Unit 12 Line and Surf'Ace Integrals

8/12/2019 Unit 12 Line and Surf'Ace Integrals

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UNIT 12 LINE AND SURF ACE INTEGRALSStructure

12.1 Introductionjeciives

Integration of a Vector12.2.1 Line Integrals12.2.2 ypes of Line Integrals12.2.3 Evaluation of Line Integrals12.2.4 Path Independence Comervative FieldsDouble Integrals12.3.1 Properties of Double Integrals12.3.2 Evaluation of Double Integrals12.3.3 Applications of Double Integrals12.3.4 Change of Variables in Double IntegralsTransformation of Double Integrals into Line Integrals Green s TheoremSurface IntegralsTransformation of Surface Integrals into Line Integrals Stoke s TheoremSummarySolutionsIAnswers

12.1 INTRODU TIONIn Unit 11 you have learnt about the differentiation of vectors and amved at the importantconcepts of directional derivative, gradient, divergence and curl. In this unit, we shall talkabout integration of vector functions. In physics and engineering, we come across manyproblems such as determining the trajectory of a particle when its equation of motion isknown or finding the potential due to an electric charge when the force acting on it isknown or to determine the total energy of electromagnetic wave flowing through a volumewhen the electric field and magnetic induction field is known. All these problem lead us tothe evaluation of a line integral, i.e., the integration of a vector with respect to a scalar orintegrals i~ivolvingcalar and vector products of vector or integration of vectors along acurve or a line.In this unit, we shall begin our discussion in section 12.2 with the definition of a lineintegral and see that it is a natural generalisation of a definite integral.Not all physical and engineering problem boil down to the evaluation of a line integral.Depending on the domain of integration we are also required to evaluate the double orsurface integrals. For instance, the area of a region inxy -plane, the volume beneath asurfacez =f (x, y) > 0) and above a region inxy -plane, the centre of gravity of adistribution of mass in the xy -plane, the moment of inertia of a mass in a region in xyplane about x and y axes are some of the geometrical and physical applicatiow of doubleintegrals. We shall discuss the concepts involving double integrals and their evaluation,along with change of variables in double integrals in section 12.3. Double integrals over aplane region may be transformed into the line integrals over the boundaly of the region andconversely. This is achieved through the Green s theorem in the plane and will form thesubject of our discussion in section 12.4. Section 12.5 is devoted to the discussion ofsurface integrals.The transformation of surface integrals into line integrals and conversely is done throughStoke s Theorem, which will form our subject matter for section 12.6. In this section, weshall also take up somd applications of the Stoke s theorem and give another physicalinterpretation of Curl.ObjectivesAfter 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,


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vectorCalculus comp ute double integrals and state the conditions uuder w hich the o rder ofintegration can be cha nged,explain the need of chan ge of variables in double in tegrals and e valuation ofdouble integrals by c hange of variables,outline methodfconditions of transfonnation of double integral into lineintegral and converjely,solve prnhlerns based on Green s theorcrn,define a surface integral and transform it into line integral,outline the representation of curl in tenns of line integral, anduse Green s theorem and Stoke s theorem to relevant physical situations andsolve problems based on them.

12 2 INTEGRATION OF A VECTORIn Unit 11,you have leanlt to differentiate a vector w.r. to a scalar. You also h io w thatintegration is the inverse process of differentiation. If F (t) and (t) be two vector functionsof a scalar variable t, connected by the relation

then we say that F (t) is integral off t) and write

This gives an indefinite integral of f(r).We c an also add an arbitrary co ~lstan t ector C on the right hand side of Equa tion (12.2) andwrite it as

This is becdused( F ) + C ) = f [.: z = ]dt

While solving problems, this constant C can be detennined using the give11 initialconditions. A lso, the dimension of C is that of F(t).You also h io w from calculus, that for scalar fu~lctio il I),he definite integral

can be defined as the limit of a sum. Here we integrate alon g xis from a to b and theintegral f is a function defined at each point between a and b.In the same manlier we can define for a sca lar variable t, the definite integral of a vecto rfunction F (1) as the limit of a sum . We c an w rite

iab( t) dt = l i ~ n [ F (tl) 6 tl F h) 6 t2 + + F (t,,) 6 t ] = 1 2 ..6 t i 0

where the ranges of integration a t b has been divided into n sub ranges correspondingto inc rem en ts Br,, 6 t2 , Bt, of the varia ble t and r,, t2 . , are values o f t lyingrespectively in these sub-ranges. Further in the limit - m as the number of sub-rangesincreases indefinitely, each of the increment t tends to zero. The sum on the right handside which its, of course, a vector sum tends to a definite finite limit, which is the definiteintegral of functionF t).A sim ple exam ple of vector integration is illustrated by a particle of unit ma ss m ovingund er gravity with constant acceleration g Let denote the position-vector of the particle attime r Th en from Newton s second law of motion, the equation of motion is


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Now if we wish to find the trajectrey of the particle during its motion, i.e., if we wish to findr in tenns of hen we integrate Equation 12.3) and obtain

where V s the initial velocity corresponding o t = 0 .A second integration yeilds

where r, is the initial position of the particle.Thus by interpretation of a vector with respect to a scalar, we could determine the trajectoryof the particle when equation of motion is known.You may note that in relations 12.4) and 12.5), and r have the dimensions of velocitydtand position respectively. Therefore, the arbitrary constant vector V n relation 12.4) hasthe dimension of velocity and r, in equation 12.5) has the dimension of position.ote that in the above example, the particle was moving with constant accelerationg. Inpractice, 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 transverse electromagnetic2 n Awave propo ~at in~nx direction may have an electric field E = En cos ct - x) k

2 n Aand a magnetic induction fieldB = o cos ct - x) k . If is the electric permitivityAand p is magnetic permeability of space, then energy flowing through a volume V per unittime is given by

If T be the time period for total energy flowing through volume V during one completecvcle of electro-magnetic oscillation, then-TTotal energy=l dt

V V 2 2 n= lo: ms2 ct x dt + B: cos ct- x dtA 2 A

( . : E - E . E and B = B . B

L i e nd urface Integrals

Tsin 4 ct x )

= V E E : + p~: . I .4 I4 - cA


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ector alculus

and

1= -V E E ~p4Note hat in the above integration for the calculation of total energy , we have usedE = E .E he dot product of vectors. It may also happen that the integration of vectorsmay involve cross product of vectors.

A A AWe may wish to evaluate s A x dr from t = 0 to t = where A = xy z j x2k and3x = t 2 , y = 2t, z = 1 .

A A AHere r = t 2 i 2 t j r 3 k

A A AA - x y i - z j + g kA A A

= 2 t 3 i t 3 j t 4 kA A A

: A x dr i j k2t3 - t 3 t 4dt 2dt 3t2dt

A A A

= - ( 3 t S 2 t 4 )d t i 4 r 5 j ( 4 t 3 2 t 4) d tk1 1 A: J A r dr = s o - ( 3 t 5 2 t 4 ) j l - 4 t5 k l 4 t3 2t4)dt0 0

t 6 t S A A 1 A= - 1 ( 1 x + 2 5 ] l o i - 1 4 g l o j + I t 4 + ~ t 5 1k

1 A 4 A 7 A= - ( T + g ) i - - j + - k

9 A 2 A 7 A= - - i - - j + - k10 3From the abov e examp le you m ust have observed that vector iutegration involving dotproduct o r crass product of vectors, essentially reduces to de termining the integral of ascala r with respect to a scalar.Recall that in a definite integrals x ) dx we integrate along x xis from a to hand theaintegral is a Eunction defined at eac h point between a and h. How ever, in many physicalproblems, you inay observe that a particle may iiot be m oving alon g a li r~ e ut alorig a curvein space, e.g., if we w ish to calculate the work don e by a force in m oving a pa rticle alongthe curve C from positionA to position B then the ordinary integration indefinite ordefinite) involving vectors will not provide the result. Such problems lead us to theconsideration of line integrals o r curve integrals.12 2 1 Line ntegralsThe concept of a line integral is a simple and natural generalization of the concep t of adefinite integral

In the line integral we do not integrate the integrand al on gx - xis from a to h; instead, weintegrate along a cu rve in plane or in space and the integrand is a function defined at thepoints of that curve. We c an define a line integml in a way sim ilar to that of a definiteintegral.


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Consider a curve in space havingA and B as the initial and tenninal points. We may ~ i o c d s u r f n u ~ n t e g n l ~represent C in parametric form a s

r ( s ) = x ( s ) i + y ( s ) j + z ( s ) k , ( a s s s b ) -where s is the arc length of andA and B correspond to s = a and s = b respectively (seeFigure 12.1)

Figure 12. 1 n directed urveC from toWe assume that r (s) is continuous and h s continuous.firstderivative different from thezero vector for all s under consideration. Then C has a unique tangent at each of its point.Such a curve C s called a smooth curve.Let w (x, y, z) be a scalar function which is defined at each point of C, and is a continuousfunction of sWe divide the curve C, fpmA to B, in an arbitrary manner inton portions. Let

Po (=A), P,, n -,, n =B)be the end points of these portions and let

be the corresponding values of s (see Figure 12.2).Let the length of each of these portions is

en we evaluate w at each point xk yk zk in Sk and consider the sum

We define the integral of w x, y, z) over curve C fromA to B to be limit of the sum as thesubdivision of C is refined so that number of subdivisions becomes very large and thelargest 6 skapproaches zero, i.e.,

whenever the limit on the right-hand side in equation (12.7) exists, we call it line integral ofw(gy,z) along C fromA to B and denote it as

Note that since w is continuous and C is smooth, the limit on the right hand side ofequation (12.7) exists and is independent of the choice of subdivision of C.Moreover, if (x,y r , w is a constant function whose value is 1, th en jw (x,y, z)gives the length of curve C fromA to B.

Figure 12 2 ub division of C


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Vector Calculus Th e concept of the line integral can be extended to vector integration an d we ge t differenttypes of line integrals depending uponi vector element 6sl 6s2 , b and performing vector addition, keeping

w (1, y, z) a s a sc alar function:ii) taking function W (r) instead of scalar function and als o taking vectorelements bl b 2 , . b Then products of W with 6sk can be either

dot products or vector products and this in turn gives rise to two types of lineintegrals.W e shall now take up various types of line integrals.12 2 2 Types of Line IntegralsThe following types of line integrals exist :i) Wh en w (x, y, z) is a scalar function defined at eac h point curve C fm mA to B, whichis divided into parts

b l , 6s . 6snalong the curve, then

define s a line integral in ordinary ca lculus.ii) When w (x, y, z) is a scalar function defined at each po int of the curve C fromA to Bwhic h is divided into n parts of vector elemellts 6sl 6s2 . . . 6sn long the curve C

and p erform vector addition, then

defines a line integral, where va lue is a vector.iii) Wh en W (x, y, z) is a vector function defined at each point of the curve C fromA to B,which is divided into n parts of vector elements 6 sl 6 st . . . 6 s along the curveC and the product of W (x k yk z k in sk in dot prod uct of vecto r, then

defines a line integral, whose value is a scalar.Moreov er, if r is a position vector of a point on cu rve C, then ds d r and in the limitwhen each sub-interval tends to be zero, a = dr herefore the last three line integralsmay also be w ritten as

curve consisting o finitelyI w ( x , y , z ) d r , S ~ ( x , y , z ) . and S w ( x , y , z ) dr

respectively.In al l the above line integrals, it is assumed that the path of integration is piecewive smooth.

Cz For a line integral over a closed path C, the sym bolf instead of S

is sometimes used in the literature.From th e definition of line integral, it follows that the following properties a re valid for lineFigure 12 3 : Formula c) integrals :


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a) s k w d r - k s w d r k Constant)C C

orientation of s the sam e in all the three integlals).c) w ds = w ds +Sw dr, where the path C is subdivided into

C l ctwo arcs C

and C,, which h ave the sam e orientation sC.Note that if the sense of integration along a curve C is reversed, the value of the line integralis multiplied by -1.Th e question is now arises is -H ow is a line integral evaluated ? We shall now answ er thisquestion.12 2 3 valuation of L i ~ entegralA line integral can be evaluated by reducing it to a definite integral. This reduction is quitesimple and is done by me am of representation of the path of integration C as follows :IfC is represented by

A

r s ) = x s ) i + y s ) j + z s ) k , a s s s b ,where s is the a n ength of C, then w e can imm ediately write

the integral on the right being a definite integral. In application s, mostly representation ofis of the formr t) = x t)I y t ) j + z t) i to s s ,,

wh ere is any parameter, or may easily be converted to this form. In this case, we can u ser [S t)] = r 4 ,

so that x [s t)] = x t) and so on.Then we can use

Here we assume that r t) and t) are continuous and , i t) 0, in agreemenassump tions mentioned in section 12 2 2 ~twith theW e now illustrate the theory discussed so far with the help of a few examples.Example 1

v lu tes (r + + z f dr, where s the a n of circular helixs n 8r t ) = c o s t i + s i n t j + 3 t k

fromA 1, 0, 0) toB 1,0 , 6 rc .

L i e andSurface Integrals


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VectorCalculus Solution

Here r ( t ) = c o s t i s i n t j 3 t k ,: ;(t) = sin t ; cos t

NOW = Zdsin2 cos2 t =dtOn C,

( ~ ~ + ~ ~ + z ~ ) ~[ c o s 2 t s in 2 r 9 t 2 1 2 = ( 1 9 t 2 ) 2AlsoA 1, 0,O) and B ( 1 , 0 , 6 n on C correspond to 0 t 2 x .Thus, we have

Using the representative of C, som etimes it may be possible for us to eliminate two of thethree independent variables in the integrand of a line integral and then we call evaluate theresultia iefinite integral in which the remaining illdependent variable is the variable ofintegration.W e illustrate this by the following examp le.ExampleEvaluate the line integral

f [ x Z Y d x ( x- z) d y x y z d z ]where C is the arc of the parabola y = x 2 n the plane = 2 frorn A (0 ,0 , 2) to

(1, 132).Solution

Since on C, y = x and z = 2 (constant),: onC, dy = 2 x a k and = 0.It follows that on C, the integral of the last term in the given integrand is zero.

1, 1 , 2 )I L

Xigure 1 2 . 4 :Path in Example 2.


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and sim ilar expressions in the o ther two cases,For sum s of these types of integrals along the sam e path C we adopt the notation

In many cases, the functionsf, g h are components fi 2?3 of a vector function

Then/

the expression in parenthesis on the right being the dot product of the vector and the unittangent vector

where r s) epresents th e path of the iiite gratio ~l f the line iiiteral. Therefo re,

where = x i d y j d z kNow iff represents a force whose point of application moves along a curve

from a pointA to a point B in space, then

represents the work done by the force in moving a particle from point A to point B alongthe curve CTh e representation in equation (12 11 ), i.e.,J . l ds emphasises the fact that the work

c fsdon e by the force is the value of the line integral alon g the curve of the tangen tialcomponent of the force fieldf.Let us consider a n examp le illustrating the work done by a force.Example :

A variable fo rc ep a cts on a particle and the particle is displaced along a given path Cin space. Find the work don e by fo rc ep in this displacement.Solution :

Let the given path be characterised by the equation r t), to s tl .Now the work done W by force p in any displacement along the path C is given hythe line integral

the integ ration being taken in the sens e of displacement.

Lime and Surface Integrals


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vectorCalculus Here a+ = - = vwhere v s the velocity of the particle. Then work W becomes

whe re to and tl are the initial and final values of t.Furthermore, by Newton s second law

m i ' = p p = m i ,w h e n m is the mass of the particle. Sub stituting this value of p in the line integral forW we get

=- Work done = gain in kinetic energy in moving from initial point to the finalpoint.

Thu s the basic law in mechanics has been reinforced through the line integralconcept.You m ay now try the following exercises.

If F = ( 3 x 2 6 y 1 4 y z j + 2 0 x z 2 k evaluatetheLineintegralJF.drfrom (0, 0, 0) to 1,l;l)along the following pathsC :( 9 x = t , y = 12 ,z =1 3,(ii) The straight line from (0, 0, 0 ) to (1,0,O); then to 1, 1 ,0 ) and then to (1,1,1),

and(iii) Th e straight line giving (0, 0,O) to (1, 1,l .

Find the work done in moving a partical once round the ci rc le 2 y = 9 in theX Y plane if the field of force, given by

In example 3 yau have noticed that the value of the line integral between the sa me endpointsA andB jbined by two different paths was the same, where as in Exa mple 4,w efound that the value of line integral between the sa m e end points joined by tw o differentpaths was unequal. Let us now study the condition under which the value of the line integralbetween two pointsA and B is independent if the path joining them.


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12 2 4 Path Independence-Conservative Fields ~ l n ~nd s W ~ ~mt gmhLet F be a vector field with com ponentsF 2 3 nd F 2 3 e continuou s throughoutsome c o ~ e c t e degion D.Consider two pointsA and B in D. Suppo se that C is any piecewise smooth curve joiningAand B given by

If there exists a differentiable functionf such that

then along C, f = f [x t), y t), z t)] s a function of t and

W e thus have

Now integratingF r along C fromA to B we get

Th e value of the integral [f B) - f (A) does not depend on the path C at all. This result isanalogue of the First Funda mental Th eorem of Integral Calculus (see Block 2), viz.,

two points of c n be joinedby broken line of finitelym ny line r segments ll ofwhich belong to D

Th e only difference is that we have f in place off x) ix This analogy sug gests thatif we define a functionf by the rule


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VectorCaleulus

then it will also be true that Vf = F 12.14)This result V f = is indeed true when the right- hand side of relation 12.13) is path-independent. ThusA necessary and sufficientcondition or the integrals F dr to be independent of rheycrbjoining thepointsA and B in some connected region is ilzat tlzere exists a differentiablefunction of such that

throughoutD where components i i 3 of vectorfield a m continuous throughoutDand then

When is a force such that the work-integral fromA to B is the same for all paths, the fieldis said to bc conservative Using the above result, we call say thatA force field F is conservaiive i f and only i it is a gradienr field i.e. F = Vj; or somedifferentiable unction fA function x, y, z) that has the property that its gradient gives the force vactor is called apotential bnction. Sometimes a minus sign is introduced, e.g., the electric intensity o afield is the negative of the potential gradient in the field.Let us consider the following example.Example 6

Find the work done when a force

moves a particle in thexy -plane from 0,O) to 1, 1 along the parabola y = x 1s thework done different when the path is the straight line y x

SolutionThe parabolay2 = x has a parametric representation

From 0,O) to 1,1), variation oft is 0 t 1.: Work done along the parabola

We can similarly find the work done when the particle move from 0,O) to 1 , l ) alongy x. In this case also we find that the work done is


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as obtained earlier. This is because, we notice that

That is, the field F is coilservative an d the work d one do es not dependYou rriay now attempt the following exercise.

If I = 2 xy + y, evaluate

where is the curvex = t y = i z = t5from t = o t = 1.

on the path followed.

I f = r y ; - + x 2 nd C i s t h e c u m e x = t y = 21 = t 3 f m m= o = 1 evaluate J x dr.

5.Suppose

F = ( e X c o s y + y z ) j m e X s i l l y ) k xy + z )IsF conservative ? If so, find f suc h that F = Of.

i em Surface ntegrals


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Vector alculus

is a part of the region andbounded means that theregion can e endo sed in acircle of sufficiently large,but tinite radius.

So far we have perfomed integration along a line or CUNe. But as we have mentionedearlier, there are many physical situation where we are reguired to find i n t e ~ lor areasand volumes. For example, determination of moment of inertia and coordinates of centre o*gravity of continuous matter. Solution of these problems involve integrals where we haveto - integration with respect to more than one variable. Such integrals a n called multipleintegrals. We shall, n he next section, discuss only double integrals, in which the integralsis a function of two variables.

12 3 DOUBLE INTEGRALSConsider a function (x, y) which is defined for all 4 ) in a closed bounded regioll ofy - lane.

The definition of double integral can be given in q~lite similar manner as that of a definiteintegral of a single variable.We subdivide the region R into sub-regions (see Figure 12.5). These sub-regions can bearbitrary or can be rectangles obtained by drawing lines parallel to x and y axes.

Y

a) Subdivis ion oPR in arbitrary manner b) Subdivision of R by drawing parallel tox andy xes

Figure 12.5We number these subdivision, which are within R, from 1 o n. In each sub-region, wechoose a point, say (x, , in the Phsub-region, and then form the sum

f ~ k , k) AAk,k - 1where A Ak is the area of the sub-regions.In a completely independent manner, we increase the number of sub-regions such that thearea of the largest sub-region tends to zero as n appmaches infinity (In the case ofrectangles as sub-regions, we can say that the length of the maximum diagonal of therectangles approaches zero as n approaches infinity). In this way, we obtain a sequence ofreal numbersf, , i . f We know that i ff (x, y) is continuous in R and region R isbounded by finitely many smooth curves, then this sequence converges and its limit isindependent of the choice of sub-divisions and corresponding points (xb y, .This limit is called thedouble integral off (x,y) over the regionR and is denoted by thesymbol S f ( * , y ) d r SSf(x ,y) dy

R R

Thus JJI(x,Y) em f (xb yk) AA,n m k- 1

The continuity of the integrand is a sufficient condition for the existence of the doubleintegral, but not necessary one, and the limit in question exists for many discontinuousfunctions as well.Pmm the de fd io n~t follows that double htegnls enjoy pmpefiies which an quite similarto Ofdefioite nfegmls of functions of a single variable. mese pmpe*ies hold for the


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Vector Chlculus

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 regions R, the double integral may be evaluated by two successive integration again,but the method will be as follows :Suppose first we draw lines parallel to y xis and that R can be described by theinequalities of the form

a s x s b , g ( x ) s y s h ( x ) (Figure 12.7(a)) (12.6)

a) b)Figure 12 7 :Evaluationo f a double integral

so that y g (x) and y (x) represent the boundary of R. Then

In this case, we first integrate the line integral

keepingx fixed, i.e., treatingx as a constant. The result of t is integration will be a functionof x, say x). IntegratingF (x) overx from a to b we then obtain the value of the doubleintegral in equation (12.17).Next, if we draw lines parallel to x -axis and the region R can be described by theinequalities of the form

c s d p b ) s s4Cv) (Figure 12.7 (b)) (12.18)then we obtain


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W e now integrate first ov erx reating y as a constant and then integrate with respect to y LhemdSurfaceIntegralsresulting the function o f y from c to d.If the region R cannot b e represented by inequalities of the type (12.16)or (1218) but canbe subdivided into finitely ma ny portions which have that property, we may integrate(x,y) over e ach positio n separately and add the results of these integrals, this will give usthe value of the doub le integral off (x, ) ove r the region R.In the equivalenceo double integrals into repeated or iterated integrals, we have assumedthat (x,y) is uniformly continuous and bounded over R and

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

is bounded and integrable rom c to d with respect toy .Iff (x,y) for discontinuities within R o r on its boundary, it may happen that the twointegrals given by equations (12.17)and (12.19)are not equal. Let us try to understand thispoint through the following example.xample :

Sho w that

P X Y dy rJ ld y ' X-y P0 0 (x+y)) 0 0 (x+y) )

olution :

1 1 1 1R.H.S. = dy P= dy X + Y - 2~ =so112 40 O(x+y) ) 0 0 (x +y)3 O X + Y ) ~ ( x +y )

1 1=so1 -- -+Y (x+y12 ldy=s: C- +L +L - L+y ( l+y y y y z ) 9 = ~ O ~ d ySometimes it may happen that w e a re required to change the order of integration in a doubleintegral for which limits are given. In suc h a case, first of all we ascertain from the givenlimits of the region R of integration. Knowing the region of integration, we then put thelimits for integration i n the reverse orders. W e illustrale it through the following exam ple.


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Vector Galculus Example :Change the order of integration in the integral

Solution :The given limits show that the regionR of integration is bounded by the curves

x 0 ; a c o s ay x t a n a and y m

Now y x tan a is a line through the origin a nd y s a circle of radiusa and centre at origin and they intersect at the point a cos a a sin a) . IIence theregion R is the shaded region O B as shown in Figure 12 8When we have to integrate with respect to x first, i.e., al ong a horizantalY strip parallel to x -axis, we see that the starting point of all the strips is

X - a m s a the same line. They all start from the line x 0, but som e of the strips endat the line OA, while other end on the circu lar arc-. Th e line of division- x u n a or dema rcatibn) is the line CA given by

..c 1 . . A acosa,asina) y a sin a

. . . Hence region OAB must be subd ivided in to two sub -regions :OAC and. CAB.

\ -A Now region OAC is bounded by the curves x 0, x y cot a , y 0and y a s in a

igure12 8 :Region The region OAC is bounded by the curves x 0 wy = a s i n a , and y = a .Hence on changing the order of integration, the given d ouble integral becomes

You m ay now try the following exercise.E6

Describe the region of integration and evaluate the following integrals :

Dou ble integrals have various geometrical and physical applications. We tak e up theseapplications in the next sub-section.123 3 pplication of Double IntegralsThe area A of a region R in the y -plane is given by the double integral


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ecall hat while defining the double integral, we had framed thesua

JI Z f ( - % , ~ k )Akk - 1n this sum, if we consider a rectangular parallelopiped with base AAk and altitudef xk k), thenf xk, k ) AA, represents the volume of the parallelopiped and hence thevolumeV beneath the surface = f (x, y) > 0) and above a regionR in the y -plane, asshown in Figure 12.9, is

l = f ( x , ~ ) d r d ~Further, iff (x, y) be the density (mass per unit area) of a distribution ofmass in thexy-plane and if s the.total mass in region R then

= f(x.y) m d ~

From mechanics, we know that the coordinatesz, of thecentreof gravity'.of the mass (of densityf (x, y) in regionR is given by

Figure 12.9Thus, the coordinates of centre of gravity of a mass is another application of doubleSimilarly, the mom ents of inertia I and Iy of the mass in R aboutx and y axes,respectively, are

1= Iy 2 f(x,y)&dy and Y= * 2 f ( x , ~ ) & d ~ ,Which are again double integrals.Let us now take up a few examples.

Find the volume of the prism whose base is the triangle in the y -plane bounded byx -axis and the lines y = x and x = 1and whose top lies in the planez = f(x,y) = 3-x -y.

For any x between 0 and 1 the variabley may very fromy Q toy x (parallels toy-axis yields these equalities), see Figure 12.10.

a) Prism with a triangularbase n the b) hkgrationl i f I->( ,dyv -piam r - 0 y - 0

Figure 12.10

LinesadSurZaa Integrals

:Double Integral asa Volume


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Vector alculus

Figure 12 11


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If we draw lines parallel to x- ax is to determine the limits of integration; it is seen ie andSurfacclntefldsthat we must divide region R into the regionR1 and R s shown in the Figure 12.12a) and then w e may calculate the area as

On the other hand, reveming the order of integration by drawing lines parallel to yaxis), the required area is see Figu re 12.12 b)).

Clearly the area given by E quation 12.21) is simpler as it is easier to calculate. Inpractice on e would bother to write the integral only in this form. Evaluation ofintegral Eqn. 21) yields

Let us take up a physical app limtio n in our next example.xample 12 :

A thin plate of uniform constant) thickness and density x, y) is boundedby y = x, y = 2 x and x xis. Find the centre of mass of the plateif 6 x ,y) - 1 + 2 x + y .Solution :

Th e line y 2 -x cuts x xis atA 2,O). The lines y =x and y = 2 - xintersect at B 1, 1).The given plate is O B as show n in Figure 12.13.By d rawing lines parallel to x -axis, the who le plate is characterized by igure 12 13 :Given Plate

R : O s x s 2 and x s y s 2 - x

Mass of the plate = M = 1 x, y) dr yR

2 2 x= Jo l y + 2 q + ~ d x2 y x


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vector ekdus

First Moment,

First Moment,

- 2 ~ 4 - 1 6 0 - 8Hence, the centre of m ss s (Z, here

- I -x m - 1 , y - = m , - 3M 6 ) M (- )


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L i e and urfaceIntegralsLet f (x, y) = 1be th density of m ss n the region

Find the centre of gravity and the m oments ofI I loSolution

The g iven region is a circle y2 1 n the first quadrantsee Figure 12.14).: Total Mass in R = rS r y

1 somx&]m = INow - x&dyM 0

Let G - z s 1 - x Z = z 2 : - 2 x d x = 2 z d z

Since the areaR is symm etrical about both the axes,

The c o o r d i ~ t e sf cen ter of gravity are

Now, M oment of Inertia,

Figure 12 14 Unit circle in the st quadrant

symmetry, 1


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ve tor elculus

i ble function i f f u, v) iscontinuous and poss s continuous first partial derivativesin som region G in th uv -

You m y now ry the following exelrcises.E 7.

Find the volume under the +y + zbounded by 2 x 3 y, y 0 and x 3. = 6 and

Find the m ss of the plate between y = x and x y , if density of plate isp = K ( X ~ + ~ ~ ) .

You may m a l l that in the case of a definite integr als f ( x )dr sometimes we have tointroduce a n ew variable of integration u in order to simplify the integration by setting

= (u )where function1 (u) is continuous and has a continuous derivative in som e interval

s u s fi such thatx ( a ) - a and x p) = b [or x a) = b, x f3) = a] ThenIn the sa me manner we often simplify the evaluation of dou ble integral by the introductionof a new variable. In next sub-section we shall show how th is new variable i s introduced.12 3 4 hange of Variables in Double IntegralsLet a region R in y -plane be transformed into a region G see Figure 12.15 in the uvplane by differentiable functionsof the form

, x = f (u, 4 , = g ( ~ 7 )so that each point % , vo in the region G corresponds t o a point

in the region R and conversely; then a function (x,y) defined in R can be thou ght of as afunction[ f ( u 7 v ) g (u , V )defined on 6

Figun 12.15 : ransformation of region in xy - lane to region in uv - plane.


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From the calculus of two variables, we have the result that all the functions involved are L i e and mtace I n t e e scontinuous and have continuous irst derivatives, then the integrand+ x, y), of the doublei n r ~ ~ r a l f l(x, )d dy, can be apressed in termsofu and v, a d r dy replaced by du dv

Rtimes the absolute value of the Jacobian of the coordinate transfonnationx= (u,v),y (u, v) given by

(whichis eitherpositive throughoutG or negative throughout G .Hence the integral ofI (x , ) overR and the integral of [ (u,v) g (u, v)] over G are related by the equation

We now take up an example to illustrate how the change of variables simplifies theevaluation of a double integral.Example 14:

Evaluate the double integral

where is the square bounded by linesy =x, y = , x -y 2, x y 2.Solution:

Shape of the square of the problem is as shown in Figure 12.16.The shape of region suggests the transformation

x y = U x - y = vThen

The Jacobian of transformation of coordinates is Figure 12 16 : egionR

Absolute value ofJ = J = 2Now, regionR in y -plane corresponds to the square s u s , 0 v s 2(see Figure 12.17).Thus,

Figure 12 17


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Vector alculus

Notice that had w e not changed the variables, evaluation of R would have to be carried outby first dividing R into two regionsR, nd R2 nd then finding the range of integrations forthe two regionsR1 nd R2 hich are different as is evident from the Figure 12.8 a) and (b)given below.

Figure 12.18 a) : hen ines parallel t x -axis am Figure 12.18 b) :When lines parallel toy - axisdrawn are drawn

Let us consider another example.Example 15

Find the mass of the plate bounded by the fou r parabola sy r 4 a r , y 2 = 4 b x x Z = 4cy and x2 = 4 d y

if the density of the plate is p = ,where k is a constant.Solution :

In this a s e , the mass of the plate =Is k g ) r dy, wherere gionA is shown inFigure 12.19 a). It is quite clear from the Figure 12.19 a), that if we wish toevaluate this integral, we shall have to subdivide the regionA in many sub-regions.

a) RegionA n y - plane b) RegionA ransformed in v - planeFigure 12.19

However, if we assume

then in uv - lane the regionA is transformed tov 4 a v = 4 6 , 4 c 4 d ,

which is a ~ c t a n g l e Ahaving sides parallel to and v axes.


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Line and Sorfna Integ~dshus he transformation givesx = u H v My = u M v

: J - 4 ) u' v M U V - 4 1 1= =V J U M V H %u%-45 9 9 3Also p - k * y = k u H v M . u W v W k u v

4d 6: Mass of plate = 1k uv - du dvJ,-&J,-&64 k= b 2 a 2 ) ( d 2- c 2 )3

We know that double integration is integration over an area in a plane. So far we have usedcartesian coordinates (x, y) for a plane area. However, in some practical problems, it isconvenient to use polar coordinates(r, 0) representing a plane. You might already beknowing the relation between cartesian coordinates (x, y) and polar coordinates (r, 0). We

-

cos 0 si n 0 If (x, y) d r dy = I (r 0, r sin 0) r dr d0

R G

is illustrated in the next example.Example 16 :

Find the polar moment of inertia about the origin of a thin plate of density p = 1bounded by the quarter circle x 2 y = 1 in the first quadrant.Solution

n-J o a o I I 2(r, 0)

I I

x = r cos 0 y = r sin 0 (Figure 12.20)

Figure 12where G is the region in r -plane corresponding regionR inxy - plane.How the change of the cartesian coordinates to polar coordinates helps in solving problems,

By definition, the polar moment of inertia about the origin is given byJJ p ( x 2 + y 2 ) k d y ,R

whereR is the quarter circle in the first quadrant (see Figure 12.21).Now, if we take

I 190)x = r c o s 0 , y = r s i n 0 , Figure 12.21 :Region inXY plane

the quarter circle in first quadrant is transformed to a rectangleG in r 0 -plane asshown in Figure 12.25 where 0' s r s 1 0 s 0 s ,.

Also x 2 y 2 = r2cos20 r2sin28 r. . I. = u ( x 2 + y 2 ) dr d y - J f r 2 . r d r d 0 o 1

R GFigure 12.22 :Region inr - tane

'52 1

- r 3 d r d 0 so d 0 = l f h d O =0 0 r o n 4 o 8


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Vector Calculus You may now try the following exercise.

Prove that the area in the positive qua drant, bounded by the curve y 4 a x ,y 2 4 b x , xy c 2 and q = d 2 i s

12.4 TRANSFORMATION OF DOUBLE INTEGRALS INTOLINE INTEGRALS GREEN S THEOREMW e may transform double integrals over a plane region, under suitable conditions, into lineintegrals over the boundary of a region and conversely. This tra~ lsfo nn atio ns of practicalinterest because it makes the evaluation of an integral easier. It also helps in th e theorywhe never w e want to switch from one type of integral to other.This transformati011can be done by ~n ea ns f a theorem known a s Green s Theorem whichis due to English mathematician, George Green (1793-1841). W e shall now state thistheorem.Green s Theorem in a PlaneLet R be a closed bound ed region in the xy -plane, where boundary C con sists offinitelymany smooth curves. Let M x, y) an d N (x, y) be finctio ns which nre continuo us an d havea M d Ncontinuous pa rti al derivatives nd very where in some dom ain containingR8 ~ a xThen

( ~ d r + ~ d ~ )[ ( z - z ) ) d r d y

the integration being taken a long the entire boundary C of R such that R is on the left as oneadvances in the direction o f integration.W e shall not be proving this theorem here as it is beyond the scop e of this course. Learnerinterested in knowing ts proof may see Appendix-I. Howeve r, we shall explain what thistheorem means and illustrate it through examples. Bu t, before that w e state the theorem invector form.Green s Theorem in Vector orm

L e t F = M i + N j + P k and r = x i + y j , t h e nF . d r M a 5 N d y

i j kAlso C u r l F - V x F a x a y a zN P


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The com ponent of Curl whit is nonnal to region R in thexy -plane is

I Hence, Green s Theorem in the plane can be written in the vector form as

where dA = d4 k = k dy is the vector normal to the region R in y -plane and is ofmagnitude I = dx dy?Green s Theorem states that the integml around C of the tangential component of F sequal to the integral over the region R bounded by C of the component of C url F that isnormal to R.

The integral over R is the flux of curl hrough R.We shall later, in section 12.6, extend this result to more general curves and surfaces in theform of a theorem known as Stoke s TheoremThere is second vector form of Green s theorem as follows

1Let - N j and letn = Unit outward vector normal curve C

= cos a ; cos 9 j= cos a i sin a

Thus

= M d x N d y .Also

Hence, Green s Theorem, which says

gives us

ineand urface Integrals


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vector CalCu1UB

nUnit 13 we shll extend result (12.24) to three-dimensional ve do r fields and call it adivergence theorem. We can thus say that Green's theorem in the plane is a twodimensional f o m of tbedivergence theorem.We now take up a few appUcationsofGreen s th or m and illustrate its impottance.I) Ama ofa plane region a Une integral o v a he boundary

Fmm the Green's theorem. we have

LetM = andN =x then fmm equation (1223,we get

The integral on the left is the areaA of the region R.Next, letM y 4= 0 hen, fmm relation (1225), w e get

Adding (12.26) and (12.27)' we get

WhereA is the area of region R enclosed by boundary C. huswe have been able toexDress the area of ree ion R in terms of a line inteeral over the boundarv. T hisinteplai

ii Area ofa plane regionin polarcoord natesLet r and 0 e the polar coordinates. We define

x r c o s 0 , y = r s i n 0 .Then x = m 0 d r r s j n 0 d 0and dy - s i n e r c o s 0 d 0@e relation (12.28) reduces to

tbeh o m lcomponentof m y v d r ieldF around the boundary ofa regDonR nwhich is continuous .ad hascmtinuwspvtlrrl derivatives, is eqwl to the doubleintegral of divergence of overR

JJmdu J x d yR

R

:resting formula haYs various applications, for eiamp le, the theory oim rta innima ers (instruments measu ring area) is based upon this formula.

= ~ ~ [ r m 6 ( s i n 0 r m s 0 d 0 ) r s i u 0 ( m 0 r s i n e d 0 ) ]2 c

= Ls rZdOc

2 2 . 2 0 2 2A - ( I - - 0 ) d 0 - -l ( I - 2 e o s 0 + m s 2 0 ) d 02 2

a form ula which iswell known in Calculus.As an application of this formula (12.29) we consider the'cordiod

r = a (1 os 0) where 0 r 8 s 2n (Figure 12.23).We find


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We now take up a few examples to illustrate the use of Green s Theorem.Example 17

Using Onen s Theorem, evaluate the integmlf - y dx x dy), whereC is theC

circumference of the circle x 2 = 1Solution

Here,M=-y, N x: f - y d x + x d y ) = ~ ~ d r d y = s J l + l ) d x d y

Circle Circle2 2 2 2x y 1 + I 1

= 2 dxdy = 2 x Area of CircleCircle2 2i y 1

P 2 X r c X 1 2

Example 8Use Green s Theorem in a plane to evaluate the integral

When C is the boundary of the surface in xy -plane enclosed by thex -axis and thesemicircley =G

SolutionHere M = 2 x 2 y 2 , N = x 2 yUsing Green s Theorem

Where is semi-circlex y = bounded by x axis.

Figure 12.24

L i e amd Surface Integrals


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Vector Calculus

(because semi-circley = bounded by x -axis is given by x ,

4= n magnitude3You may now try the following exercises.E 10.

Use Green s Theorem to evaluate[ ( x 2 * y ) ( x 2 y 2 ) d y ] .

Where C is the boundary of the square y = * , x = *

E 11.~valuatef [ 2 ly ) r + x 2y + 3 ) dy] around the boundary C of the regionY 2 = 8x x = 2.

In Sections 12.3and 12.4 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 aplane. For example, potential due to changes distributed on surfaces, centre of gravity ofcurved lamina, area of the surface out from the bottom of the paraboloid z = x 2 + y 2 bythe planez = 1,etc. This gives rise to the concept of surface integrals, which we shall takeup in the next section.

12 5 SURF CE INTEGRALSThe concept of a surface integral is a natural generalization of the concept of a doubleintegral considered in Section 12.3.There we integrate over a region in a plane and here weintegrate over a piecewise smooth surface in space.The definition of a surface integral is parallel to that of a double integral. Here we considera portion S of a surface. We assume that S has finite area and is simple, i.e., S has no pointsat which it intersects or touches itself. Letf (x, y, z be a function which is defined andcontinuous on S. We sub-divide S into n parts S,, s;, . ,S, of areasMI,AA2, ...,AArespectively. Let P(xb yb zk e an arbitrary point in each part Sk Let us form the sum


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and Surface Integralsn

Jn = (xk9 Yk, zk) AAk (12.30)x -Now w e let tend to infinity in such a way that the largest part out of S,, S, , ,S, shrinksto a point. Then the infinite sequence J, , J2 , . ,J, has a limit wh ich is independent of ;hechoice of subdivisions and points P Thi s liinit is called the Surface Integral of F (x, y, z)o ve rs and is denoted by

J J F ( x , Y, z ) d S .S

J J f ( x , Y, z ) d S = lim 2 f ( ~ ~ . ~ ~ , z k )Ak, (12.31)S n + m k -

provided the limit exists.Evaluation of Surface Integral :To evaluate the surface integral

J J f ( x , Y, z ) & .S

we inay reduce it to a double iiltegral as follows :surface S can be represeiited in the parametric form a sr (u, V) = x (u, V) y (u, v) z (u, v) k , (12.32)

where u and are tw o il~depe nde nt eal variables, called the parameters of therepresentation.Hence r (u, v) is the position vector of the points of S and its tip ranges o v e r sas (u, v) varies in some region R in uv -pl ane (see Figure 12.25). S u r f a c e s i n

r (u. I ) sp a c eTo ea ch (u,, v,) in R, there correspon ds a point of S w ith position vectorr (u,, v,). H ence , region R is the image of S in uv -plan e. In order that surface I Yhave certain geo met ric properties, we assume that r u, v) is continuous andhas continuous first partial derivativesru ] and rv - ) i n a vl Romain of uv -plane which includes the region R and R is simplyconnected and bounded. I4Froin equa t ioi~12.32), we have d r = ru du rv dv. v laneThus, h e a r elelnent of surface S is give11 by Figure 12.25 : Parametric representat ion o f asurface

d s ' = d r . d r= ( r u d u r v d v ) . ( r u d u r ,d v)= r u . r u d u 2 2 r u . r v d u d v r v . r v d v 2 , (U ~u . AV= , T a b 2 2 F d u d v G d v 2 , (1 2.33)

/which is quadratic differential fo rm and is called the First Fundamental form m JofS. H e r e E = r u u , F = ru , and G = r,. r, ( ' 4 vv (U A U. V)In the d efinition of surface integral, we are su b- dividing S illto parts S,, SZ, ,Sn, l e t A be the area correspo ndin g to one of the parts in uv lane. Then the Figure 12.26 :Areasmallest parallelogram in Figu re 12.26 h as area; (by the definition of vectorA A - Iru A U x r , ~ v = r x r A U A V , (12.34)


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s b r alculus and it is called the elementof area.Hence, if R is the region corresponding to S in uv -plane , then

Wh ere the right-hand side is now is double integral ov er a plane area.W e know that

Thu s we may also write: 55 f x, Y, z) d S -55f [ x u, v), y u, u, t,) dE G - Z u dv 12-37)If a surface S is given by

z = Y) ,W e may set x = u, y = v and then parametric representation of S can be witten as

r u, V) = u i v j g u, v) kThus

ru = i gU k and r, = j g , k

2 2Hence r , x rvl = EG - F = g , g ,Thus, if S is represented by z = g x, y), then w e have

Wh ere R is now the image of S in xy plane.W e can also write the surface integral 12.38) in terms of the no rmal to the surface S asfollows.Suppose that V F represents normal to the surface S, F x, y, z) = C constant). Let regionarea A be the projection of surfaceS on a plane. Let ; be a normal to the regionA and y bethe angle between the normal to the surface (VF and normal to its uroiection n

Also if A S is an element of surface S surrounding arbitrary point on S and A A is itsprojection on th e planes, thenA S c o s y I = AA


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

Let us now take up examples, from various physical situations to illustrate the evaluation ofsurface integrals.Example 19 :

Find the moment of inertia I of a homogeneous spherical laminaS : X ~ + ~ ~ + Z ~ - ~ ~

of mass M about z xis.Solution :Let a mass of density x, y, z) be distributed over the surfaceS then moment of

inertiaI of the mass with respect to a given axisL is defined by the surface integral

WhereD is the distance of the point x,y, z from axisLSince in our case, spherical lamina is homogeneous, thus s constant. Also area ofspheres = 4 x a 2 .

. Hence= Mass per Unit Area

The parametric repre~~ nta tio nf the spherex2 y z 2 = a isr ,, , a cos u cos v i a cos u sill v a sill 1:

r,, = -asinu cosv i asinu sin v j acosu k

rv = a cos u sin v i a cos u cos v2 2 2 2 2 2 2 2: = r,,. v = a sin u cos v a sill u s in v a a cos u = a 2

ru .rv a2s inu cosu cosv s inv a 2s in u s invco su cosv

If areaA is the image of surfaceS in uv -plane, thendA = Ir,,xrvI dudv

ineno da c e Integrals


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VectorCalculus= a2cosu du .dv

Further, the square of the distance of a piiilt (x, y, 7) from z axis is= x 2 + y 2 = a 2 cos2u cos2v + a2cos 2u sin2v = a2cos2u

Hence, we obtain$$ p D 2 d s =A x2(u,V) + y2(u, v)] d~s 4 n a 2 ,

M v 2z= U ( a2 co s 2 u ) . a2 co s u dudv4 n a 2 -- /2 u o

M a 4 IY= -$* cos3u v I du =4 a. 2 . -a2 2 = M a 22 3 3

Let usnow take up an example from geomet~y.

Example 2Find the area of the surface out from the bottom of the paraboloid z =x +y by theplanez = 1.

SolutionThe projection of the area of the surface of paraboloid z = x +y out froin thebotlom by the plane z = 1 on xy -plane is the disk

x 2 y 2 $ 1Now surface area

= $$ d~S= $1'41 g + g dr dy

R y

= $$x2+y2s$*2+4~2+1 drdYX

Figure 12 27 :Area of parabolicsurf ce

Letx = rcos8, y = r s i n 8I 3 en f o r x 2 + y Zs 1 , s 5 1 and 0 s 8 5 2 n:. Required surface area

= so2so1m= jo2: ( 4 9 + l Y h Ih d B = (5n- l) j : dB

r - 0 2

= 5 6 1)6


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You m ay now try the following exercises.E 12

Th e electrostatic potential at (0, 0 a) of a charge ofhemispheres : x 2 + y 2 z 2 - a2 z 0 is

Evaluate U

Liae nd urface Integrnls

constant density a n

d.9

Find the area of the upper cap cut from the sphere x y z = 2 bythe cylinderx y = 1 (see Figure 12.28)Hint Take surface s = 4 2 x 2 - y 2 its projection on xy -plane asdisk y 1and use polar coordinates to evaluate double integral.

Figure 12 28

In Section 12.4, we had sh ow n that double integrals over a plane region can be transformedinto line integrals ov er the boundary curve of the region. We shall generalize this result andwe shall now c ons ider the correspond ing problem in the case of a surface integral in thenext section.

12 6 TRANSFO RM ATION OF SURFACE INTEGRAL INTOLINE INTEGRAL S TOKES THEOREMTransform ation of su rface integrals into line integrals and coilversely is done with the helpof a theorem known as Sto ke's the ore m, given by Gorge Gabriel Stokes (1819-1903), anIrish mathe matician and physicist, wh o made imp ortant contribution to the theory of infiniteseries and s everal bran ches of theoretical physics. Stok es theorem is an extension ofGreen 's theorem in vector form to surfaces and curves in three dimensions. Unde r suitablerestrictions(i) on the vector(ii) on the boundry curve C nd(iii) on the surface S bounded by C.Stok e's th eorem co nnects line integral to a surface integral.In Stoke's Theorem , we require that the surfaceS is orienta ble. By orientable, we m eanthat it is possible to consisten tly assign a unique direction, called positive, at each point of S,

Aand that there exists a unit ilonnal pointing in this direction. As we move about over theAsurfaceSwithout touching its bou ndary, the direction cosines of the unit vector n should

Avary continuo usly and whe n we return to the straight position, n should return to its initialdirection.


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ectorCalcduf W e now state Stoke s theorem.Stoke s TheoremLet S be apiecewise smooth oriented surface in space and let the boundary of S be apiecewise smooth simple closed curveC Let F (x,y z) bea continuous vector fincti onwhich has continuous irstpartial derivatives in a domain in space which contains S. Then

wherea n US where s a unit normal vector of S, so that (Curl F ) I; is thecomponent of Curl F in the direction of he integration around C is taken in thedirection of integration onS andF, is the component of F in the direction of the tangentvector of C.If we represent curve C in the form (s),whe re the arc-lengths increases in thedirection of integration, the n unit tangent vector is

and, therefore, if F = F 1 i + F , j + F 3 k ,thenThus F,ds F 6 ds Flak F 2 4 F3dzaLet n cos a i cos fl j cos y k be the outw ard unit normal vec tor of S.If w e use the re pm enta tion of Curl in terms of right- handed Cartesian coordinates, thenformula (12.39) in Stoke s T heorem may be w ritten as

Where a p y are direction cosines of unit normal to surfac e S.W e shall not prove this theorem here. Learner interested in know ing the proof of theth eore m m ay refe r to ~ p ~ e n d i x2.2.The re are many consequences and applications of Stoke s theorem. W e shall, however, takeup a few of these. consequenc es and applications.Consequences and Applications of Stoke s Theorema) Green s Theorem in the Plane as a Special Case of Stoke s Theorem

Let F N j be a vector function which is continuou sly differentiable in adomain in the ry-plane. Let ry-plane contain a simplyco nnec ted closed region S whoseboundary C is a piece-wise sim ple smooth curve. Then k is the direction of unitoutword normal to S.


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Now

Furthemlore, F ds = M k + N dy.Now fonnula (12.39) o fXstoke 's heorem yields

Usi ng above va lues in this case, we get

wh ich is in aggrem ent with the result of the Gree n's theorem in the plane (refer Section12.4).This sho ws that Green's th eo xm in the plane is a special case of Stoke's theorem.

b Physical In ter pre ta t ion of the CurlStoke's theorem provides a physical interpretation of the Curl.Let U first define the term Circulation.Let V x,y, 2) be a continuous differentiable vectorfunction in a domain containing asurfaceS bounded by a closed curve C. If V be the velocity field of a moving fluid ofderlsily p, then tlte integral

measures the extent to whiclr the corresponding luid motion is a rotation around thecontocrr C and is called tlte circulation.By Stoke 's theorem , circulation is also equal to the flux of curl pV ) through a surface SspanningC. Thus

Next, let us fix a point P and a direction u at P.Let C he a circle of radius 6, with centre atP hose plane is normal to u. If curl p V ) is continuous at P, then (by mean valuetheorem for surfac e and double iotegrals) the average value of u component of curl pV )ove r the circ ulatio ~l isc bounded by C approaches the u -colnponent of curl p V ) at P as6 - 0 ,

i.e. [ [ ~ u r l ( p ~ ) ]u I p = lim L J J ~ u r l ( p ~ ) . u d s6-0 rcbZ S

i e arnd Surface ntegrals

Now in ou r case, the plane of curve C is normal to u, area of S r ti2 = , ay sothat d = dA. Let F = pV. nThus, fmm equations 12.41) and (12.42), we get

[ C u r l F ( P ) ] , = lim ~ J J C ~ ~ I F . ; ~ A6-0

i.e., tlte component of the curl in tlte positive normal direction can be regarded as thespec. & ci rc ul at io n circulationper unit area)of he flow in the surface at thecorresponding point.

igure 12 29 nterpretof curl


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VectorCaIcutus c Evaluat ion of Line In tegra l by Stoke's Theorem :You may no tice that evaluation of line integral by Stoke's theorem leads to a lot ofsimplification in many problems.aw e v a l u a t e l ~ ,r where is the circle x y 4, z - 3 oriented in the

counbrclockw ise sense as viewed from the origin and with respect to right- handedcoordinates, and

We ca n take the plane circular disk x + y 4 in the plane z = - 3 s a surfaceSbounded by C. Then in the Stoke's theorem point in the positive z -direction, so thatAn k

AHence, (Curl F is simply the compo nent of Curl in the k direction. Now withz = -3 has the com ponentsF = y, F2 = - 27x and F3 = 3y 3. Thus

Hence, integral over S in Stoke's theorem equals - 28 times the area of the disk S,which is 4 GThus, IF, -28 x 4 n -112x -352

You will appreciate the simplification due to Stoke's theorem in this case. Th ecomputation s will be long and difficult if you evaluate the given line integral directly,starting from parametric represelltation of C , calc ulati~ ig nit tangent vector t of curve C in

Athe direction of integration, findingF = F and then integrating with respect to arclength s of C.It should be remembered that fo r Stoke's theore m, th e surfa ce S is assumed to be a nope n surf ace . The significance of the statement will become clear from the followingresult.d l l ~ u r l S will Zero over any Closed Surface S :

Since S is any closed surface, we cut open the surfaceS by a plane and let S1 and S2denote the lower and upper positions of S. Let C denote the common bounding curvefor both these positions. Then,

It may be noted that the normal to lowe r portion S1and the normal to upper portion Swill be in opposite sen ses so that the direction of integration ab out S1 s reverse of thedirection of integration about S2.Using Stoke's theorem and result 12.43) bove, we get

~ u ~ i ~ . d s5 dr + f ~ d r 0S due to surfaceS due to surface S2

which proves the required result.Let Now take up some example to illustrate the use of Stoke's theorem in numericalproblems.

xample 21:Let S be the poltion of the paraboloid z = 4 - y2 that lies above the plane z 0.Let C be their bounding cu rve of intersection. If


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Lineand Sur aa lotegrds

verify Stoke's theoremSolution

The curve C is the circle 4 in the -lane. Along C, whem z 0 andr x i dyj, we have

NOW, J ~ - & - ( - ~ d r + ~ d y )

Now for surface integral, we have

VY . - . a .ax ~z - y z + x - ( x+ y )

For the positive un t nonnal on the surfaces : ( x , Y , z ) - z - 4 + 2 + Y 2 - ~

we takeThe projection onS on xy-plane is the region 9 and for the element of surface, we take

- J 2 h d ~ (Since odd power of x or integrate to zerox2+y2 4 over the interior of fhe circle)

area


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Vector alculus

= 2 ~ x 2 ~xHence, Stoke's theorem is verified.

- You may now attempt a few exercises.E14

Verify Stoke's theorem for the functionF = ~ i ^ - x y ~

integrated round the square in the plane z = 0 nd bounded by the lines x 0,y = 0,x = a a n d y = a .

E16~ e t be the outer normal of the elliptical shell

~ : 4 x ~ + 9 ~ ~ + 3 & - 3 6 ,r Oand let F y i - 3+ 2+y2)3R sin (e*) 2 .Use Stoke's theorem to fin the value of

(cu r l FS

E17Let Sbe the region bounded by the ellipse

c : 4 2 + 1 4 in the plane z =and le t ; -$ nd ~ - ~ ? i ^ + 2 r j + t % .Use Stoke's theorem to find the value of f F .

A fluid of constant density p rotates round the z-axis wi th velocity V = x i - y i ),where is a positive constant. If F = pV, find curl F nd show its relation to specificcirculation.


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vectorCakdus

2.8l

where k is the area of ktb ub-regior of R, exists and is the doub le integral off (x, y) ove r the region R and i s denoted by

iff (x, y) is uniformly continuous and bounded over R and f x,y) dy isY -A x ))bounded and integrable from a to b with respect to x a long wi th5 f (x,y drx - pG . )being bounded and integrable from c to d with respect toy.

Transform ation of double integrals into line integrals is based on Green 'stheorem in the plane.Iff (x, y, z) is defined and continuous on a surface S then the limit

where Sk is the area of kt subd ivision of S exists and is the Surface Integral of x, , z ver the surface S and is denoted by

Transformatioil of surface integrals into line integrals is based on Stoke'sTheorem.If Vbe the velocity field of a moving fluid of density p then

is the circulationof fluid around curve CComponent of the curl in the positive normal direction can b e regarded as thespecific circulation (or circulation density) of the flow in the surface at thecorresponding point, i.e.,

[ url P)], lim - F, d6 4 0

where F = pV.SOLUTIONS ANSWERS

(i) Along the curve Cx = t , y = t = , z=C3

: d r - d t , d y - 2 d t , d ~ = 8 d t


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Lint and d a a ntegrals

- 3 - 4 6 = 5ii) Here

E2~ e r e ~ = 2 r - ~ - z ) ? + ~ + ~ - i ? ) ~ + 3 ~ - 2 ~ + 4 ~ ) ~The parametric equation of circle y2 9 in the XOY-plane is g iven

x = 3 c o s 0 , y = 3 s i n 0 , z - OHere

Now work done by force F in moving a particle once round the cillcleXOY-plane

2 n9 9 m s 0 sin 0)

E3Here cp wZz g Y


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vector alculus

E4ere


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Liue and Surfam Intrgra

= 8 os y +yz)du + xz 8 sin y)dy + xy + z dz

which is true.Here F is conservative and F = Vf here = c o s y + xyz +

7 J

a) For 1 + y 2 ) 4 dr he region of integration is y = 0 = Lr = 00

x = 1, which is shaded portion in the Figure

Now,

b) For 1 - ry)dy dr he region of integration is x = 0 x = I ,x - 0 y - xparabola y =x2 and line y =x, w hich is shaded portioil in Figure.

The base triangle inxj-p lane is hounded by3 = 3 y , y = O a n d x = 3

he mouired volume 1s uilder the plallc


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H e n density of plate =K x2 y2). The plate lies between y 2 nd =y2.For the plate, range of integration i s x = 0 to 1 and y t o y =2Hence, m ass of the plate

Th e required area is in the positive quadrant, bounded by the curve sZ 4ux = 4bx, y = c xy = d2

Let this area be transformed to uv-plane, using the transform ation

s o that the transformed area is the rectangleA having sides parallel to u and v axesgiven by


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ime ud d o a n ~ s


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Vector Calculus 2-J ( t r . r n + t r . S ) d xx 0

I 28161a 4J5 -5 5

The surfaceS y2 3= a2, z z 0 can be represented byr = a s i n u c o s v i a s i n u s i n v j a c o s u k

c : r , - a c o s u c o s v c a c a s u s i n v j - a s i n u krv= - a s i n u s i n v i + a s i n u c o s v j

: = r,, r cos2u cos2 v a2 os2u sin2 v a2sin2u = a22 2F = r; r, -- a c o s u s i n u c o s v s i n v + a c o s u c o s v s i n v s i n u = O

= r, rv= sin2 u sin2 v a2sin u cos2 v = a2 in2 uIf areaA is the image of surface S in uv-plane, then

d4 = Ir,, x rv( u dv= W d u d v

= a2 in u du dvFurther,*2+y2+ 2+a)2= asinucosv)2+ asinusinv)2+ acosu a ) 2

= a 2 s i n 2 u c o s 2 ~ 2 s i n 2 u s in 2 v a 2 c o s Z ~ a 2 Z u Z c o= 2a2 2a2 cos u = 4a2 cos2 2

: u-JJ I dsv x2 +y 2+ (Z+.)Za2 sin u du dva J J z a c o s u 1 2

2 s i n h / 2 c a s ~ / 2 ~ ~ ~ ~2 c o s u / 2SI2= a a l - o ~ v - o s i ndu dv :s is P hemi-sphere).

n12= o a J u - o s i n y . i h

- 2 r c a a ( 2 - a .

Surface area -JJ= J J m d x d y , where z=g(x,y)=\/2-I-J

Sis the surface of cylinder ndR is its projection on xy-plane taken as y 2 1.


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L i e andSurface Integrals

et x = r cos 8, y = r sin 8,then ~ ~ + ~ ~ s l + O s r s ln d O s 8 k 2 n

Zrc: Required surface area =J r dr d80 0 r 0 4 2 3

E14HereThe curve C is the boundary o f the square in the plane z = 0 and bounded by thex = O , y = O , x = n , y = a .Here dr = dx i^ d y j

NowX a Y 0= J (x2dx y dy) J ( 2 k - T Y d y )0 0I x2dx xy dy +J (2 dx y dy)x a Y 0

a t y a a t x 0

For surface integral

= J x o x 2 d x - o J Y 0 y d y + j a $ u x + ~0 o ~ x - o ~ Y )

N o w positive unit normal to the plane z = 0 is = k.Also elem ent of area dS = dx dy.Thus J J ~ u r l ~ - ; d S = - j - yd rd yx 0 y 0

lines

a.a )


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Hence

which verifies Stoke's Theorem.

E15Here F = ~ V = ~ W ( ~ Q - ~and

a a = ~ ( P W + ~ W ) = ~ P W ~url ax ay az

The work done by a force equal to is IF drC

If C lies in a plane parallel to the xy-plane, then Stoke's theorem gives

Here S is the area ellclosed by C, which is a circle of radius a

Thus(Cud F ) 2 p F. circulation density.n u

Thus component of.Curl in the direction normal to the plane of V is equal tocirculation density.E16

The elliptic shellS is4x2+?y2+362=36,z z 0

One parametric representative of the ellipse C at the base of the shell isx = 3 c os t , y = 2 s i n t , O s t s 2 n .


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Here y f + x 2 j + 2+y2)3 sin (e*) iBy Stoke's theorem

= y f 3 + 1 y2)3' in r e ) ( f + d y j+ dz L)ellipse2 2r 9y r 6= [y dx +x2 dy (x2 y2)3 sin(eG)dr]ellipse

2= 2 sin (- 3 sin r)dt + 9 cos' r(2 cos r)df1-0

Ir9 sin 3r- t + -+ -2 3 + 3 sin t l o

l7Here

For region S bounded by ellipse C 4 1 +y2= 4 in plane z = 1,n = k: yStoke's Theorem

( '.'area of given ellipse = n m'1.2 = 2n)


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Vector alculus APPENDIX 12 1PROOF OF GREEN S THEOREMWe first prove the theorem when C is a simple closed curve in the xy-plane such that a lineparallel to either axis cuts C in at most two points. The special can be represented in thefonn of

Thus curve C consists of lower portion y u x) and upper portiony v x) and both theportions extend fromx a to x = b Figure 12.l a)).

Figure 12.1 Special region for Greeo s Theorem

Hence, using relation 17 of Section 12.3.2), we get

Now integrating the inner integral, we get

Substituting this into relation 1), we get

Sincey u x) represents the oriented curve C, fromx a to x b and y v x) representsthe oriented curve C2 fromx = b to x = a, thus the integrals on the right may be written asline integrals over C, and C2, and therefore,

and

From equatio. ~2) , 3) and 4), we get


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Similarly by taking strips parallel tox-axis (refer Figure 12.I b)) and using relations 19) ofSection 12.3.2.we obtain

N(X,Y)&Fmm Equations (5) and 6),we obtain

which complete the pmof of Green's Theorem in the plane for special regions which areenclosed bv s im~lemooth curves.

=J N[qb),yl - ~ b b ) , y l l y[ N[qO.),Y dy + ; N W ) , Y I ~ Y

Next we consider that the region R itself is not a special region, but can be subdividedfinitely many special regions (see Figure 12.II).

Figure 12 11 Multiply-connected region R

the theorem for the most general region R satisfying the co2itions Gthe theorem,wapproximateR by a region of the type just considered and then use a limiting pmcess.

into

In this case, we apply the theorem to each subregion and then add the results. he left-handmembers add up to the integral overR, which the right-hand members add up to the lineintegral over C (which now consists of outer boundaryC and inner boundary C2) plusintegrals over the cuzves intmduced for sub-dividingR. Each of these later integrals occurstwice, once taken in each direction. Hence these two integrals along curves intmducedcancel each othe~ ther and we are left with the line integral overC. Hence the theorem.The pmof thus covers all regions which are of interest in engineering problems. To prove

ine and urfaa Integrals

must


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istandSmiau Integrals

irtue of relation (3)=IF x, y, z by virtue of relation 2).

Thus w e have proved relation (I).Using the representatioon of surfac esS as y = g x, z and x = h v ,z , and reasoning exactlyas above, we obtain

(4)

(5)

By adding I) , (4) and (5 ), we obtain the formula given by equation (12.38).This proves Stoke s Th eorem for a surfaceSwhich ca n be represented simultaneously inexplicit fonn s z x, y), y g x, z), x = h x,y).W e may immediately extend our results to a s ur fa ce s which can be decomposed intofinitely many pieces, eac h of which is of the type considered above.It should be noted that a person m oving 011 C with his head in the positive direction of thenormal to surface S, should keep the surface on the left.


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UNIT 3 VOLUME INTEGRALStructure

13.1 IntroductionObjectives

13.2 Triple Integral13.2.1 Definition13.2 .2 Properties of Triple Integrals13.2.3 Volume13 .2. 4 Evaluation of Triple integrals13.2.5 Physical Applications of Triple Integrals

13.3 Transformation of Volume Integrals into Surface Integrals13.3.1 Gauss Divergence Theorem13.3.2 Consequences and Applications of Divergence Tbeorem13 .3. 3 Integral Definitio ns of Gradient, Div erg enc e and Curl13.3 .4 Physical Interpretation of Divergence Theorem13.3.5 Modelling of Heat Flow13.3.6 Green s Theorem and Green s Formula13.3.7 AB asic property of Solutions of laplace s Equation

13 4 Irmtational and Solenoidal Vecto. Fields Revisited13.5 Summary

13 1 INTRODUCTIONI11 the previous unit, you have studied about line integrals, double integrals a ~ ~ durfaceintegrals. In the process you have learnt to transform d ouble integ rals and surfaceintegral into line integrals. You had learnt that line integral is the generalization of asingle integral and a su rface integral is a sort of gen eralization of a doub le integral. Inthis unit, we shall give another generalization of double integral called tr ip le in teg ralso r vo lume in teg ra l s.W e shall first of all define triple integrals in Sec tion 13.2 wherein we s hall also givethe properties and evaluation of suc h integrals. This section will be closed w ith som ephysical applications of triple integrals.W e had made use of integral transform theorems Gree n s Theorem and Stoke sTheorem - in the last unit. In this unit, we shall discu ss another important intergraltransform theorem, known as Ga uss Divergen ce The ore m, w hich helps in thetransformation of volume integral to surface intergal and conv ersely. Divergen ceThe orem has many important consequences and various applications, som e of whichhave been discussed in Se ction 13.3.We had earlier discussed solenoidal vector fields and irrotational vector fields inUnit 11 With ou r knowledge o f vector calculus, we have revisited these co ncepts inSection 13.4 where we have now given integral form conditions for vector fields to hesolenoidal and irrotational. The summ ary of the results discussed in this unit ispresented at the end of this unit.

bjectivesAfter go ing through this unit, you should be ab le to

Unit 12 Line and Surf'Ace Integrals

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Transcript of Unit 12 Line and Surf'Ace Integrals