problems based on Greens theorem,

define a surface integral and transform it into line integral,

outline the representation of curl in terms of line integral, use Greens theorem and Stokes theorem to relevant physical situations

and solve problems based on them,

define and evaluate triple integral, and

learn the method and conditions under which a volume integral can betransformed into a surface integral.

7.2 INTEGRATION OF A VECTOR

In Unit 6, you have learnt to differentiate a vector w.r. to scalar, You also know that

integration is the inverse process of differentiation. If F (t) andf (t) be two vectorfunctions of a scalar variable t, connected by the relation

( ) ( )d

t tdt

=F f . . . (7.1)


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Line and Surface

Integralsthen we say that ( )tF is integral of ( )tf and write

( ) ( )f t dt t= F . . . (7.2)this gives an indefinite integralof ( )tf .

We can also add an arbitrary constant vector Con the right hand side of Eq. (7.2) and

write it as

( ) ( )t dt t = +f F CThis is because

( ( ) ) ( )d

t tdt

+ =F C f 0d

dt

= C

While solving problems, this constant Ccan be determined using the given initial

conditions. Also, the dimension of C is that of ( ).tF

You also know from calculus, that for scalar function ( ),f x the definite integral

( )b

a

f x dx

can be defined as the limit of a sum. Here we integrate along axis x-axis from ato band

the integralfis a function defined at each point between aand b.

In the same manner, we can define for a scalar variable t, the definite integral of vector

function ( )tF as the limit of a sum. We can write

1 1 2 20

( ) lim ( ) ( ) ( ) , 1, 2, ,i

b

n nt

a n

x dt t t t t t t i n

= + + + = F F F F

where the ranges of integration a t b has been divided into nsub rangescorresponding to increments

1 2 , , ,

nt t t of the variable t and

1 2, , ,

nt t t are values of

t lying respectively in these sub-ranges. Further in the limit ,n as the number ofsub-ranges increases indefinitely, each of the increment t tends to zero. The sum on the

right hand side which is, of course, a vector sum tends to a definite limit, which is the

definite integral of function ( )tF .

A simple example of vector integration is illustrated by a particle of unit mass moving

under gravity with constant accelerationg.Letrdenote the position-vector of the particle

at time t.Then from Newtons second law of motion, the equation of motion is

2

2

d

dt=

rg . . . (7.3)

Now if we wish to find the trajectory of the particle its motion, i.e., if we wish to findr in

terms of t then we integrate Eq. (7.3) and obtain

d

dt=

0

rgt + V . . . (7.4)

where0

V is the initial velocity (corresponding to 0t= ). A second integration yields

21

2

t t= + +0 0

r g V r . . . (7.5)

where0

r is the initial position of the particle.

Thus by interpretation of a vector with respect to scalar, we could determine the

trajectory of the particle when equation of motion is known.


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Engineering MathematicsYou may note that in Eqs. (7.4) and (7.5),

d

dt

randrhave the dimensions of velocity and

dimension of velocity and0

r in equation (7.5) has the dimension of position.

Note that in the above example, the particle was moving with constant accelerationg. In

practice, it may happen that the force acting on the particle or the system may also be

function (scalar or vector) of the scalar variable. For example, a transverseelectromagnetic wave propagating inx-direction may have an electric field

0

2 cos ( )ct x= E E k

. If E is the electric permitivity and is magnetic permeability

of space, then energy flowing through a volume Vper unit time is given by

( )2

V= +2 2U E B

If Tbe the time period for total energy flowing through volume V during one complete

cycle of electro-magnetic oscillation, then

Total energy0

Tdt=

U

0( )

2

T Vdt= + 2 2E B

0( )

2

T Vdt= + E E B B

2 2 2 20 00 0

2 2 cos ( ) cos ( )

2 2

T TV Vct x dt ct x dt = + E B

2 2( and )= = E E E B B B

2 2 20 0 0

2 cos ( )

2 2

TV VE B ct x dt = +

2 20 0 0

1 ( ) 1 cos 4 ( )

4

TV E B ct x dt

= + +

2 20 0

0

sin 4 ( )

1 ( )4

4

T

ct x

V E B t

c

= + +

2 20 0

1 4 4( ) sin ( ) sin4 4 4

V T cT x xc c

= + + +

E B

Notethat in the above integration for the calculation of total energy, we have used2E E E,

the dot product of vectors. It may also happen that the integration of vectors

may involve cross product of vectors.

We may wish to evaluate drA from 0t= to 1t= where 2 xy z x= +A i j k and2 3, 2 , .x t y t z t= = =

Here 2 3 2t t t= + +r i j k

2 (2 2 3 )t t dt = + +dr i j k

and 2 xy z x= +A i j k


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Line and Surface

Integrals 3 3 4 2t t t= +i j k

3 3 4

2

2

2 2 3

d t t t

t dt dt t dt

=

i j k

A r

5 4 5 3 4 (3 2 ) 4 (4 2 )t t dt t dt t t dt = + + +i j k

1 1 15 4 5 3 4

0 0 0

(3 2 ) 4 (4 2 )dr t t dt t dt t t dt = + + + + A i j k 1 1 16 5 6

4 5

000

2 3 2 46 5 6 5

t t tt t

= + + +

i j k

1 2 4 7 2 5 6 5

= + +

i j k

9 2 7 10 3 5

= +i j k

From the above example you must have observed that vector integration involving dot

product or cross product of vectors, essentially reduces to determining the integral of a

scalar with respect to scalar.

Recall that in a definite integral ( ) ,b

af x dx we integrate alongx-axis from a to band the

integralfis a function defined at each point between a and b. However, in many physical

problems, you may observe that a particle may not be moving along a line but along a

curve in space, e.g., if we wish to calculate the work done by a force in moving a particle

along the curve C from positionA to positionB,then the ordinary integration (indefinite

or definite) involving vectors will not provide the result. Such problems lead us to the

consideration of line integrals or curve integrals.

7.2.1 Line Integral

The concept of a line integral is a simple and natural generalization of the concept of a

definite integral

( )b

af x dx . . . (7.6)

In the line integral we do not integrate the integral alongx-axis from a to b; instead, weintegrate along a curve in plane or in space and the integrand is a function defined at the

points of that curve. We can define a line integral in a way similar to that of a definiteintegral.

Consider a curve C in space havingA andBas the initial and terminal points. We mayrepresent Cin parametric from as

( ) ( ) ( ) ( ) ,s x s y s z s= + +r i j k ( )a s b

where sis the arc length of CandAandBcorrespond to s = a and s = b respectively

(Figure 7.1).


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Engineering Mathematics

Figure 7.1 : The Directed Curve CfromAtoB

We assume that ( )sr is continuous and has a continuous first derivative different from

the zero vector for all sunder consideration. The Chas a unique tangent at each of its

point. Such a curve C is called a smooth curve.

Let ( , , )w x y z be a scalar function which defined at each point of C, and is a

continuous function of s.

We divide the curve C, fromAtoB,in an arbitrary manner into nportions.

Let0 1 1( ), , , , ( )

n nP A P P P B= =

be the end points of these portions and let

0 1 2( ) ( )

ns a s s s b= < < < < =

be the corresponding values of s(Figure 7.2)

Let the length of each of these portion is

1 2 3 , , , . . . ,

ns s s s

Then we evaluate wat each point ( , , )k k k

x y z in k

s and consider the sum

1

( , , ) n

k k k k k

x y z s=

w

Figure 7.2 : Sub-division of C

We define the integral of ( , , )w x y z over curve CfromA toB to be limit of the sum as

the subdivision of Cis refined so that number of subdivisions becomes very large and the

largest k

s approaches zero, i.e.,

1

( , , ) lim ( , , ) n

k k k k n k

C

w x y z ds w x y z s =

= . . . (7.7)

whenever the limit on the right-hand side in Eq. (7.7) exists, we call it line integral of( , , )w x y z along C formA toBand denote it as


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Line and Surface

Integrals( , , ) .

C

w x y z ds

Note that since wis continuous and Cis smooth, the limit on the right hand side of

Eq. (7.7) exists and is independent of the choice of subdivision of C.

Moreover, if ( , , ) 1,w x y z w= is a constant function whose value is 1, then

( , , )

C

w x y z ds

gives the length of curve C fromA toB.

The concept of the line integral can be extended to vector integration and we get different

types of line integrals depending upon.

(i) vector element , , . . . ,1 2

n

s s s and performing vector addition, keeping

( , , )w x y z as a scalar function.

(ii)

taking function W(r) instead of scalar function and also taking vector

elements , , . . . ,1 2

n

s s s . Then products of Wwith k

s can be either dot

products or vector products and this in turn gives rise to two types of line

integrals.

We shall now take up various types of line integrals.

7.2.2 Types of Line Integral

The following types of line integrals exists :

(i) When ( , , )w x y z is a scalar function defined at each point curve C fromA

toB,which is divided into parts

, , . . . ,1 2

n

s s s

along the curve, then

1

( , , ) lim ( , , )n

k k kn

kC

x y z x y z

=

= kw ds w s

defines a line integral in ordinary calculus.

(ii) When ( , , )w x y z is a scalar function defined at each point of the curve C

fromAtoB.which is divided into nparts of vector elements

, , . . . ,1 2

n

s s s along the curve Cand perform vector addition, then

1

( , , ) lim ( , , ) n

k k k k n

kC

x y z x y z s

=

= w ds w

defines a line integral, where value is a vector.

(iii) When ( , , )x y zW is a vector function defined at each point of the curve C

fromAtoB, which is divided into nparts of vector elements

, , . . . ,1 2

n

s s s along the curve Cand the product of ( , , )k k k

x y zW in

k

s in dot product of vector, then

1

( , , ) . lim ( , , ) .n

k k kn

kC

x y z x y z

=

= kW ds W s

defines a line integral, whose value is a scalar.

Moreover, ifris a position vector of a point on curve C,then ds ~ dr and

in the limit when each sub-interval tends to be zero, ds = dr ; therefore the

last three line integrals may also be written as


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Engineering Mathematics( , , ) , ( , , )

C C

x y z x y z w dr W dr and ( , , )C

x y z W dr

respectively.

Figure 7.3 : Formula (c)

In all the above line integrals, it is assumed that the path of integration is

piecewise smooth.

For a line integral over a closed path C, the symbol

(instead of )C C

is sometimes used in the literature.

From the definition of line integral, it follows that the following properties are valid for

line integrals :

(a)

C C

k wds k wds= (kConstant)

(b) ( )

C C C

f g ds f ds g ds+ = +

(orientation of Cis the same in all the three integrals).

(c)

1 2

,

C C C

wds w ds w ds= + where the path Cis subdivided into two arcs 1C

and2

C , which have the same orientation as C.

Notethat if the sense of integration along a curve Cis reversed, the value of the lineintegral is multiplied by 1.

The question is now arises is How is a line integral evaluated? We shall now answerthis question.

7.2.3 Evaluation of Line Integral

A line integral can be evaluated by reducing it to a definite integral. This reduction is

quite simple and is done by means of representation of the path of integration Casfollows :

If Cis represented by

( ) ( ) ( ) ( ) , ,t x s y s z s a s b= + + r i j k

where sis the arc length of C, then we can immediately write

( , , ) [ ( ), ( ), ( )] ,

b

af x y z ds f x s y s z s ds= C . . . (7.8)

the integral on the right being a definite integral. In applications, mostly representation ofCis the form


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Line and Surface

Integrals ( ) ( ) ( ) ( ) ,t x t y s z s= + +r i j k

0 1,t t t

where tis any parameter, or may easily be converted to this form. In this case, we can use

[ ( )] ( ),s t t=r r

so that [ ( )] ( )s t t=x x and so on.

Then we can use1

0

( , , ) [ ( ), ( ), ( )] ,t

t

dsf x y z ds f x t y t z t dt

dt=

C

. . . (7.9)

where 2 2 2ds

y zdt

= = + + r r x . . . (7.10)

Here we assume that ( )tr and ( )tr are continuous and ( ) 0,t r in agreement with theassumption mentioned in Section 7.2.2.

We now illustrate the theory discussed so far with the help of a few examples.

Example 7.1

Evaluate 2 2 2 2( ) ,

C

x y z ds+ + where Cis the arc of circular helix

( ) cos sin 3t t t t = + +r i j k,

from (1,0,0)A to (1,0,6 ).B

Solution

Here ( ) cos sin 3t t t t = + +r i j k,

( ) sin cos 3t t t= + +r i j k

Now 2 2sin cos 9 10ds

t tdt

= = + + = r r

On C,

2 2 2 2 2 2 2 2 2( ) [cos sin 9 ] (1 9 )x y z t t t t+ + = + + = +

Also (1,0,0)A and (1,0,6 )B on Ccorrespond to 0 2 .t

Thus, we have

( ) ( )

2 222 2 2 2

0 1 9C

ds

x y z t dtdt

+ + = +

( )2 4 2

010 1 81 18t t dt

= + +

25 3

0

10 81 185 3

t tt

= + +

3 58110 2 6(2 ) (2 )5

= + +

506400Using the representative of C, sometimes it may be possible for us to eliminate two

of the three independent variables in the integrand of a line integral and then we

can evaluate the resulting definite integral in which the remaining independent

variable is the variable of integration.


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Engineering Mathematics We illustrate this by the following example.

Example 7.2

Evaluate the line integral

2[ ( ) ]

C

x ydx x z dy xyz dz+ +

where Cis the arc of the parabola 2y x= in the plane 2z= fromA(0, 0, 2) toB(1, 1, 2).

Solution

Since on C, 2y x= and 2z= (constant),

on C, 2dy x dx= and 0.dz=

It follows that on C, the integral of the last term in the given integrand is zero.

Figure 7.4 : Path in Example 7.2

Thus,12 2 2

0( ) ( 2) 2

C

x y dx x z dy xy z dz x x dx x x dx + + = +

2( and 2 on )y x z C= =

. ( )1

5 3 21 4 2

00

2 4 2 45 3 2

x x xx x x dx= + = +

1 2 172

5 3 15= + =

Let us now take up an example in which we consider integration of a line integral over

different paths will the same end points.

Example 7.3

Let Cbe the line segment fromA(0, 0, 1) toB(1, 1) and let 2( , ) .f x y x y= +

Evaluate ( , )

C

f x y ds when

(i) Cis characterized by , , 0 1.x t y t t= =

(ii)

Chas parametric representation sin , sin , 0 .2

x t y t t

= =


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Line and Surface

IntegralsSolution

(i) If , , 0 1,x t y t t= = then

2 22

1

0

( , ) ( )

C

dx dyf x y ds t t dt

dt dt

= + +

=

1

2 32

0

1

0

5 2( ) 1 1 22 3 6t tt t dt + + = + =

(ii) Ifx= sin t,y= sin t,

02

t , then

2 22( , ) ( )

C C

dx dyf x y ds x y dt

dt dt

= + +

/ 2 2 2 2

0(sin sin ) cos cost t t t dt = + + / 2 2

02 (sin cos sin cos )t t t t dt = +

/ 22 3

0

sin sin2

2 3

t t= +

5 2

6=

In this case, you may notice that even though paths described by Cwere different,

the value of the line integral is the same. Is this always true?

Let us examine this by considering yet another example.

Example 7.4

Evaluate

2 2 2( ) ( ) ( )

C

x y dx y z dy z x dz + +

where Cis the curve from origin Oto the point (1,1,1)A

(i)

along the straight line OA

(ii) along the curve 2 3, , , 0 1.x t y t z t t= = =

Solution

(i) Equation of line OA, joining O(0, 0, 0) andA(1, 1, 1) and .x y z= =

For path OA,

2 2 2( ) ( ) ( )

C


1 2 2 2

0( ) ( ) ( ) ,x x dx x x dx x x dx = + +

( and , so that and )y x z x dy dx dz dx= = = =

13 2

0

13

3 2 2

x x= =


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Engineering Mathematics(ii) Along the curve 2 3, ,x t y t z t= = =

2, 2 , 3 .dx dt dy t dt dz t dt = = =

We get 2 2 2( ) ( ) ( )

C


1 2 2 4 3 6 2

0( ) ( ) 2 ( ) 3t t dt t t t dt t t t dt = + +

1 16 5 9 4

0 0

2 36 5 9 4

t t t t = +

1 1 1 12 3

6 5 9 4

= +

29

60=

In this example, the integrands and the end points of the paths of integration are the

same, but the values of the line integrals are different. This illustrates the important fact

thatIn general, the values of a line integrals of a given function depends not only on the

end points but also on the geometric shape of the path of integration.

In many applications, the integrands of the line integrals are of the form

( , , ) , ( , , ) or ( , , ) ,f x y z f x y z f x y zdx dy dz

ds ds ds

where ,dx dy

ds dsand

dz

dsare the derivatives of the functions occurring in the parametric

representation of the path of integration. Then we simply write

( , , ) ( , , )

C C

f x y z ds f x y z ds= dx

ds

and similar expressions in the other two cases.

For sums of these types of integrals along the same path C, we adopt the notation

( )

C C C C

f dx g dy h dz f dx g dy h dz+ = + +

In many cases, the functions,f, g, hare components1 2 3, ,f f f of a vector function

1 2 3 f f f= + +f i j k

Then1 2 3 1 2 3

,dx dy dz

f dx f dy f dz f f f dsds ds ds

+ + = + +

the expression in parenthesis on the right being the dot product of the vector fand the

unit tangent vector

,d dx dy dz

ds ds ds ds= + +

ri j k

wherer(s) represents the path of the integration of the line integral. Therefore,

1 2 3( )

C C

df dx f dy f dz f ds

ds+ + =

r . . . (7.11)


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Line and Surface

Integrals

C

f d= r . . . (7.12)

where .d dx dy dz= + +r i j k

Now iffrepresents a force whose point of application moves along a curve

( ) ( ) ( ) ( ) ,t x t y t z t = + +r i j k a t b

from a pointAto a pointBin space, then

C

f d r

represents the work done by the force f is moving a particle from point A to point B along

the curve C.

The representation in Eq. (7.11), i.e.,

C

dds

ds

rf emphasizes the fact that the work

done by the forcefis the value of the line integral along the curve of the tangential

component of the force fieldf.Let us consider an example illustrating the work done by a force.

Example 7.5

A variable forcepacts on a particle and the particle is displaced along path Cin

space. Find the work done by forcepin this displacement.

Solution

Let the given path Cbe characterized by the equation0 1

( ), .t t t t r

Now the work done Wby forcepin any displacement along the path Cis given by

the line integral

,

C

d= W p r

the integration being taken in the sense of displacement.

Hered

d dt dt dt

= =r

r v

where vis the velocity of the particle. Then work Wbecomes

1 1

0 0

,t t

t t

drdt dt

dt= = W p p v

where0

t and1t are the initial and final values of t.

Furthermore, by Newtons second law

,m m= = r p p v

where mis the mass of the particle. Substituting this value of pin the line integral

for W, we get

1 1

0 0

t t

t t

d mW m dt dt

dt 2

= =

v v v v

1 1

00

2 2| |2

t t

tt

d m mdtdt 2

= = v v

Work done = gain in kinetic energy in moving from initial point to thefinal point.


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129

Line and Surface

Integralsd

fdt

= r

We thus have

d f d = F r r

df dt

dt

= r

dfdt

dt=

Now integratingF .dralong CfromA toB,we get

2

1

t

tC

dfdr dt

dt = F

2

1

[ { ( ), ( ), ( )}]t

td f x t y t z t =

2

1( ( ), ( ), ( )) ttf x t y t z t=

2 2 2 1 1 1( ), ( ), ( ) ( ), ( ), ( )f x t y t z t f x t y t z t =

( ) ( )B

C A

d f d f B f A = = F r r

The value of the integral [ ]( ) ( )f B f A does not depend on the path Cat all. This resultis analogue of the First Fundamental Theorem of Integral Calculus (see Block 1), viz.,

( ) ( ) ( )

b

af x dx f b f a =

The only difference is that we have f d r in place of ( )f x d r . This analogy suggeststhat if we define a functionfby the rule

( , , )( , , )

x y z

Af x y z d

= F r . . . (7.13)

then it will also be true that

f =F . . . (7.14)

This result f =F is indeed true when the right-hand side of Eq. (7.13) is pathindependent. Thus

A necessary and sufficient condition for the integralB

Ad F r to be independent of the

path joining the points A and B in some connected region D is that there exists a

differentiable function of such that

f f ffx y z

= = + +

F i j k

throughout D, where compounds1 2 3, ,f f f of vector fieldF are continuous throughout

D and then

( ) ( )B

Ad f B f A = F r

WhenFis a force such that the work-integral fromA toBis the same for all paths, the

field is said to be conservative.Using the above result, we can say that


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Engineering Mathematics A force fieldF is conservative if and only if is a gradient field, i.e., ,f= F for somedifferentiable function f.

A function ( , , )f x y z that has the property that its gradient gives the force vector Fis

called a potential function. Sometimes a minus sign is introduced, e.g., the electric

intensity of a field is the negative of the potential gradient in the field.

Let us consider the following example.

Example 7.6

Find the work done when a force

2 2 ( ) (2 )x y x xy y= + +F i j

moves a particle in thexy-plane from (0, 0) to (1, 1) along the parabola 2 .y x= Isthe work done different when the path is the straight liney =x?

Solution

The parabola 2y x= has a parametric representation

2, .y t x t= =

From (0, 0) to (1, 1), variation of tis 0 1.t

Work done along the parabola

2 2 ( ) (2 ) ( )

C C

d x y x xy y dx dy = = + + + F r i j i j

2 2[( ) (2 ) ]

C

x y x dx xy y dy= + +

1 4 2 2 2

0[( ) 2 (2 ) ]t t t t dt t t t dt = + +

16 4 2

0

1 1 1 2

3 2 2 3t t t= =

We can similarly find the work done when the particle move form (0, 0) to (1, 1)

alongy=x. In this case also we find that the work done is

(1,1)3 2 2 2

(0,0)

1 1 1 1 1 1 21

3 2 2 3 2 2 3x x xy y

= + = + =

as obtained earlier. This is because, we notice that

3 2 2 21 1 1grad3 2 2x x xy y

= +

F

That is, the fieldFis conservative and the work done does not depend on the path

followed. You may now attempt the following exercise.

SAQ 2

(a)

If 2 22 ,xy z x y = + evaluate

grad ,

C

d

r

where Cis the curve 2 3, ,x t y t z t= = = from 0t= to 1.t=


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Line and Surface

Integrals(b) In 2 xy z x= +A i j k and Cis the curve 2 3, 2 ,x t y t z t= = = from

0t= to 1t= , evaluateC

d A r.

(c) Suppose ( cos ) ( sin ) ( )x xe y yz xz e y xy z= + + + +F i j k

IsFconservative? If so, findfsuch that f= F .

So far we have performed integration along a line or a curve. But as we have mentioned

earlier, there are many physical situation where we are required to find integral for areas

and volumes. For example, determination of moment of inertia and coordinates of center

of gravity of a continuous matter. Solution of these problems involve integrals where we

have to do integration with respect to more than one variable. Such integrals are called

multiple integrals. We shall, in the next section, discuss only double integrals, in which

the integrals is a function of two variables.

7.3 DOUBLE INTEGRALS

Consider a function ( , )f x y which is defined for all (x,y) in a closed bounded regionR

ofxyplane.

The definition of double integral can be given in quite in quite a similar manner as that

of a definite integral of a single variable.

We subdivide the regionRinto sub-regions (see Figure 7.5). These sub-regions can bearbitrary or can be rectangles obtained by drawing lines parallel to xandyaxes.

(a) Sub-division ofRin Arbitrary Manner (b) Sub-division ofRby Drawing Parallel toxandyaxes

Figure 7.5

We number these sub-division, which are withinR,from 1 to n. In each sub-region, we

choose a point, say ( , )k k

x y in the thk sub-region, and then form the sum

1

( , ) ,n

n k k k k

F f x y A

=

=

wherek

A is the area of the thk sub-regions.

In a completely independent manner, we increase the number of sub-regions such that the

area of the largest sub-region tends to zero as napproaches infinity (In the case of


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Engineering Mathematics rectangles as sub-regions, we can say that the length of the maximum diagonal of the

rectangles approaches zero as napproaches infinity). In this way, we obtain a sequence

of real number1 2 7, , . . . , , . . .f f f We know that if ( , )f x y is continuous inRand

regionRis bounded by finitely many smooth curves, then this sequence converges and

its limit is independent of the choice of sub-divisions and corresponding points ( , )f x y .

This limit is called the double integral of ( , )f x y over the regionRand is denoted bythe symbol

( , ) or ( , )

R R

f x y d f x y dx dy A

Thus

1

( , ) = lim ( , )n

k k kn

kR

f x y d f x y

=

A A

The continuity of the integrand is a sufficient condition for the existence of the double

integral, but not a necessary one, and the limit in question exists for many discontinuous

functions as well.

From the definition, it follows that double integrals enjoy properties which are quite

similar to these of definite integrals of functions of a single variable. These properties

hold for the sums from which integrals are defined. We shall state some of these

algebraic properties that are useful in computations and applications.

7.3.1 Properties of Double Integrals

Letfand gbe the function ofxandy, which are defined and are continuous in a regionR,

then

(i) ( , ) ( , )

R R

k f x y d k f x y d = A A (kis a constant)

(ii) [ ]( , ) ( , ) = ( , ) ( , )R R R

f x y g x y d f x y d g x y d A A A

(iii) ( , ) 0

R

f x y d A if ( , ) 0f x y onR

(iv) ( , ) ( , )

R R

f x y dA g x y dA if ( , ) ( , )f x y g x y onR

(v)

1 2

( , ) ( , ) ( , ) ,

R R R

f x y dA f x y dA f x y dA= +

Which hold whenRis the union of two non overlapping regions1

R and2

R as shown in

Figure 7.6 and is called domain-activity property.

Figure 7.6

Furthermore, there exists at least one point0 0

( , )x y inRsuch that


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Line and Surface

Integrals(vi)

0 0( , ) ( , ) ,

R

f x y dA f x y= A

whereA is area ofRand this is called mean-value theorem for double integrals.

Let us now take up the evaluation of the double integrals.

7.3.2 Evaluation of Double Integrals

If ( , )f x y is continuous on the rectangular regionRgiven by,a x b ,c y d

then by drawing lines parallel toxandyaxes, the double integrals over rectangles can

always be calculated as integrated (or reported) integrals and

( , ) ( , ) ( , )

d b b d

R y c x a x a y c

f x y dA f x y dx dy f x y dy dx

= = = =

= = . . . (7.15)

This means that we can evaluate a double integral by integrating one variable at a time

(treating the other variable as constant), using the integration techniques we already

know for function of a single variable.

Further, we may calculate the double integral over a rectangle by integrating in either

order (as per our convenience)

Consider the following example.

Example 7.7

Calculate

( , ) ,

R

f x y dA

where 2( , ) 1 6f x y x y= and : 0 2,R x 1 1.y

Solution

Using Eq. (7.15), we can write

1 2 2

1 0( , ) (1 6 )

y xR

f x y d x y dx dy= =

= A

03

1

10

63y

x

xx y dy

= =

=

1

1(2 16 )y dy

=

12

1

2 162

yy

=

(2 8) ( 2 8) 4= =

Reversing the order of integration gives the same answer (you may check it for yourself).

Let us now consider double integral for bounded non-rectangular regions.

In such regionsR, the double integral may be evaluated by two successive integration

again, but the method will be as follows :

Suppose first we draw lines parallel toy-axis and the thatRcan be described by theinequalities of the form

,a x b ( ) ( )g x y h x (Figure 7.7(a)) . . . (7.16)


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(a) (b)

Figure 7.7 : Evaluation of a Double Integral

so that ( )y g x= and ( )y h x= represent the boundary ofR. Then

( )

( )( , ) ( , )

b h x

x a y g xR

f x y dA f x y dy dx= =

= . . . (7.17)

In this case, we first integrate the line integral

( )

( )( , )

h x

y g xf x y dy

= ,

keepingxfixed, i.e., treatingxas a constant. The result of this integration will be a

function ofx, say ( ).xF Integrating ( )xF overxform ato b, we obtain the value of the

double integral in Eq. (7.17).

Next, if we draw lines parallel tox-axis and the regionRcan be described by the

inequalities of the form

, ( ) ( )c y d p y x q y (Figure 7.7 (b)) . . . (7.18)

then we obtain

( )

( )( , ) ( , )

d q y

y c x p yR

f x y dx dy f x y dx dy= =

= . . . (7.19)

we now integrate first overx(treatingyas a constant) and then integrate with respect to y

resulting the function ofyfrom cto d.

If the regionRcannot be represented by inequalities of the Eqs.(7.16) or (7.18), but can

be subdivided into finitely many portions which have that property, we may integrate

( , )f x y over each position separately and add the results of these integrals, this will

give us the value of the double integral of ( , )f x y over the regionR.

In the equivalence of double into repeated or iterated integrals, we have assumed that

( , )f x y is uniformly continuous and bounded overR and

( )

( )( ) ( , )

h x

y g xx f x y dy

== F

is bounded and integrable from a to b with respect to x along with

( )

( )( ) ( , )

g x

x p yG y f x y dx

==

is bounded and integrable form c to d with respect to y.

In ( , )f x y for discontinuities withinRor on its boundary, it may happen that the two

integrals given by Eq. (7.17) and (7.19) are not equal.

Let us try to understand this point through the following example.

Example 7.8

Show that


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Line and Surface

Integrals1 1 1 1

3 30 0 0 0( ) ( )

x y x ydx dy dy dx

x y x y

+ +

Solution

L.H.S. =1 1 1 1

3 30 0 0 0

2 ( )

( ) ( )

x y x x ydx dy dx dy

x y x y

+=

+ +

11 1 1

3 2 20 0 00

2 1 1

( ) ( ) ( )y

x xdx dy dx

x yx y x y x y =

= + = +

++ + +

11 1

2 20 00

1 1 1 1 1 1

1 1 2(1 ) ( 1)

xdx dx

x x x xx x

= + + = = =

+ ++ +

R.H.S.

1 1 1 1 1 1

3 3 2 30 0 0 0 0 0

( ) 2 1 2

( ) ( ) ( ) ( )

x y x y y ydy dx dy dx dy dx

x y x y x y x y

+ = = =

+ + + +

11 1 1

2 2 2 20 0 00

1 1 1 1

1( ) (1 ) (1 )

y y ydy dy dy

x y y yx y y y y

= + = + + = + ++ + +

1 1

12 2

= =

L.H.S R.H.S

Sometimes it may happen that we are required to change the order of integration in a

double integral for which limits are given. In such a case, first of all we ascertain from

the given limits of the regionRof integration. Knowing the region of integration, wethen put the limits for integration in the reverse orders. We illustrate it through thefollowing example.

Example 7.9

Change the order of integration in the integral

2 2cos

0 tan( , )

a a x

xf x y dy dx

Solution

The given limits show that the regionRof integration is bounded by the curves

0;x= cos x a=

tan y x= and 2 2y a x=

Nowy = xtan is a line through the origin and 2 2y a x= is a circle of

radius aand center at origin and they intersect at the point ( cos , sin )a a .

Hence the regionRis the shaded region OABas shown in Figure 7.8.


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Figure 7.8 : RegionR

When we have to integrate with respect toxfirst, i.e., along a horizontal stripparallel tox-axis, we see the starting point of all the strips is the same line. They all

start form the line 0,x= but some of the strips end at the line OA,While otherend on the circular arcAB. The line of division (or demarcation) is the line CA

given by

sin y a=

Hence region OABmust be subdivided into two sub-regions : OACand CAB.

Now region OACis bounded by the curves 0,x= cot , 0x y y= = andsin y a=

The region OACis bounded by the curves 2 20, , sin ,x x a y y a= = =

and .y a=

Hence on changing the order of integration, the given double integral becomes

2 2sin cot

0 0 sin 0( ) ( , )

a x y a a y

y x y a xf x y dx dy f x y dx dy

=

= = = =+ + +

You may now try the following exercise.

SAQ 3

Describe the region of integration and evaluate the following integrals :

(i)1 2 2 2

0 0(1 )

xx y dy dx+ +

(ii)2

1 2

0(1 )

x

xxy dy dx

Double integrals have various geometrical and physical applications. We take up these

application in the next sub-section.

7.3.3 Application of Double Integrals

The areaAof a regionRin thexy-plane is given by the double integral

R

A dx dy=

Recall that while defining the double integral, we had framed the sum

1

( , )n

n k k k k

F f x y A

=

=


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Line and Surface

IntegralsIn this sum, if we consider a rectangular parallelepiped with base

kA and altitude

( , )k k

f x y , then ( , )k k k

f x y A represents the volume of the volume of the

parallelepiped and hence the volumeVbeneath the surface ( , ) ( 0)z f x y= > and abovea regionRin thexy-plane, as shown in Figure 7.9, is

( , )R

V f x y dx dy=

Figure 7.9 : Double Integral as a Volume

Further, if ( , )f x y be the density (mass per unit area) of a distribution of mass in the

xy-plane, ifMis the total mass in regionR,then

( , )

R

M f x y dx dy=

From mechanics, we know that the coordinates ,x y of the center of gravity of the mass

(of density ( , )f x y in regionRis given by

( , ) ( , )and

( , ) ( , )

R R

R R

f x y dx dy y f x y dx dyx y

f x y dx dy f x y dx dy= =

Thus, the coordinates of center of gravity of a mass is another application of double

integrals.

Similarly, the moments of inertiax

I andy

I of the mass inRaboutxandyaxes,

respectively, are

2 ( , )x R

I y f x y dx dy= and 2 ( , ) ,y RI x f x y dx dy= Which are again double integrals

Let us now take up a few examples.

Example 7.10

Find the volume of the prism whose base in the triangle in the xy-plane bounded

byx-axis and the lines y x= and 1x= and whose top lies in the plane

( , ) 3 .z f x y x y= =

Solution

For anyxbetween 0 and 1, the variable may vary from 0y= to y x= (parallelstoy-axis yields these equalities), as shown in Figure 7.10.


24/99


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139

Line and Surface

Integrals5 4 1

12 2 2

= + =

Thus, the value Vobtained in two ways is equal, as it should be.

We now determine the area of a given regionR.

Example 7.11

Find the area of regionRenclosed by the parabola2

y x= and the line 2y x= + Solution

If we draw lines parallel tox-axis to determine the limits of integration; it is seen

that we must divide regionRinto the region1

R and2

R as shown in the

Figure 7.12(a) and then we may calculate the area as

1 2

1 4

0 1 2

y y

y x y y x yR R

A dA dA dx dy dx dy+

= = = =

= + = +

. . . (7.20)

(a) (b)

Figure 7.12 : Area in Example 7.11

On the other hand, reversing the order of integration (by drawing lines parallel to

y-axis), the required area is (see Figure 7.12(b)).

2

2

2 2

1

x

x y xR

A dA dy dx+

= =

= = . . . (7.21)

Clearly the area given by Eq. (7.21) is simpler as it is easier to calculate. In

practice one would bother to write the integral only in this form. Evaluation of

integral (Eq. 7.21) yields

2

22 32 22 2

1 11

( 2 ) 22 3

x

xx x

x xA y dx x x dx x

+

= =

= = + = +

4 8 1 1 94 2 .

2 3 2 3 2

= + + =

Let us take up a physical application in our next example.

Example 7.12

A thin plate of uniform (constant) thickness and density ( , )x y is bounded by

, 2y x y x= = and axis.x Find the center of mass of the plate if

( , ) 1 2 .x y x y= + +

Solution


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Engineering Mathematics The line 2y x= cuts axisx atA(2, 0). The lines y x= 2y x= intersectatB(1, 1).

The given plate is OABas shown in Figure 7.13.

Figure 7.13 : Given Plate

By drawing lines parallel tox-axis,the whole plate is-characterized by

: 0 2R x and 2x y x

Now,

Mass of the plate ( , )

R

M x y dx dy= =

2 2

0(1 2 )

x

x y xx y dy dx

= =

= + +

22

2

02

2

x

xy x

yy xy dx

==

= + +

2 22 2

0

(2 )2 2 (2 ) 2

2 2x

x xx x x x x dx

=

= + +

22 2 2 2

0

12 4 2 (4 4 ) 2

2 2x

xx x x x x x x dx

=

= + + +

2 2

0(4 4 )

xx dx

==

22

0

4.8 84 4 8 0

3 3 3

xx= = =

First Moment,

2 2

0 0 ( , ) (1 2 )

x

x x yR

M y x y dx dy y x y dy dx

= = = = + +

22 2 3

2

02

2 2 3

x

xy x

y y yx dx

==

= + +

2 2 32 2 3 3

0

(2 ) 1

(2 ) (2 )2 3 2 3x

x x x

x x x x dx=

= + +


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141

Line and Surface

Integrals2 3

2 2 2 3 2 3

0

1 1(4 4 ) (4 4 ) (8 12 6 )

2 3 2 3x

x xx x x x x x x x x dx

=

= + + + + +

2 2 2

0

14 22 2

3 3x x x dx

=

2

2 42

0

14 2 23 3 3 4

x xx x=

28 16 8 84

3 3 3 3

= =

First Moment,

2 2

0 ( , ) (1 2 )

x

y x y xR

M x x y dx dy x x y dy dx

= = = = + +

22

2

20 ( 2 ) 2

x

xy x

yx x y x dx

==

= + +

32 2 2 2

0( 2 ) (2 ) (4 4 ) ( 2 )

2 2x

x xx x x x x x x x dx

=

= + + + +

3 32 2 2 3 2 2 3

02 4 2 2 2 2

2 2x

x xx x x x x x x x dx

=

= + + +

22 4

2 3

00

4 4(4 4 )

2 4x

x xx x dx

== =

2 4 16 8= =

Hence, the center of mass is ( , )x y where

8

31,

8

3

xM

xM

= = =

( 8)

38

3

yM

yM

= = =

Example 7.13

Let ( , ) 1f x y = be the density of mass in the region

2: 0 1 ,y x R 0 1x

Find the center of gravity and the moments of0

, , .x y

I I I

Solution

The given region is a circle 2 3 1x y+ = in the first quadrant (Figure 7.14).

Total Mass inR

dx dy=R

21 1

0 0

xdy dx

=


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142

Engineering Mathematics 1 2

01 x dx=

/ 2 2

0

cos

4d= =

Figure 7.14 : Unit Circle in the 1stQuadrant

Now

21 1 1 2

0 0 0

1 4 4

1

x

Rx x dx dy x dy dx x x dxM

= = =

Let 2 2 21 1x z x z = = 2 2x dx z dz =

0

30 2

11

4 4 4( )

3 3

zx z dz= = =

Since the areaRis symmetrical about both the axes,

4

3x y= =

The coordinates of center of gravity are

4 4( , ) ,

3 3x y

=

Now, Moment of Inertia,

21 12 2

0 0

x

x

R

I y dx dy y dy dx

= =

1 2 3/ 2

0

1(1 )

3

x dx= (let sin x= cos dx d= )

/ 2 4

0

1 cos .

3 16d= =

By symmetry,

.16x y

I I= =

Also0

0.3927.

16 16 8x yI I I= + = + =

You may now try the following exercise.

SAQ 4

Prove that the area in the positive quadrant, bounded by the curve 2 4 ,y ax= 2 24 ,y bx xy c= = and 2xy d= is


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Line and Surface

Integrals2 21 ( ) log .

3

bd c

a

You may recall that in the case of a definite integral ( ) ,b

af x dx sometimes we have to

introduce a new variable of integration uin order to simplify the integration by setting

( )x x u=

where function ( )x u is continuous and has a continuous derivative in some interval

u such that ()x a= and ()x b= [or () , () ]x b x a= = . Then

( ) [ ( )] .

b

adxf x dx f x u dudu=

In the same manner we often simplify the evaluation of double integral by the

introduction of a new variable. In next sub-section we shall show how this new variable

is introduced.

7.3.4 Change of Variables in Double Integrals

Let a regionRinxy-plane be transformed into a region G(Figure 7.15) in the uv-plane by

differentiable functions of the form

( , ),x f u v= ( , )y g u v=

so that each point 0 0( , )u v in the region Gcorresponds to a point

0 0 0 0 0 0 0 0( , ) ( , ), ( , ) ( , )x u v f u v y u v g u v = =

in the regionRand conversely; then a function ,x y( ) defined inRcan be thought of asa function

[ ], , ,f u v g u v ( ) ( )

defined on G.

Figure 7.15 : Transformation of RegionRinxy-plane to Region Gin uv-plane

From the calculus of two variables, we have the result that if all the functions involved

are continuous and have continuous first derivatives, then the integrand ,x y( ) ,of the

double integral ( , )

R

x y dx dy , can be expressed in terms of u and v, and dx dy


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Engineering Mathematics replaced by du dv times the absolute value of theJacobain of the coordinate

transformation x = f (u,v), ( , )y g u v= given by

( , )

( , )

x x

u vx y

y yu v

u v

= =

J

(which is either positive through out G or negative G ).Here the integral ,x y( ) over R

and the integral of [ ], , ,f u v g u v ( ) ( ) over G are related by the equation

( , ), ( , , ( , )]

( , )R R

x yx y dx dy f u v g u v du dv

u v

( ) = [ )

We now take up an example to illustrate how the change of variables simplifies the

evaluation of a double integral.

Example 7.14

Evaluate the double integral

2 2

R

x + y dx dy( )

whereRis the square bounded by lines , , 2, 2y x y x x y x y= = = + =

Solution

Shape of the squareRof the problem is as shown in Figure 7.16.

Figure 7.16 : RegionR

The shape of regionRsuggests the transformation

,x y u+ = x y v =

Then1

( ),2

x u v= + 1

( )2

y u v=

The Jacobian of transformation of coordinates is

1 1

( , ) 1 1 12 2

1 1( , ) 4 4 2

2 2

x yJ

u v

= = = =

Absolute value of1

.2J J= =

Now, regionRinxy-plane corresponds to the square 0 2,u 0 2v (seeFigure 7.17)


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Line and Surface

Integrals

Figure 7.17

Thus,

2 22 2 2 2

0 0

1 1 1( ) ( ) ( )

2 2 2u vR

x y dx dy u v u v du dv= =

+ = + +

23

2 2 22 2 2

0 0 00

1 1( )

4 4 3u v u

vu v du dv u v du

= = =

= + = +

23

2 2

00

1 8 1 2 4 4 82

4 3 2 3 3 2 3 3u

uu du u

=

= + = + = + =

Notice that had we not changed the variable, evaluation ofRwould have to be

carried out by first dividingRinto two regions1

R and2

R and then finding the

range of integrations for the two regionsR1andR2which are different as is evident

from the Figures 7.18(a) and (b) given below.

(a) When Lines Parallel to x-axis are Drawn (b) When Lines Parallel to y-axis are Drawn

Figure 7.18

Let us consider anther example.

Example 7.15

Find the mass of the plate bounded by the four parabolas

2 2 2 24 , 4 , 4 and 4y ax y bx x cy x dy= = = =

if the density of the plate is ,kxy= where kis a constant.

Solution


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Engineering MathematicsIn this case, the mass of the plate ( ) ,

A

k xy dx dy= where regionAis shown in

Figure 7.19(a). It is quite clear from Figure 7.19(a), that if we wish to evaluate this

integral, we shall have to subdivide the regionAin many sub-regions

(a) RegionAinxy-plane (b) RegionATransformed in uv-plane

Figure 7.19

However, if we assume

2 2

, ,x y

u vy x

= =

then in uv-plane the regionAis transformed to

4 , 4 , 4 , 4 ,v a v b u c u d = = = =

which is rectangleA having sides parallel to uand v axes.

2 / 3 1/ 3x u v=

1/ 3 2 / 3y u v=

1/ 3 1/ 3 2 / 3 2 / 3

2 / 3 2 / 3 1/ 3 1/ 3

2 / 3 1/ 3( , ) 4 1 1

( , ) 9 9 31/ 3 2 / 3

u v u vx yJ

u v u v u v

= = = =

Also 2 / 3 1/ 3 1/ 3 2 / 3 k xy ku v u v k uv= = =

Mass of plate4 4

4 4

1

3

d b

v c u ak uv du dv

= ==

2 2 2 264( ) ( )

3

kb a d c=

We know that double integration is integration over an area in plane. So far we

have used Cartesian coordinates ( , )x y for a plane area. However, in some

practical problems, it is convenient to use polar coordinates ( , )r representing a

plane. You might already be knowing the relation between cartesian coordinates

( , )x y and polar coordinates ( , )r . We write

cos , sin x r y r= = (Figure 7.20)

Thencos sin ( , )

sin cos ( , )

rx yJ r

rr

= = =

and ( , ) ( cos , sin )

R G

f x y dx dy f r r r dr d=

Where Gis the region in r-plane corresponding regionRinxy-plane.


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Figure 7.20

How the change of the cartesian coordinates to polar coordinates helps in solving

problems, is illustrated in the next example.

Example 7.16

Find the polar moment of inertia about the origin of a thin plane of density = 1

bounded by the quarter circle 2 2 1x y+ = in the first quadrant.

Solution

By definition, the polar moment of inertia about the origin is given by

2 20

( ) ,

R

I x y dx dy= +

whereRis the quarter circle in the first quadrant (Figure 7.21).

Figure 7.21 : Region inXY-plane

Now, if we take

cos , sin ,x r y r= =

the quarter circle in first quadrant is transformed to a rectangle Gin r-plane as

shown in Figure 7.22, where 0 1,r 0 .2

Figure 7.22 : Region inr

-plane

Here( , )

( , )

x yJ r

r

= =


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Engineering MathematicsAlso 2 2 2 2 2 2 2cos sin x y r r r+ = + =

2 2 20

( )

R G

I x y dx dy r r dr d= + =

14

/ 2 1 / 2 / 23

0 0 0 00

1

4 4 8r

rr dr d d d

= == = = =

SAQ 5

(a) Find the volume under the plane 6x y z+ + = and above the triangle inthexy-plane bounded by 2 3 , 0x y y= = andx= 3.

(b) Find the mass of the plate between 3y x= and 2 ,x y= if density of plate is2 2 ( ).K x y= +

7.4 TRANSFORMATION OF DOUBLE INTEGRALS INTO

LINE INTEGRIALS GREENS THEOREM

We may transform double integrals over a plane region, under suitable condition, into

line integrals over the boundary of a region and conversely. This transformation is ofpractical interest because it makes the evaluation of an integral easier. It also helps in the

theory whenever we want to switch from one type of integral to other.

This transformation can be done by means of a theorem known as Greens Theorem

which is due to English mathematician, George Green (1793-1841). We shall now state

this theorem.

Greens Theorem in a Plane

Let R be a closed bounded region in the xy- plane, where boundary C consists of

finitely many smooth curves. LetM (x, y) andN (x, y) be functions which are

continuous and have continuous partial derivatives

y

Mand

x

Nevery where in

some domain containingR.

Then

( )C

R

dx dy dx dyx y

+ =

N M

M N . . . (7.22)

the integration being taken along the entire boundary CofRsuch thatRis on the

left as one advances in the direction of integration.

We shall not be proving this theorem here as it is beyond the scope of this course.

Learner interested in knowing its proof may see Appendix-I. However, we shall

explain what this theorem means and illustrate it through examples, But, beforethat we state the theorem in vector form.

Greens Theorem in Vector Form


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Line and Surface

IntegralsLet M N P= + +F i j k and ,x y= +r i j then

d M dr N dy = +F r

Also Curl

y z

M N P

= =

i j k

F Fx

y z z x x y

= + +

P N M P N Mi j k

The component of CuriFwhich is normal to a regionRin thexy-plane is

( )x y

N M

F k

Here, Greens Theorem in the plane can be written in the vector form as

(Curl ) ( )

C R R

d dx dy d = =

F r F k F A . . . (7.23)

where dA= d dx dyA k = k is the vector normal to the regionRinxy-plane and

is of magnitude .d dx dy=A

Greens Theorem states thatthe integral around C of the tangential component

of F is equal to the integral, over the region R bounded by C, of the component

of Curl F that is normal to R.

The integral overRis the flux of curlFthroughR.

We shall later, in Section 7.6, extend this result to more general curves and

surfaces in the form of a theorem known as Stokes Theorem.

There is second vector form of Greens theorem as follows :

Let N M= F i j and let

=n Unit outward vector normal curve C

cos cos (90 )= + +i j

cos sin = i j

dy dx

ds ds= i j

Thus ( ) dy dxds N M dsds ds

=

F n i j i j

dy dxds ds

ds ds= +N M

.dx dy= +M N

Alsoin plane in plane

(dis ) ( ) ( )xy xy

Mx y

= = +

F F i j Ni j

=

x y

N M

Hence, Greens Theorem, which says


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( )

C R

M dx N dy dx dyx y

+ =

N M

gives us

C R

ds dx dy = F n F . . . (7.24)

In other words, Greens theorem also states thatthe normal component of any

vector fieldFaround the boundary of a regionR, in whichFis continuous

and has continuous partial derivatives, is equal to the double integral of

divergence ofFoverR.

In next section, we shall extend result (7.24) of three-dimensional vector fields and

call it a divergence theorem. We can thus say that Greens theorem in the plane is atwo dimensional from of the divergence theorem.

We now take up a few applications of Greens theorem and illustrate its importance.

Area of a Plane Region as a Line Integral Over the Boundary

From the Greens theorem, we have

+

R C

dx dy dx dyx x

= N M

M N . . . (7.25)

Let 0=M and = x,N then, from Eq. (7.25), we get

R C

dx dy x dy= . . . (7.26)

The integral on the left is the areaAof the regionR.

Next, letM= y,N= 0, then, from Eq. (7.25), we get

R C

A dx dy y dx= = . . . (7.27)

Adding Eqs. (7.26) and (7.27), we get

1( )

2 CA x dy ydx=

whereAis the area of regionRenclosed by boundary C. Thus we have been able

to express the area of regionRin terms of a line integral over the boundary. Thisinteresting formula has various applications, for example, the theory of certain plan

meters (instruments measuring area) is based upon this formula.Area of a Plane Region in Polar Coordinates

Let r and be the polar coordinates. We define

cos, sin .x r y r= =

Then cos sin dx dr r d =

and sin cos dy dr r d = +

Eq. (7.28) reduces to

1

[ cos

(sin

cos

) sin

sin

]2C

A r dr r d r dr r d

= +


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Integrals21 ,

2C

r d=

a formula which is well known in Calculus.

As an application of this formula (7.29), we consider the cordiod

(1 cos )r a= where 0 2 (Figure 7.23)

Figure 7.23 : Cordiodr=a(1 cos

)

We find2 22 22 2

0 0(1 cos ) (1 2cos cos )

2 2

a aA d d= = + 2

2

0

11 2cos (cos2 1)

2 2

ad

= + +

23

2a=

We now take up a few example to illustrate the use of Greens Theorem.

Example 7.17

Using Greens Theorem, evaluate the integral ( ),C

y dx x dy + where Cis the

circumference of the circle 2 2 1x y+ = .

Solution

Here, ,y= M x=N

2 2 2 2Circle Circle

1 1

( ) ( ) ( ) (1 1)C

x y x y

y dx x dy x y dx dy dx dyx y

+ = + =

+ = = +

2 2Circle

1

2 2 Area of Circle

x y

dx dy =

+ =

=

22 1=

2 .=

Example 7.18

Use Greens Theorem in a plane to evaluate the integral

2 2 2 2

(2 ) ( ) ,C x y dx x y dy + + +

where Cis the boundary of the surface inxy-plane enclosed by thex-axis and the

semicircle 21 .y x=


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Engineering Mathematics Solution

Here 2 22 ,x y= +M 2 2x y= +N

Using Greens Theorem

2 2 2 2(2 ) ( )C

x y dx x y dy + + +

R

dxdyx y

=

N M

whereRis semi-circle 2 2 1x y+ = bounded by aaxis.

(2 2 )

R

x y dx dy=

21 1

1 0(2 2 )

x

x yx y dy dx

= =

=

Figure 7.24

(because semi-circle 21y x= bounded byx-axis is given by

21 1, 0 1x y x )

211 12 2 2

1 102 2 1 (1 )

x

x yxy y dx x x x dx

= = = = +

1

2 3/ 2 3

1

(1 ) 1 1 41 1

33 3 3 32

x xx

= + = + + =

4

3= in magnitude

You may now try the following exercises.

SAQ 6

(a) Use Greens Theorem to evaluate

2 2 2( ) ( ) ,C

x xy dx x y dy + + +

where Cis the boundary of the square 1,y = 1x=


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Line and Surface

Integrals(b) Evaluate 2 2( 2 ) ( 3)

C

x xy dx x y dy + + around the boundary Cof the

region 2 8 , 2.y x x= =

In Sections 7.3 and 7.4, we have been discussing integrals over plane areas. However,

there are many situations in which we may have to consider the areas, which may not lie

in a plane. For example, potential due to changes distributed on surfaces, center of

gravity of a curved lamina, area of the surface out form the bottom of the paraboloid2 2

z x y= + by the plane 1,z= etc. This gives rise to the concept of surface integrals,which we shall take up in the next section.

7.5 SURFACE INTEGRALS

The concept of a surface integral is a natural generalization of the concept of a double

integral considered is Section 7.3. There we integrate over a region in a plane and here

we integrate over a piecewise smooth surface in space.

The definition of a surface integral is parallel to that of a double integral. Here we

consider a portion Sof a surface. We assume that Shas finite area and is simple,i.e., S

has no points at which it intersects or touches itself. Let ( , , )f x y z be a function which

is defined and continuous on S.We sub-divide Sinto nparts 1 2, , nS S S of areas

1 2, , ,

nA A A respectively. Let ( , , )

k k kP x y z be an arbitrary point in each part

kS .

Let us form the sum

1

( , , )n

n k k k k x

J f x y z A

=

= . . . (7.30)

Now we let ntend to infinity in such a way that the largest part out of1 2, , ,

nS S S

shrinks to a point. Then the infinite sequence1 2, , ,

nJ J J has a limit which is

independent of the choice of subdivisions and points .k

P This limit is called the Surface

Integral of ( , , )F x y z over Sand is denoted by

( , , ) .

S

F x y z dS

Thus1

( , , ) lim ( , , ) ,n

k k k k n k

S

F x y z dS f x y z A =

= . . . (7.31)

provided the limit exists.

Evaluation of Surface Integral

To evaluate the surface integral

( , , ) ,

S

f x y z dS

we may reduce it to a double integral as follows :


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Engineering Mathematics A surface Scan be represented in the parametric formas

( , ) ( , ) ( , ) ( , ) ,u v x u v y u v z u v= + +r i j k . . . (7.32)

where uand vare two independent real variables, called parameters of the

representation.

Hence ( , )u vr is the position vector of the points of Sand its tip ranges over Sas( , )u v varies in some regionRin uv-plane (Figure 7.25).

To each0 0

( , )u v inRthere corresponds a point of Swith position vector

0 0( , )u vr . Hence, regionRis the image of Sin uv-plane. In order that surface

have certain geometric properties, we assume that ( , )u vr is continuous and has

continuous first partial derivativesu u

=

rr and

v v

=

rr in a domain of

uv-plane which includes the regionRandRis simply connected and bounded.

From Eq. (7.32), we have = .u v

d du dv+r r r

Figure 7.25 : Parametric Representation of a Surface

Thus, linear element of surface Sis given by

2ds = d d r r

( ) . ( )u v u v

= du dv du dv+ +r r r r

2 22 . ,u u u v v v

= du du dv dv + + r r r r r r

= 2 22 ,E du F du dv G dv+ + . . . (7.33)

which is quadratic differential from and is called the First Fundamental form of

S. Here ,u u u v

E = r r F r r = and .v v

G = r r

In the definition of surface integral, we are sub-dividing Sinto parts

1 2, ,. . . , ,

nS S S Let A be the area corresponding to one of the parts in uv

plane. Then the smallest parallelogram in Figure 7.26 has area; (by the definition

of vector product).

,u v u v

A u v u v = = r r r r . . . (7.34)

and it is called the element of area.


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Integrals

Figure 7.26 : Area

Hence, ifRis the region corresponding to Sin uv-plane, then

[ ]( , , ) ( , ), ( , ), ( , )S R

F x y z ds f x u v y u v z u v dA=

[ ]( , , ) ( , ), ( , ), ( , ) u vS R

F x y z ds f x u v y u v z u v du dv= r r . . . (7.35)

where the right-hand side is now double integral over a plane area.We know that

2 2 2( ) ( ) ( )u v u u v v u v

E G F + = r r r r r r r r . . . (7.36)

Thus we may also write

[ ] 2( , , ) ( , ), ( , ), ( , )S R

f x y z dS f x u v y u v z u v E G F du dv= . . . (7.37)

If a surface Sis given by

( , ),z g x y=

We may set ,x u y v= = and then parametric representation of Scan be written as

( , ) ( , )u v u v g u v= + +r i j k

Thus u u

g= +r i k and v v

g= +r j k

21 ,u

E g= + ,u v

F g g= 21v

G g= +

Hence 2 2 21u v u v

EG F g g = = + +r r

Thus, if Sis represented by ( , ),z g x y= then we have

[ ]22

( , , ) , , ( , ) 1 ,

S R

g gf x y z dS f x y g x y dx dy

x y

= + + . . . (7.38)

whereRis now the image of Sinxy-plane.

We can also write the surface integral (7.38) in terms of the normal to the surface,

Sas follows.

Suppose that F represents normal to the surface , ( , , )S x y z C =F (constant).

Let region areaAbe the projection of surface Son a plane. Let n be a normal to

the regionAand be the angle between the normal to the surface ( )F and

normal to its projection ( )n .

Then cos = F n F n


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or1

cos

= =

F n F

F n F n

Also if S is an element of surface Ssurrounding arbitrary point on Sand A isits projection on the planes, then

cos S A =

1

cos S A =

AS

=

F

F n

Hence the surface integral can be written in terms of double integral as

[ ]( , , ) ( , ), ( , ), ( , )

S S

f x y z dS f x u v y u v z u v dA

= =

F

F n

Let us now take up example from various physical situations to illustrate the evaluation

of surface integrals.

Example 7.19

Find the moment of inertiaIof a homogeneous spherical lamina

2 2 2 2:S x y z a+ + =

of massM andz-axis.

Solution

Let a mass of density ( , , )x y z be distributed over the surface S, then moment of

inertiaI. of the mass with respect to a given axis Lis defined by the surface

integral

2 ,

S

I D ds=

whereDis the distance of the point ( , , )x y z from axisL.

Since in our case, spherical lamina is homogeneous, thus is constant. Also area

of sphere 24 .S a=

Here = mass per Unit Area

24 a

= M

The parametric representation of the sphere 2 2 2 2x y z a+ + = is

( , ) cos cos cos sin sin

u va u v a u v a u= + +r i j k

sin cos sin sin sinu

a u v a u v a u= +r i j k

cos sin cos cosv

a u v a u v= +r i j

2 2 2 2 2 2 2 2 2sin cos sin sin cosu v= a u v a u v a u a = + + =E r r

2 2sin cos sin sin sin cos cos 0u v

= a u v v a u v u v = =F r r


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Line and Surface

Integrals2 2 2 2 2 2 2 2cos sin cos cos cos

v v= a u v a u v a u = + =G r r

If areaAis the image of surface Sin uv-plane, then

u vdA du dv= r r

2EG F du dv=

2 2 2cos 0a u a du dv=

2 cosa u du dv=

Further, the square of the distance of a point ( , , )x y z fromz-axis is

2 2 2 2 2 2 2 2 2 2 2cos cos cos sin cosD x y a u v a u v a u= + = + =

Hence, we obtain

2 2 2

2

( , ) ( , )4S r

MD ds x u v y u v dA

a

= +

/ 2 22 2 2

2

/ 2 0

( cos ) cos4

u u

Ma u a u du dv

a= =

=

4 2 / 2 / 223 3

02 / 2 0cos 2 cos

24

M a M au v du u du

a= =

2 22 22

2 3 3

M a M a= =

Let us now take up an example from geometry.

Example 7.20

Find the area of the surface out from the bottom of the paraboloid 2 2z x y= + bythe plane 1z= .

Figure 7.27 : Area of Parabolic Surface

Solution

The projection of the area of the surface of paraboloid 2 2z x y= + out from thebottom by the plane 1z= onxy-plane is the disk

2 2 1x y+

Now surface area


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SdS=

2 21xy

x yRg g dx dy= + +

2 2

2 2

1

4 4 1

x y

x y dx dy

+

= + +

2 2

Hence, ( , )z g x y

x y

=

= +

Let cos ,x=r cos y=r

Then for 2 2 1, 0 1x y r+ and 0 2

Required surface area

2 1 2

0 04 1 r r dr d = +

1

2 3/ 2 3/ 22 2

0 0

0

1 (4 1) (5 1)

38 122 r

rd d

=

+ = =

(5 5 1)

6=

You may now try the following exercises.

SAQ 7

(a) The electrostatic potential at (0, 0, )a of a charge of constant density

on hemisphere 2 2 2 2: ,S x y z a+ + = 0z is

2 2 2( )dS

x y z a=

+ + + U

Evaluate U.

(b) Find the area of the upper cap cut from the sphere 2 2 2 2x y z+ + = by the

cylinder 2 2 1x y+ = (Figure 7.28)

(Hint : Take surface as2 22 ,z x y= its projection onxy-plane as

disk : 2 2 1x y+ and use polar coordinates to evaluate doubleintegral.)

Figure 7.28


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Line and Surface

IntegralsIn Section 7.4, we had shown that double integrals over a plane region can be

transformed into line integrals over the boundary curve of the region. We shall generalize

this result and we shall now also consider the corresponding problem in the case of a

surface integral in the next section.

7.6 TRANSFORMATION OF SURFACE INTEGRAL INTO

LINE INTEGRALS STOKES THEORM

Transformation of surface integrals into line integrals and conversely is done with the

help of a theorem known as Stokes theorem,given by Gorge Gabriel Stokes

(1819-1903), an Irish mathematician and physicist, who made important contribution to

the theory of infinite series and several branches of theoretical physics. Stokes theorem is

an extension of Greens theorem in vector from to surfaces and curves in three

dimensions. Under suitable restrictions

(i)

on the vectorF,

(ii)

on the boundary curve C, and

(iii) on the surface Sboundary by C.

Stokes theorem connects line integral to a surface integral.

Stokes Theorem, we require that the surface Sis orientable. By orient able, we mean

that it is possible to consistently assign a unique direction, called positive, at each point

of S,and that there exists a unit normal n pointing in this direction. As we move about

over the surface Swithout touching its boundary, the direction cosines of the unit vector

n should vary continuously and when we return to the straight position, n should return

to its initial direction.

We now state stokes theorem.

Stokes TheoremLet S be a piecewise smooth oriented surface in space and let the boundary of S be

a piecewise smooth simple closed curve C. Let ( , , )x y zF be a continuous vector

function which has continuous first partial derivatives in a domain in space which

contains S. Then

(Curl )

S C

dS d = F F r

,

C

d= tF r . . . (7.39)

where ,dS dS =n where n is a unit normal vector of S, so that (CurlF). n is the

component of CurlFin the direction of n ; the integration around C is taken in the

direction of integration on Sandt

F is the component ofFin the direction of the

tangent vector of C.

If we represent curve Cin the form ( ),s=r r where the arc-lengths increases in thedirection of the direction of integration, then unit tangent vector is

,dr dx dy dz

ds ds ds ds= + +i j k

and, therefore, if 1 2 3 ,F F F= + +F i J k

then1 3 3

dr dx dy dzF F F

ds ds ds ds= = + +

tF F


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Engineering MathematicsThus

1 3 3

d dx dy dzds F F F

ds ds ds ds= = + +

t

rF F

Let cos cos cos = + +n i j k be the outward unit normal vector of S.

If we use the representation of Curl in terms of right-handed Cartesian coordinates,then formula (7.39) in Stokes Theorem may be written as

3 32 1 2 1cos cos cos

S

F FF F F F dS

y z z x x y

+ +

1 2 3,

C

F dx F dy F dz= + + . . . (7.40)

where , , are direction cosines of unit normal n to surface S.

We shall not prove this theorem here. Learner interested in knowing the proof ofthe theorem may refer to Appendix-II.

There are many consequences and applications of Stokes theorem. We shall,

however, take up a few of these consequences and applications.Consequences and Applications of Stokes Theorem

(a) Greens Theorem in the Plane as a Special Case of Stokes Theorem

Let M N= +F i j be a vector function, which is continuouslydifferentiable in a domain in thexy-plane. Letxy-plane contain a simply

connected closed region S whose bounding Cis a piece-wise simple smooth

curve. Then k is the direction of unit outward normal to S.

(Curl ) (Curl ) (Curl ) =n

= =

N M

F F n F kx y

Further more, .t

F ds dx dy= +M N

Now formula (7.39) of Stokes theorem yields

(Curl )n t

S C

ds ds= F F

Using above values in this case, we get

( ),

S C

dx dy dx dyy

= +

N M

M Nx

which is in agreement with the result of the Greens theorem in the plane(refer Section 7.4).

This shows that Greens theorem in the plane is a special case of Stokes

theorem.

(b) Physical Interpretation of the Curl

Stokes theorem provides a physical interpretation of the Curl.

Let us first define the term Circulation.

Let ( , , )x y zV be a continuous differentiable vector function in a domain

containing a surface S, bounded by a closed curve C. If Vbe the velocity

field of a moving fluid of density, then the integral

C

dds

ds

rV or

C

ds tV


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Line and Surface

Integralsmeasures the extent to which the corresponding fluid motion is a rotation

around the contour C and is called thecirculation.

By Stokes theorem, circulation is also equal to the flux of curl ( )V

through a surface Sspanning C. Thus

( ) ( ) Curl ( )C C S

dds ds ds

ds = = t

rV V V n . . . (7.41)

Next, let us fix a point Pand a direction uat P.Let Cbe a circle of radius ,with center at P, whose plane is normal to u. If curl ( )V is continuous at

P, then (by mean value theorem for surface and double integrals) the

average value of ucomponent of curl ( )V atPas 0,

i.e., [ ]2 0

1Curl( ) lim Curl ( )

PS

ds

= V u V u . . . (7.42)

Figure 7.29 : Intercept of Curl

Now in our case, the plane of curve Cis normal to u, area of 2 ,S A= = say so that .dS dA= Let .=F V

Thus, form Eqs. (7.41) and (7.42), we get

[ ] 01

Curl ( ) lim CurlnS

P dAA= F F n

0

1lim

tC

dsA

= F

i.e., the component of the curl in the positive normal direction can beregarded as thespecific circulation (circulation per unit area) of the flow in

the surface at the corresponding point.

(c) Evaluation of Line Integral by Stokes Theorem

You may notice that evaluation of line integral by Stokes theorem leads to a

lot of simplification in many problems.

Let us evaluate ,t

C

ds F where Cis the circle 2 2 4, 3x y z+ = =

oriented in the counterclockwise sense as viewed from the origin and withrespect to right-handed coordinates, and

3 3 .y xz zy= + F i j k

We can take the plane circular disk 2 2 4x y+ in the plane 3z = as a

surface Sbounded by C.Then n in the Stokes theorem point in the positive

z-direction, so that n = k.

Hence, (Curl )nF is simply the component of CurlFin the k direction.

NowFwith 3z = has the components1 2

, 27F y F x= = and3

33 .F y= Thus,


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Engineering Mathematics2 1(Curl ) (Curl ) 27 1 28

n k

F F

x y

= = = =

F F

Hence, integral over Sin Stokes theorem equals 28 times the area of the

disk S,which is 4.

Thus, 28 4 112 352t

C

F ds = =

You will appreciate the simplification due to Stokes theorem in this case.

The computations will be long and difficult if you evaluate the given line

integral directly, starting from parametric representation of C,calculating

unit tangent vector t of curve Cin the direction of integration, finding

t

F = F t and then integrating will respect to arc length sof C.

It should be remembered that for Stokes theorem, the surface Sis

assumed to be an open surface.The significance of the statement will

become clear from the following result.

(d) CurlC

dS F will be Zero Over any Closed Surface SSince is any closed surface, we cut open the surface Sby a plane and let

1S and

2S denote the lower and upper positions of S.Let Cdenote the

common bounding curve for both these positions. Then,

1 2

Curl Curl Curl

C S S

dS dS dS = + F F F . . . (7.43)

It may be noted that the normal to lower portion1

S and the normal to upper

portion2

S will be in opposite senses so that the direction of integration

about 1S is reverse of the direction of integration about 2S .

Using Stokes theorem and result (7.43) above, we get

Curl 0C CS

d d d = + = F S F S F r

1(due to surface )S

2(due to surface )S

which proves the required result.

Let now we take up some example to illustrate the use of Stokes theorem in numerical

problems.

Example 7.21

Let Sbe the portion of the paraboloid 2 24z x y= that lies above the plane0.z=

Let Cbe their bounding curve of intersection. If

( ) ( ) ( ) ,z y z x x y= + + +F i j k

verify Stokes theorem.

Solution

The curve Cis the circle2 2

4x y+ = in thexy-plane C, where 0z= and dr dx dy= +i j, we have

( ) ( ) ( ) 0d z y dx z x dy x y = + + + F r


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Line and Surface

Integrals ,y dx x dy= + on C

Now, ( )C C

dr ydx xdy = + F

on C, 2cos , 2sin , 0 2x y= =

2

0

[ 2sin ( 2sin ) (2cos ) (2cos )]C

dr d d = + F

2

0

4 8d= =

Now for surface integral, we have

Curl 2 2 2

( )

x y z

y z x x y

= = + +

+ +

i j k

F i j k

z

For the positive unit normal on the surface

2 2: ( , , ) 4 0S f x y z z x y= + + =

we take2 2

2 2

4 4y 1

gradf x y

gradf

+ += =

+ +

i j kn

x

The projection on Sonxy-plane is the region 2 2 4x y+ and for the element ofsurface area dS, we take

222 21 4 4 1

z zdS dx dy x y dx dy

x y

= + + = + +

Thus, Curl

S

F ndS

2 2

2 2

2 24

2 2 ( 2 2 2 ) 4 4y 14 4y 1

.x y

x xdxdy

+

+ + = + + + + + +

i j k

i j k xx

2 2 4

( 4 4 2)x y

x y dx dy+

= + +

2 2 4

2

x y

dx dy

+

=

(Since odd power ofxoryintegrate to zero over the interior of the circle)

2 2 4

2

x y

dxdy

+

=

= 2 area of circle 2 2 4x y+

22 2 8= =

Hence, Stokes theorem in verified.


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Engineering Mathematics You may now attempt a few exercises.

SAQ 8

(a) Verify Stokes theorem for the function

2 x xy= F i j

integrated round the square in the plane 0z= and bounded by the lines0, 0,x y x a= = = and .y a=

(b) A fluid of constant density rotates round thezaxis with velocity

( ),w x y= V j i where wis a positive constant. If ,=F V find curl F and show its relation to specific circulation.

(c) Let nbe the outer normal of the elliptical shell

2 2 2: 4 9 36 36, 0S x y z z+ + =

and let 2 2 2 3/ 2 ( ) sin ( ) .xyz

y x x y e= + + +F i j k

Use Stokes theorem to find the value of

(Curl )

S

ds F n

(d) Let Sbe the region bounded by the ellipse

2 2: 4 4C x y+ = in the planez =1

and let =n k and 2 2 2 .x x z= + +F i j k

Use Stokes theorem to find the value of .

C

d F r

7.7 TRIPLE INTEGRALS

In section 7.1, the concept of a line integral was introduced. Section 7.2 was devoted tothe properties and evaluation of double integrals, in which the integrand is a function oftwo variables. A natural generalization of double integrals was provided in Section 7.5 in

terms of surface integrals, where though the integrand may be a function of threevariable, but it is defined on a surface only and it is still evaluated as a double integral.

However, there are many physical and geometrical situations, where the integrand maybe a function defined in a region in space and integration may have to be carried out withrespect to all th

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