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[Pages:16]Multiple Integrals

5

Example 3 Solution:

a a2 ?x2

Evaluate dydx

00

a a2 -x2

a

dydx = ( y)0 a2-x2 dx

00

0

a

= a2 - x2 dx

0

=

a2

sin-1

x

+

x

2 a 2

= a2 sin-1(1) 2

= a2 = a2 22 4

a a2 - x2

0

Example 4 Solution:

Evaluate

1

0

1+x2

dydx

0 1+ x2 + y2

a2

-

x2

dx

=

a2

sin-1

x

+

x

2 a a

a2 - x2

( ) 1

0

1+ x 2

dydx

1 1+x2 =

0 1+ x2 + y2 0 0

dy

dx

1+ x2 + y2

1 1+x2 =

dy

dx

( ) 0 0

1+ x2

2

+

y

2

1

=

0

1 1+

x2

tan-1

y 1+ x2

1+x2 0

dx

( ) ( ) ( ) 1

=

1

tan-1 1 - tan-1 0 dx

0 1+ x2

=

1

1

dx

4 0 1+ x2

( ) =

4

log

x+

1+ x2

1 0

Q

( ) =

4

log

1+

2

- log1

( ) =

4

log

1+

2

Q

dx x2 + a2

tan-1

x

a

( ) dx

= log x + 1+ x2

1+ x2

 y

2 9y dy 9 y 2

2

27

1

21 2

Notice that in Example 1 we obtained the same answer whether we integrated with respect to y or x first. In general, it turns out (see Theorem 4) that the two iterated integrals in Equations 2 and 3 are always equal; that is, the order of integration does not matter. (Thisiss6imilar to Clairaut's Theorem on the equality of the mixed partial derivativEensg.)ineering Mathematics

The following theorem gives a practical method for evaluating a double integral by

exp4re.3ssingEitVaAs LanUiAteTraEtedTiHntEegDraOl (UinBeLitEherINorTdeEr)G. RAL (REGION FORM)

Fubini's theorem:

e Italian mathema943), who proved a theorem in 1907. But unctions was known n Augustin-Louis lier.

If4f isFucboinntii'snuToheuosroenmthIef rfecistacnognlteinuous on the rectangle

R x, y a x b, c y d , then

yy y y y y f x, y dA b d f x, y dy dx d b f x, y dx dy

ac

ca

R

MMoorree ggeenneerraallllyy,, tthhiiss iissttrruueeiiffwweeaassssuummeeththaattf fisibsobuonudneddedonoRn ,Rf,isf diisscdoisnctoinnutionu-s only on a finite number of sumouosotohnclyurovnesa, fianditethneuimtebraetreodfinsmteogorathlsceuxrivste.s, and the iterated integrals exist.

C A(x)

s Fubini's mation of

Example 1

( ) Evaluate the double integral x - 3y2 dydx Where R

{( ) } The proofRof=Fubxi,nyi's|T0heoxrem2,is1tooy

give an intuitive indication of why it is

dif2fic,ult true for

to include in this the case where f

book, but we can at least x, y 0. Recall that if

f iSs oploustiitoivne:, then we can the solid S that lies above

interpret the R and under

double integral the surface z

xfxRx,f

x, y.

y dA as the volume V of But we have another for-

( ) ( ) mula that we usexd-f3ory2vodluydmxe=in2 C2 hxap-t3ery26,dnyadmxely,

y

R

01

y =

2

0

xy

-V y3 12

adbxAx

dx

2

( ) ( ) { } where Ax is the area of a cr=oss-se2cxti-on8 o-f Sx -in1thedxplane through x perpendicular to the

x-axis.

( ) ( ) is z

From Figure 1 you can s0ee that

f x, y, where x is held co2nstant = x-7

0

Ax

Aadnxdx=cdisf xthx-y,e2y7ar2deday02.

under the curve Therefore

C

whose

equation

y and we have

=

1 2

25 -

49

c

= -12.

c y

( ) yy y y y Example 2

R

f

Exv, ayludaAte

Vysin

R

abxAy dxyddxx,

R =

bd

a[1, c2]f

?x[,0y,d].y

dx

d

Solution:

( ) ( ) y

A similar argument, using cross-se2ctions perpendicular to the y-axis as in Figure 2, shows

that

y sin xy dydx,= y sin xy dxdy

R

01

( ) yy y y R

f x, =

y dAcos -y

d

xyc

2b f x, a dy

y

dx

dy

0

y 1

2

= - cos ( xy) dy earning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole

t any suppressed content does not materially affect the overall learning experience.

or in part. Due to electronic Cengage Learning0reserves

trhigehrtisg,hstotmoeretmh1iordvepaardtdyitcioonntaelnctomntaeynbt eatsaunpyprteimsseediffsruobmsethqeueenBtoroigkhatsndre/ostrriecCtihoanpsterer(qsu)i.rEe diti.torial

review

has

= (-cos 2y + cos y) dy 0

=

-

sin 2

2

y

+ sin

y 0

= 0.

x x

1, 2) 2)

x x

Using the same methods that were used in establishing 3 , we can show that

c

0

x

5 5

yy y y f

fx,xy,yd AdA

d

d c

h

h2 y

f 2h1yy

fx,xy, yd

d x

x d

d y

y

yy y y D

c h1 y

D

y d

where D is a type II region given by Equation 4. where D is a type II region given by Equation 4.

x=h?( y) D x=hTM( y)

Multiple Integrals

7

0

x

( ) vpvarEaXEbAXoMAlaMPsLPyEELE1xa1mE2pxvEla2velaau3nlaudtaeytexxxDxD1xExvax2lu2y2.ayted AdD,Aw, wxhe+hre2eryeDDdisyidsthxteh, WerergheiegorineonDbobiuosnutdhneedderedbgybioytnhtebhoeunded

c

by

the

parabolas

y

=

2x2

&

SpnOaoSnwrtLOoaeeUtLbeTtcUohIaTtOlahnaItNOaswttNhyTtreSThihTtoehreehelerug2epeptixgapoiaro2ianraonaarbnaDn:bobdoD,loalsy,alskassesiknitenictnt1ethcetreehrsdrseeysecdiexc=ntcit2wnt.F1wwhiF+ghehiuegnxenru2ne2r2xe28x2,x82i2,s=isa11at1y+tpyxepx2e,I2x,trI2hte,hragteathigtoaiisintos,in,bxsxu,2b2txu=n2t 1on,to1ast,1oast,yoxstpyox=epxeI-I1IrI.erWg1eig.o1eiWn.onWneaoFSnteIaoeGdnmUtdheRatEyt p7theeIIrreeggiioonnsD, sketched in

we can writFeigure 4.1D,Disatyx,pxye,yIregi1on1buxtxnot1,a1,t2yx2p2xe2IIyreygio1n1andx2xw2e can write

SgiiSgvniiecvnseecsethteheloySlwoi=nwecr1eebr+tobhuxoe2nu,ldnoEadwqrayeurryaistbiiosoynuyn3d2garix2vy2xeia2ssnaydn=dth2tehxeu2 puapnpedpretbrhoebuounupdnpaderaryrbyiosiusynydar1y1isx

2x,

2, Equation 3 Equation 3

(_1, 2)

yy y y xx2y2yd AdA

1

1 1

1 1 x 2

2x2

x2x

x2y2yd

d y

y d

d x

x

yy y y D

1 2x2

D

yy [ [ ] ]

1 1

1x

1

x y

yy

2y

2 y1x2

d yy12xx22

y2 x 2

d x

x

y y=1+ (1, 2)

D y=2

y

1

11xx11x

2x2

11x

2x222 xx2x22x2

2x22x222d

d x

x

y1

_1

1

x

y

1

113x34x

4 x

3x

3 2x22x

2 x

x11d xdx

y1

3 3x

5x55x

4x 4 4 2

54

2 x

3x33x

2x 2 2 x

32

1

x11131251352

FIGURE 8 Figure 4.1

constant not onl We also cons

4

where h1 and h2 Using the sam

5

where D is a

v EXAMPLE 1

parabolas y 2 SOLUTION The p note that the reg we can write

Since the lower gives

yy x

D

4.4 CHANGE OF ORDER OF INTEGRATION ights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has

tesdRceosnetrevnetdd. oMesaynontomt baetecrioaplliyeda,fsfceacnt ntheed,oovrerdaullplleicaartneidn,ginexwpheroileenocer.inCepnagrta.gDeuLeetaorneilnecgtrroesneicrvreisghthtse, rsiogmhtetothriredmpoavretyadcodnittieonntaml caoynbteenstuaptparneyssteimd efriofmsutbhseeeqBueonotkriagnhdt/sorreesCtrhicatpiotenrs(sr)e.qEudirietoirti.al review has content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

As stated earlier, in the double integral with constant limits, the order of integration is immaterial, provided

the limits of integration are changed accordingly. But in case of double integral with variable limits, the

limits of the integration changes with the change in the order of the integration. The new limits are obtained

by drawing a rough sketch of the region of integration. Sometimes in changing the order of integration, it

is required to split up the region of integration, and the given integral is expressed as the sum of number of

double integrals with the changed limits. The change of order of integration often makes the evaluation of Copyright 2012 Cengage Learning. All Rights Reserved. May not be copied, scanned, or dupl

double integrals easier.

deemed that any suppressed content does not materially affect the overall lear

Y

Example 1

Change the order of integration in

? ?a

0

a y

x dx dy x2 + y2

,

and hence

evaluate

the

same.

x=0

Solution: From the limits of integration, it is clear that the region

of integration is bounded by x = y, x = a, y = 0 and y = a. Thus,

the region of integration is DOAB (see Fig. 4.2), and this region is divided into horizontal strips. To change the order of integration,

O

y=a B

x=y

x=a

y=0 A

X

Fig. 4.2

8

Engineering Mathematics

divide the region of integration into vertical strips. The new limits of integration become y varies from 0 to x and x varies from 0 to a.

? ? ? ? ? a a x dx dy

0 y x2 + y2

=

a 0

x 0

x dy x2 +

dx y2

=

a 0

x

?1 ?? x

tan -1

y x

?

x 0

dx

? = a0 p4 dx = p4 [x]0a

=

pa 4

Example 2

Change the order of integration in the following integral and evaluate:

4a 2 ax

4a 2 ay

? ? dy dx = ? ? dx dy

0 x2 4a

0 y2 / 4a

Solution: From the limits of integration, it is clear that first the integration is to be performed w.r.t. y, which varies from y =

4

ay

Y Q P?

y = 4a

y2 = 4ax

A(4a, 4a)

Q? x = 4a

2

=

(x2/4a) to y = 2 ax , and then w.r.t. x, which varies from x = 0

to x = 4a. Thus, we have to first integrate along the vertical strip PQ which extends from a point P on the parabola y = x2 /(4a)

P

x

X

O

(i.e., x2 = 4ay ) to the point Q on the parabola y = 2 ax

(i.e., y2 = 4ax) . Then the strip slides from O to A (4a, 4a), the point of intersection of the two parabolas. To change the order

Fig. 4.3

of integration, divide the region of integration OPAQO into horizontal strips PQ which extend from P on the parabola y2 = 4ax , i.e., x = y2 /(4a) to Q on the parabola

x2 = 4ay, i.e., x = 2 ay . Then this strip slides from O to A (4a, 4a), i.e., varies from 0 to 4a. Therefore,

? ? ? ? ? ? 4a 2

ax

4a 2

dy dx =

ay

dx dy

=

4a

x

2 y2

ay / 4a

dy

4a

=

? ?2

0 x2 4a

0 y2 / 4a

0

0?

ay

-

y2 4a

?

dy

? = ?2

?

a

y3/2 3/ 2

-

y3 12a

?

4a 0

= 4 a (4a)3/2 - 64a3 = 4 a 8a3/2 - 16a2 = 32a2 - 16a2 = 16a2

3

12a 3

3

3

3

3

Multiple Integrals

Example 3

a 2x

a a2-x2

Express as a single integral ? ? x dy dx + ? ? x dy dx and evaluate it.

00

a2 0

9

Y

Y

A y=x

R1

R2

O

M

X

y=x

R1 O

X

(i)

(ii)

Fig. 4.4

a 2x

a a2-x2

Solution: Let I1 = ? ? x dy dx and I2 = ? ? x dy dx

00

a2 0

Let R1 and R2 be the regions over which I1 and I2 are being integrated, respectively and are shown by the

shaded region in Fig. 4.4(i).

Also from Fig. 4.4(ii), it is clear that R = R1 + R2

\

I = I1 + I2 = ?? x dx dy

R

For evaluating I, change the order of integration and then take an elementary strip parallel to the x-axis from

y = x to y = a2 - x2 , i.e., the circle x2 + y2 = a2 . Thus,

a/ 2 a2 - y2

a/ 2 ? a2 - y2

I= ?

?

x dx dy =

?

? ?

?

x dxdy

0y

0? y

?

? ? a/

=

2

x2

02

a2- y2 y

dy =

1 a/

2

(a2 - y2 - y2 )dy

20

=

1 2

??a2 ?

y

-

2 y3 3

?

a/ 0

2

= a3 32

0

Engineering Mathematics

Example 4

1 2-x

Change the order of integration in I = ? ? xy dy dx , and hence evaluate the same. 0 x2

Solution: From the limits of integration, it is clear that the integration is to be performed first w.r.t. y which varies from y = x2 to y = 2 - x and then w.r.t. x which varies from x = 0 to x = 1. The shaded region in Fig. 4.5 is the region of integration. Divide this region into vertical strips. To change the order of integration, divide the region of integration into horizontal strips.

Solving y = x2 to y = 2 - x , we get the coordinates of A

as (1, 1). Draw AM ^ OY. The region of integration is divided

into two parts, OAM and MAB.

For the region OAM, x varies from 0 to y and y varies from 0 to 1. For the region MAB, x varies from 0 to (2 ? y) and y varies from 1 to 2. Therefore,

x y=2

Y B(0, 2)

x=1

x=0 M

y = x2 y = 1 A(1, 1)

O y=0

X

Fig. 4.5

1 2-x

1y

2 2- y

? ? xy dy dx = ? ? xy dx dy +? ? xy dx dy

0 x2

00

10

? ? ? ? =

1 0

y

? ? ?

x2 2

?

0

y

dy

+

2 1

y

? ? ?

x2 2

?

2- y 0

dy

=

1 2

1 0

y2dy

+

1 2

2 1

y(2

-

y)2 dy

? =

1 2

? ? ?

y3 3

1

?0

+

1 2

2 1

(4 y

-

4y2

+

y3 )dy

=

1 6

+

1 2

? ?2

y2

?

-

4 y3 3

+

y4 4

2

?1

=

1 6

+

1 2

?? ????

8

-

32 3

+

4^?~

-

? ??

2

-

4 3

+

1^ 4?~

?

=

3 8

Example 5

Change the order of integration in the double

2a 2ax

integral ? ? f (x, y)dy dx . 0 2ax-x2

Solution: From the limits of integration, it is clear that the integration is to be performed first w.r.t. y which varies from

y = 2ax - x2 to y = 2ax and then w.r.t. x which varies from x = 0 to x = 2a. To evaluate the given integral, take the elementary

strip parallel to the y-axis and its lower end is on y = 2ax - x2 ,

Y

I II (a, a)

P(2a, 2a) III

X

O

(2a, 0)

Fig. 4.6

Multiple Integrals

1

i.e., the circle x2 + y2 - 2ax = 0, and the upper end on y = 2ax , i.e., the parabola y2 = 2ax . Then the strip is moved parallel to itself from x = 0 to x = 2a . Thus, the shaded portion between the parabola and the circle is the region of integration. To change the order of integration, first integrate w.r.t. x and then w.r.t. y. The elementary strip is taken parallel to the x-axis. To cover the whole shaded area the region has to be divided into following three parts, as shown in Fig. 4.6. Region I: The strip extends from the parabola y2 = 2ax , i.e., x = y2 2a , to the straight line x = 2a . Then the strip is taken parallel to itself from y = a to y = 2a to cover the region I. Thus, the part of double integral

2a 2a

in this region is given by I1 = ? ? f (x, y)dx dy . a y2 2a

Region II: The strip extends from the parabola y2 = 2ax , i.e., x = y2 2a to the circle x2 + y2 - 2ax = 0 ,

i.e., x = a - a2 - y2 . Then the strip is taken from y = 0 to y = a to cover the region II. Thus, the part of the

a a- a2-y2

integral in this region is given by I2 = ? ? f (x, y)dx dy . 0 y2/2a

Region III: The strip extends from the circle x2 + y2 - 2ax = 0 , i.e., x = a + a2 - y2, to the line x = 2a. The

strip is taken from y = 0 to y = a to cover the region III. Thus, the part of the integral in this region is given

a 2a

by I3 = ? ? f (x, y)dx dy. Therefore, 0 a+ a2-y2

2a 2ax

2a 2a

a a- a2-y2

a 2a

? ? f (x, y)dy dx = ? ? f (x, y)dx dy + ? ? f (x, y)dx dy + ? ? f (x, y)dx dy

0 2ax-x2

a y2/2a

0 y2/2a

0 a+ a2-y2

EXERCISE PROBLEMS

4.4.1. Calculate the iterated integral

42

( ) (i) 6x2 y - 2x dydx 00

Ans: 222

( ) 3

2

(ii) y + y2 cos x dx dy

-3 0

Ans: 18

4

(iii)

1

2

1

x

y

+

y

x

dy

dx

Ans: 21 In 2 2

2

4.4.2. Calculate the double Integral

{ } (i) sin (x - y) dA, R = R

(x, y) | 0 x 2 , 0 y 2

Ans:

9 4

(ii) R

xy 2 1+ x2

dA,

R

= 0,

1 x-3,

3

Ans:

1 2

(1-

cos

1)

(iii) ye-xydA, R = 0, 2?0, 3 . R

Ans: 0

Engineering Mathematics

4.5 EVALUATION OF DOUBLE INTEGRALS (POLAR FORM)

Introduction

( ) 2 r2

To evaluate the double integral f r, dr d over the region R bounded by the curve

1 r1

r = r1, r = r2 & the straight line = 1, = 2. We first integrate w?r to r (keeping constent) between the limits r1 & r2 & then integrating the new expression w?r to between the limits 1 & 2

2 r2

f (r, ) drd = f (r, ) dr d .

R

1 r1

Example 1

Evaluate ?? r3 dr dq , over the area bounded between the circles r = 2cosq and r = 4cosq.

Solution: The shaded region in Fig. 4.7 is the region of integration R. Here, r varies from 2cosq to 4cosq and q varies from ?p/2 to p/2. Therefore,

q

=

p 2

Y r = 2 cos q

P

r = 4 cos q Q

?? ? ? ? p / 2 4cosq

r3 dr dq =

r3dr dq

R

-p / 2 2cosq

=

p /2

r4

4 cosq

dq

-p /2 4 2cosq

O

q

X

? = p /2 1 (256cos4 q -16cos4 )dq

-p /2 4

q=

p 2

Fig. 4.7

 ................
................

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