Multiple Integrals - Weeblydigirit.weebly.com/.../16653588/em-ii_multiple_integrals.pdf · 2019. 8. 14. · Multiple Integrals 1. Introduction b You know that an integral I f(x) dx

Multiple Integrals

1. Introduction b

You know that an integral I f(x) dx has been defined as the limit of the sum as

a

11 b

J f(x) dx= a

11 � oo r� l f(tr) Or The idea can be extended further to define of functions of two

and three independent variables as follows :

Double Integral : Definition :

Let f (x, y) be a continuous and singlevalued function of two independent variables

R x, y defined on the region R of area A bounded

by a closed curve C. Let the region be divided into n sub-intervals in any manner (e.g. by drawing horizontal and vertical lines) into

sub-regions Rb R2, ... R11 of areas &AI> 'OA2, ...

0 •

'OA11• Let P (xr, Yr) be any point inside the r th Fig. (9.1) sub-region of area 'OAr. We now form the sum

f(xJ, YJ) M1 + f(xz , Yz) &Az + ····· + f(xn, Yn) '6 A n . n

i.e. L f(xn Yr) 0 A r r= I •... (1)

We now increase the number of sub-regions such that the area of each

sub-region becomes smaller and smaller. The limit of the sum (1), when it exits, as n tends to infinity and the area of each sub - interval tends to zero is

called the double integral off(x, y) over the region A and is denoted by

I JA f(x, y) dA .... (2)

Thus, n

lim L Jf f(x, y) dA = n -too _ f(Xr>Yr) 'OAr A OA-->or-1

.... (3)

Engineering Mathematics - II (9-2) Multiple Integrals 2. Evaluation of Double Integral :

The double integral as defined above can be evaluated by successive single integrations as follows :

y

If A is a region bounded by the curves y = f 1 (x) , y = h (x}, x =a, x = b. Then IJ f(x,y)dA= t

{J12(x) f(x,y)dy } dx A a j1 (x)

where the integration w.r.t. y is performed first by treating x as constant.

Consider the area bounded by two simple curves y = !1 (x) and y= h (x); and the ordinates at x = a, and x = b.

Now, consider a strip parallel to the y-axis. On this stripy varies from y = !1 (x) toy = h (x). If the strip is moved parallel to itself so that it will sweep the shown area then x varies from a to b.

Fig. (9.2) Now, it can be shown that

b { /2 (x) } IJ f( x, y)dA= J J f( x, y)dy dx A a f1 (x)

Note : The order of integration can be understood from context. 1 y

Ex.l: Find J0 J0 xye-x2 dx dy

1 1 1 2 Sol. : I = J o y o dy = -2 J o ( y e-Y -y ) dy

1 _

_ .!. _ Z.:] - 2 -2 2 0 4 4e 1 X

Ex. 2: Evaluate I I xy (x+ y) dy dx. 0 x2

Sol.: I= J j (x2y+xy2) dy dx = + xy3I dx 0 2 0 2 3 X �

= J( x4 + x4- x6 x7) dx= [�· xs- x7- xs l 0 2 3 2 3 6 5 14 24

Engineering Mathematics - II (9-3) Multiple Integrals

3 =-----=-

6 14 24 56

a .; Ex. 3 : Find I I 2 � x 2 -y 2 dx dy (S.U. 1988) 0 0

a Sol. : I = J J 2- y 2 ) -x 2 dx dy 0 0

= ( ["I 2 _ y 2 ) _ x 2 a2 -- i -I ( x d + �n

y 2 a2-i fa 1t 1t [ y3 ]a 1ta3

= o 2 · 2 dy = 4 a 2 y - 3 0 = 1 � 1 Ex. 4 : Find I I 7 7 dy dx 0 o 1+x-+y-

1 � 1 Sol. : I = I I 2 2 dy dx o o y +(l+x)

[ [ r 1 -1 y dx = o + x2 tan + x2 o 1t 1 dx

= 4 Jo = f [ tog(x+ =f log(1+v'2).

I dxdy Ex. 5 : Evaluate J I o o 1-x -y Sol. : Integrating w.r.t. x, treating y constant i.e. 1 -l = a 2 say

1 )12 dx dy 1 = Io Jo

I [ ] =

I 0 sin- I ( dy


= · =f� dy 1t ( ]I 1t = 4 yo=4.

I I x Ex. 6: Evaluate f_1 f0 x113 y

-I/2 (1-x-y) 112 dy dx Sol. : Let 1 -x =a. Then inner integral,

I-x a II= f y- 1/2 ( 1 -X- y) 1/2 = f y -I/2 (a- y) 1/2 dy 0 0 Now , put y =at, dy = adt.

I :.It =fo a-112t-112all2(1-t)ll2a dt

I�IYz = a f � t- 1/2 ( 1 - t) 1/2 dt = a B , = a 12 =a

- 1-1� . 2!� = na where a = 1 -x

1! 2

=fl xll3 1ta dx =.K. fl xii3(1 x)dx .. I - I 2 2 -1 - 1!. Jl [ 1/3- 4/3) = 1!. [x 4/3- X 7/3 ]I - 2 _ 1 X X dx 2 4/3 7/3 1 = I [ t (l)- t

Engineering Mathematics - II (9·5)

aVJ 1 [1t ] =f · ·xdx o 4

Multiple Integrals

7t aVJ x 1t [ ]a{J =4 f 2 2 dx"=-4 o x +a o 1t 1t = 4 [2a - a] = 4 a .

Ex. 8: Evaluate J1 dy dx 0 0 2 2 a +x +y I dy dx Sol. : I == J 0 J 0 (a 2 + x 2 ) + y 2 I I y

== tan-1 fo a' + x' ( + x'

_ _:JI dx (.:-o) dx _ _:JI - 4 0 " 4 - 4 0 2 =�[log( +x2 ) I ==�[log( I+�) -log (a) J

Ex. 9: Evaluate s; s; e-x2(1+ Y2>x dx dy.

Sol. :We put x2 (1 +

i) = t :. 2.:r (1 + y2) dx = dt When x = 0, t = 0; when x = oo, t = oo.

oo foo -1 df dy . _ e · 2 ··I- fo o 2(1 + y )

Engineering Mathematics - II (9-6)

foo 1 [ l]oo = -e dy 0 2(1 + y2) 0 foo 1 [

J = 0-1 dy 0 2(1 + y2)

= _!_ Joo �=.!.[tan I Y]oo 201+y2 2 0

1 7r 7r =-·-=-2 2 4

Ex. tO: Evaluate J; d.xJ�e-xay dy.

Multiple Integrals

Sol. : Since the limits are constants and integration with respect toy first then

with respect to x leads to a complicated integral, we reverse the order of integration.

1 t(l/a)-1 :.dx=-· dt

a ylla

W hen x = 0, t = 0; when x = oo, t = oo.

:.I= f dyf""e- 1· 1 ·t(lla) l dt 0 0 a·ylla = _!_ Jl y -1/ady. Joo e -1. t (lla)-ldt = _!_ J I y

a o 0 a o

= .!.[ y -(II a)+ 1 ] 1 ·I.!. a (-lla)+1 a

0

II I a =

la-1

[By definition of ln]

Ex. 11 : Sketch the area of integration and evaluate

2 I I 2xzyzdxdy 1

Sol.: We have

:. � = 2-y i.e. y- 2 =- .1-.

Engineering Mathematics - II (9-7) The curve is a parabola with vertex at (0, 2)

as shown in the figure. Andy varies from 1 to 2.

2 :. I = J J 2x 2 y 2 dx dy I y

2 :2 J J 2x2 y2dx dy I 0

Multiple Integrals

X

2 [ y 4 2 = 2 J 2y 2 L dy = - J y 2 (2 -y ) 3/2 dy

I 3 0 3 I Fig. (9.3)

Putting 2 -y = t, dy = - dt When y = 1 , t = 1; When y = 2, t = 0

4 0 :. I = 3 J - (2- t) 2 · t 312 dt I

4 I 4 I = 3 J (4 4t + t2) t3n dt = 3 J (4t 312- 4t 512 + t 112) dt 0 0 - .1_ [4 · 1_ t 512 _ 4 . 1_ t 7/2 + 1. t 912] I - 3 5 7 9 0 - ± [�- � + 1. ) = 856 - 3 5 7 9 945

Jl 1 x-y

Ex. 12 : Show that dx J dy 0 0

Jl

Jl x-y f. dy dx

o o (x + y)3 I I 2t-(x+ y)

Sol.: l.h.s.= J dx J ( + )3 dy 0 0 X y - dx - dy

I I[ 2x 1 ] - Jo Jo (x+y)3 (x+y)2

I [ -x I ]I

= J o (x + y)2 +

(x + y) o dx

f [ -X 1 1 1 ] =

o (x + 1 )2 +

x + I + x- x dx

Jl dx -[ I ]

I 1 1

= . o (x + 1 )2 x + 1 0 = -2 + I = 2

(S.U.1985)


r.h.s. - J 1 dy f x + y- 2y dx - 0 o (x + y)3

- dy --- dx 1 I [ 1 2y ] - Io Jo (x+ y)2 (x+y)3 1 [ 1 y ]

1 -- +-- dy = J o (x + y) (x + y)2 0

Il [ 1 y 1 1 l

= 0 -1 + y + (1 + y)2 + y - y dy

f 1 dy= [ 1+1y ] ol

= 21

-1=- 21 = 0 (1 + y)2 :. l.h.s. t= r.h.s.

Exercise -I

Evaluate the following integrals

7tl2 3(1 - cos t) 1. J0 I0 x2 sin t dxdt

Multiple Integrals

7tl2 I It 3. J 0 n/2 cos (x + y) dxdy J

2 X 1 4. 1 J 0 X 2 + y 2 dy dx

2 5. J0 J0 xydydx J

l X 6. 0 J0 (x2 + i> xdydx

Jl X

8. 0 I 0 ex+ Y dy dx (S.U. 2005) a

J I 2 9. X y dy dx (S.U. 2005) 10. xy dy dx 0 0 0 0

faJ2..[; 2 11. 0 0 X dy dx (S.U. 2003)

[ Ans. : (I) 9/4; (2) 4096/15; (3)- 2; (4) (1t log 2) /4; (5) 2/3;

(6) 4/15 ; (7) 64/3; (8) (e- 1)2; (9) a2 /15; (10) 2a4 /3; (11) a4 17 ].

3. Double Integral in Polar Co-cordinates : Let the region be defined by the curves r = fi (8), r = h (8), 8 = a,

8 = f3.


Then the double integral can be

evaluated as follows :

f11 f z f(r, 6)dr d6 { f (9) }

a fi (9) Fig. (9.4) where integration with respect to r is

performed first by treating e as constant. Ex. 1 : Evaluate

Sol.: I

1t/4 2 8 f f r dr de o o (l+r2)2

1tl4 I [ l 2 8 =f -2 d e o + r 0 __ I- de 1 1tl4 [ I 1 - 2 I o 1 + cos

1t/4 1 [ 1 ) = I 0 2 I -2 sec 2 e d e

I 1 1 1t 1 1 [ ]1t/4 [ ] = 2 e-2tane o =2 4-2 =s (1t-2).

1t/2 a cos e Ex. 2 : Evaluate f f r 2- r 2 dr · de 0 0

rt/2 1 [ acose

Sol.:/= fo -3 (a2-r2)3/2]o de 1 1t/2[ ] = - 3 I0 (a2-a2c0s2e)312_(a2)3/2 de 1 rt/2[ ] = -3 fo (a 2 sin 2 e )3/2_ (a 2)3/2 de 1 1t/2

=-3 J0 (a3sin3e-a3)de

3 1t/2 = f 0 (I -sin 3 e) de

J1tl2[ 1 l = � Jo 1 + 4 (sin 38 --3 sine ) de

(S.U.1985)


= [ e + ! (-t cos 30 + 3 cos e) ] :'2 a3 (1t 2) a3

=3 2-3 =(31t-4)18. rm rn

(a) To prove that B (m, n) =; lm + n (S.U. 1986, 87)

Proof : To prove the property we assume a property of double integral. If the limits of integration are constants and F (x) and (y) are respectively functions of x and y only then the double integral can be expressed as the product of two integrals i.e.

b d b d I I F (x) · (y) dx dy = I F ( x) dx ·I ( y) dy.

a c a c

Now consider second form of Gamma function. We have

oo 2 Ioo 2 rm·rn = 2· I0 e-x ·x2m-1·dx·2· o e-y ·y2n-1dy.

Ioo Ioo 2 2 = 4 0 0 e (x + Y ). x2m-l.y2n-1 dxdy

We now change the integral on r.h.s. to polar form by putting

X = r COS 0 , y = r sin 0 and dxdy = rd 0 dr. Since x and y change from 0 to oo the region is the entire first quandrant.

To span this region in polar coordinates r must vary from 0 to oo and e from 0 to n/2.

n/2 00 2 :. rm rn = 4 I 0 I 0 e

r • r 2m- I cos e 2m- I . r 2n-1 . sin e 2n- I . r drd e. = 4 I

n/2 r e r 2 . r 2 (m + n)- I cos e 2m- I . sin e 2n- I dr d e 0 0

00 n/2 = 2 . I e r 2 • r 2 (m + n ) -1 dr . 2 I cos e Zm - I . sin e 2n -1 . d e

0 0 But now the first is a Gamma function and the secend is a Beta function.

:. fm fn = im + n · B (m, n) rm rn :. B (m, n) = lm + n

I1 m n m! n! Ex.1: Prove that 0

x (1-x) dx = (m + n + 1)! where m, n are positive integers.

Sol. : By definition of the Beta function. 1

I0 xm(l-x)ndx=B(m+l,n+l)

(S.U. 1985, 86, 87)

•••• (i)


Now, by the above result

lm+T·In+T �(m+1,n+1)= lm+n+2

But lm+T = m!, In+ 1 = n! and lm+n+2 = (m+n+ 1)!

Combining (i), (ii) and (iii), we get

II m n m! n! o x (1-x) dx = (m +n + 1)!

Multiple Integrals

•••• (ii)

•••• (iii)

Exercise - II Evaluate the following integrals.

7t asina 1. I0 I0 drde

I7tl2

I a (1 +sin a)

3. 0 0 cos e dr de

n/4 2 a dr de 5. Io Io

In

I a (1 +cos a)

7. 0 0 drde

7t/2 a cos a 2. I o I o r sin e dr de

7tl2 a cos a 4. I0 I0 ? drde

7t/2 a sin 9 6. Io Io r drde

7tl2 2a cos a 8. I0 I0 ? sinedrde

7t a (I cos 8) 9. I0 I0 2n? sine dr de (S.U. 1998, 2004)

[ Ans. : (1) na2/4; (2) a2 /6 ; (3) 5a3/4; (4) 5a3/18; (5) (1t- 2)/8; a 3 [ 7t 2 ] 3 2a3 32na 3 [9 ) (6) 2 2- 3 ; (7) 4 1ta2; (8) T ; (9) B 2' 1 .]

4. Triple Integral The concept of double integral of a functionf(x, y) over a given region

in x-y plane can be extended a step further to define triple integral.

Consider a function f (x, y, z) defined over a finite region V of three dimensonal space. Let the region be sub-divided into n sub-intervals avl , 3V2, ... 3Vn. LetP (xro Yro Zr) be a point in the rth sub-interval. We now form the sum

n

� f(xr> Yr• Zr) 3Vr r=l


The limit of the above, when it exists, as n tends to infinity and the volume of each sub-region tends to zero is called the triple integral ofj(x�;z) over the region V and is denoted by

J J I v f (x, y, z) dV Thus, J J fv f(x, y, z) dV

n lim

= n�oo f(x ,y ,z)OV OA-.o r=l r r r r

(a) Evaluation of Triple Integral. The triple integral can be evaluated by successive single integrations as

follows : fx=b fy=z(X) rz=fl(x,y) x=a y= t (x) z=ft(x,y) f(x,y,z)dzdydx

where the integration w. r. t. z is performed first by treating x and y constants then the integration w.r.t. y is performed treating x constant and finally integratio,n is performed w.r.t.x

. Jl dz dy dx Ex. 1 : Fmd

(S.U. 1988) 00 0 1 2 2 2

-x -y -z

I 1 I� [ { z Sol. : I = 0 0 I dydx -X 2-y 2 0

(�) d dx=�JI[ o o 2 Y 2 o Yo = I� -X 2 . tit

I 7t [X � 1 . I ] 7t2 = 2 2 1 -x 2 + 2 sm - x 0 = 8 . J rc/2 I a sin 6 J (a 2- r 2) Ia Ex. 2 : Evaluate 0 0 0 r dz dr de

Sol.: I= J;/2 I;sin6 r[z]�az-r2)tadrde Irt/2 JasinO r = - (a 2-r2) drd9 o o a rt/2 1 [ 2 2 4 ] a sin 6 I - � _!._ d9 = o a 2 4 0

(S.U.1985)

!2

Engineering Mathematics - II (9-13) Multiple Integrals J

rc/2 1 [ a4 a4 ] = - -- sin2 e - - sin4 e de o a 2 4 a3 f

rt/2 a3 f

rc/2 = - sin2 e de - - sin4 e de 2 0 4 0

a3 1 1t a3 3 1 1t 5 3 = 2 '2 ·2--;r··.;r-2·2 = 64 1ta

f2 8-xLy2 Ex. 3 : Evaluate I I dz dy dx -2 X2+3y2

Sol.: 2 [

]8-xLy2 dydx I I z 2+ 3y2 I = -2 x

2 X 2)/2 = I J [2 ( 4 -X 2) - 4y 2] dy dx -2 [ ] = f � 2 2 ( 4-X 2) y- t ( y 3) dx = J 2 4 y'2 ( 4 -X 2)3/2 dx -2 3

Put X= 2 sin e ' dx = 2 cos e d e

(S.U.1986)

- 64 v'2 J rc/2 cos 4 e d e = 128 v'2 J rc/2 cos 4 e de - 3 -rc/2 3 0 = 128 .fi .i.L2: = s.fin

3 4 2 2 f2 J4-x J 3 Ex. 4 : Evaluate 0 x (3xl2)-y dz dy dx. J4-x 3 Sol.: I= x [ z ](Jx/2)-y dx dy

I: e-x( 3- +y) dxdy 2 3x y2 [

= fo Jy-2 y+T x dx

(S.U. 2003)

{ 3x 1 [ 2 2 J} = 3(4-x-x)-2(4-x-x)+2" (4-x) -x dx

Engln�erlng Mathematics - II (9-14)

= J:(l2 6x 6x+3x2 +8-4x)th = J:(20 16x+3x2)dx =[ 20x 8x2 +x2)! = 16.

JaJa-xJa-x-y 2 Ex� 5 : Evaluate 0 0 0 x dz dy dx. I a I a-x 2 a-x-y Sol.: I = 0 0 [x z]0 . dy dx

Multiple Integrals

(S.U.1987)

=I: I:-x x> dydx= I:[ x>y-x> r dx =! J;x 2 (a- x) 2 dx= � J; (a 2x 2 2ax 3+x 4 ) dx = .!. [a2£ 2a �+ £]a =� 2 3 4 3060

Ex. 6: Evaluate J; I; J: (yz + zx + xy) dz dy dx.

Sol.: I = I: +x +X}'< I dydx - ] dydx .. o 2 2 axy

2y2 + a 2xy + axy2 ]a dx 4 2 2 0

= J: [ a4 4 + a23 X+ a23 X ] tU = J: [ a4 4 + a3 X r tU [a4 x21 a aS aS 3 = 4x+a3 T o = 4 + T == 4 aS.

Jlog2 Jx Jx+y + +

Ex. 7 : Evaluate 0 0 0 ex Y z dz dy dx.

Jlog2 Ix [ ]x+y Sol.:/= 0 0 ex+y ez 0 dydx [

) ·' = 0 ex+y ex+y-} dydx

(S.U. 1�87, 90, 2000)


= J�og2 J:[e2 (x .. y)_e(x+y) ] dydx = e2x· - -ex·eY dx J

log 2 [ e 2y ]x 0 2 0

J Jog 2 [ e 4x 2; e 2x ) [ e 4x e 2x e 2x ] log 2

- --e :c- - +ex dx= - - - - - +ex - o 2 2 8 2 4 0

- i -� +2 J -(i-�-�+l) = i· Jl

Jz

Jx+::

Ex. 8 : Evaluate __ 1 0 x _ z (x + y + z) dy dx dz (S.U.1986, 88, 2003)

1 [ 2 lx+z Sol. : I = I J z (x + z) y + L dx dz -1 0 2 x-z =fl r[

Engineering Mathematics - II (9-16) IX h = 0 a2 3 (1 +cosS)2da

a2h Jx

= 3 0 (1 + 2 cos a+ cos2 0) d a

= I; ( 1 + 2 cos a + 1 + 2e ) de = (a+ 2 sin + J:

a2h 3n na2

Ex. 10: Evaluate I: I: I:x+2y ex+y+z dz dy dx. Sol.: I = J: I: I:x+ly ex+y (ez) dz dy dx

2 x [ ]2x+2y = Io Io ex+y ez o dydx = I: I: ex+y [ e2x+2y- eo J ely dx = I: I:[e3x+3y -ex+y Jdy dx

= dx

= J: (•" -•"]-•'(•' -•"J} x

= J:[ •'"'

-e:zx +e'

Multiple Integrals

Engineering Mathematics - II

e12 e6 e4 4 18 9 2 9

(9-17)

f2JyJx+y Ex. 11 : Evaluate (x + y + z)dz dx dy. 0 0 x-y

[ 2 ]x+y

Sol.: I= J: J; xz+yz+ dxdy x-y

J2 JY[ (x + y)2 = 0 0 x(x+y)+y(x+y)+ x(x-y)

Multiple Integrals

(x-y)2 ] -y(x-y)- dxdy

=JJ �+�+�+l+-+�+-2 Y[ x2 l

0 0 2 2

-x2+�-�+l--+�-- dxdy x2 l] 2 2

= J: J;[4�+2i1dxdy

= J:[2x2y+2�2I dy

= J:[2l+2 y3]dy

= J:(4y3)dy=[i1 =16

Exercise - Ill Evaluate the following integrals.

J1 dx J2 d I2 1. 0 0 Y 1 yz dz (S. U.1988, 94, 97) [ Ans. : 1.]

2. I� r: J� + y ex+y+'dz dydx

(S.U. 1987, 90) [ Ans.: t (ef!- 6e4 + 8e2-3)]

Jl J

1 J

l x 3. 0 y. 0 xdzdxdy (S. U. 1991, 93, 97) [ Ans. ]

IaJ

a xJ

a-x-y ( aS] 4. 0 0 0 (� + l + i> dz dydx (S. U.1995) Ans.: 20

Ie J

logyJ

ex [ 1 ] 5. 1 1 1 log z · dz dx dy (S. U. 2001, 05) Ans. : 4 (

e 2- 3e + 13)


1 1-x x+y 6. J0 J0 J0 ex dz dy dx (S.U. 1993, 95, 97, 2005) [Ans.: t 1

f4 J 2Vz . 4z -:-?-

7. dy dxdz 0 0 0 (S.U. 1997,2003, 04, 06) [Ans.: 81t] 1 1-x I -x- y 1

s.Jo Jo Jo (x+y+z+l)3 ·dz·dy·dx (S.U. 1996, 97)

[ Ans. : t ( log 2 - j) ]

[Ans.: a43]

(S.U. 1998) [ Ans. : a6 4 ]

[ 8 19 ] Ans. : 3 log 2 - 9

(S.U. 1998) [ Ans.: 3a7/4]

(S.U. 2002)

[ Ans.: e - 3 e

42a + e a- i]

5. Change of the order of Integration : y

y=d

X= cl> 1 (y)

It is sometimes more easy to evaluate

integral w.r.t. one variable than that w.r.t. the other

variable.

Now, in the region of§ 2 if we consider a strip parallel to the x-axis then on this strip

x changes from «


Sol. : The region of integration is given by x = y i.e. a line through the origin

y and x = 2 + 2y i.e. (x- 2)2 = 4 - 2y i.e. (x- 2)2 = 2 (2 - y) a parabola with vertex at (2, 2) and opening downwards, y = 0 i.e. , the x-axis andy= 2 i.e. a line parallel to the x-axis.

If we change the order of integration y varies from y = x to 2y = 4 - (x- 2) 2 i.e. y = 2x- � /2

x andx varies from 0 to 2.

Fig. (9.6) 2 2x x2/2 :.I= I0 Ix f(x,y)dxdy Ex. 2 : Change the order of integration

a y+a I t dx dy

Sol. : The region of integration is given by

x = 2 y 2 i.e. � + i = a2 (a circle). x = y +a ( a straight line), y = 0 (the x-axis), andy =a (a line parallel to the x-axis,). If we

have to change the order of integration then Fig. (9.7)

the region is to be divided into two parts, ABD and BCD.

In the region ABD, y varies from L x 2 to a and x varies from 0 to a. In the region BCD, y varies from x-a to a and x varies from a to 2a.

a 2a a Jx-a f(x,y)dydx

Ex. 3 : Change the order of integration a a21x

f0 t f(x, y) dy dx y

X

Fig.(9.8)

Sol. : The region of integration is given by y = x, 2

the line through the origin, y =ax i.e. xy = a2, a rectangular hyperbola, x = 0, the y-axis and x = a, a line parallel to the y-axis.

If we have to change the order of integration, the region is to be divided into two parts OAB and above the line AB.

In the region OAB, x varies from x = 0 to x = y andy varies from 0 toy= a. In the region above AB, x varies from x = 0 to x = a2 I y andy varies from a to oo.


JooJ

a21y :.1= of(x,y)dxdy+ a 0 f(x,y)dxdy

Ex. 4 : Change the order of integration 1tl2 2a cos a

J J f(r, 8) dr de y

0 0 Sol. : Since r = 2a cos e, ? = 2ar cos e i.e. :l- + l = 2 ax i.e. (x- a)2 + l = a2 is a circle, the region of integration is the upper half of this

circle. e varies from 0 to 7t/2

To change the order of integration we Fig. (9.9)

consider a circular strip as shown in the figure. On this strip e varies from 0 to cos-1 (r/2a) and this strip moves from r = 0 to 2a.

1 I • -J

2aJ

cos (r2a)/ e dad .. I - 0 0 (r, ) r.

Ex. 5 : Change the order of integration

JaJ

2a-x 2 f (x,y) dy dx. 0 x Ia

2 Sol. : The region of integration is given by y = xa i.e. a parabola, y = 2a- x

y i.e. x + y = 2a i.e. a line, x = 0 i.e. they-axis and x = a i.e. a line parallel to the y-axis. If we have to change the order of

integraion, the region is divided into two parts

OACand CAB.

In the region OAC, x varies from 0 to x yay and y varies from 0 to a ( The point of

2 Fig. (9.10) intersection A of the parabola y = xa and the

line x + y = 2a is A (a, a)). In the region CAB, x varies from 0 to 2a-y, andy varies from a to 2a. Hence,

a Vi 2a 2a - y :.1 = J0 J0 f(x,y)dxdy + Ja J0 f(x,y)dxdy

Ex. 6 : Change the order of integration a cos a -x2

J0 J f(x,y)dy dx. x tan a Sol.: The region of integration is given by y 2_x 2 i.e. :l- + l = a2, a circle, y = x tan a. , a straight line, x = 0 the y-axis and x = a cos a. , a line parallel to the y-axis.


If we have to change the order of integration,

the region is divided into two parts OAB and BAC.

At the point of intersection A we have y ==a cos a. tan a.== a sin a.

Multiple Integrals

ij

8 l'O II

I h . OA . f 0

n t e region B, x vanes rom to

y I tan a and then y varies 0 to y == a sin a..

In the region ABC, x varies from 0 to a 2 y 2 0 Fig. (9.11) and then y varies from y == a sin a. to y == a.

fasinu Jycota Ja -l :.1= f(x,y)dxdy+ . f(x,y)dxdy. a 0 asmu 0


f�a f� f (x,y) dy dx. Sol.: The region of integration is given by y == x2 i.e. i == 2ax-x2 i.e.

(x- a)2 + i == a2 i.e. a circle with centre at {a, 0) and radius a ; y = i.e. l = 2ax i.e. a parabola; x = 0 i.e. the y-axis; and x == 2a, the

line parallel to the y-axis .

If we have to change the order o f

integration , the region i s divided into three parts

ABC, EBD and OCE.

Fig. (9.12) We first note the coordinates of the points of intersection of the curves. Cis (a, a), A is (2a, 0), B is (2a, a), Dis 2a),

E is (a, Vf· a). We also note that since (x- a)2 + i = a2 i.e. x =a± 2- y 2 forthe arc OC, x =a- 2-y 2 and for the arc CA, x =a+ 2 y 2 .

In the region A CB, x varies x = a + a 2 - y 2 to x = 2a andy varies from 0 to a.

In the region OEC, x varies from x = i I 2a to x =a- 2-y 2 andy varies 0 to a.

In the region EBD, x varies from x = l I 2a to x = 2a andy varies from a to 2a. Hence

a 2a fa fa -y2

:. I= f f 2 f(x,y)dxdy 0 a + 0 y I 2a 2a 2a

+f f2 f(x,y}drdy a y I 2a



0 f (x,y) dy dx.

Sol. : The region of integration is given by y = L x 2 i.e. x2 + l = a2, a circle, y = x + 3a, a straight line, x = 0, the y axis and x = a, a line parallel to the y-axis.

If we have to change the order of integration

Multiple Integrals

F(a, 4a)

D(a, 3a)

Fig. (9.13)

the region is divided into three parts ACB, CDEB and DEF.

In the region ACB, x varies a 2-y 2 to a andy varies from 0 to a.

In the region CDEB, x varies from 0 to a andy varies from a to 3a. In the region DEF, x varies from y - 3a to a and y varies from 3a to 4a

Hence,

a a 3a a :. I= f f f(x,y)dxdy 0 a 0

4a a + f f f(x,y) dx dy 3a y- 3a

Ex. 9 : Change the order of integration of k

J J f (x,y) dx dy. 0 Sol. : The region of integration is given by

x= - k + k 2-y2 i.e. (x + k) 2 + l =�i.e. the circle with centre(- k, 0) and radius k; x = k +

k2-y2 i.e. (x- k)2 + l = k2 i.e. the circle with centre (k, 0) and radius k; y = 0, the

x-axis and y = k, the line parallel to the x-axis.

y

(- k, 0) (k, 0) (2k, 0)

Fig. (9.14)

Since, x= - k + k2-y2 is the arc OHB andx= k+ k2-y2 is the arc EGD the region of integration, is BCDGEF OHB.

If we have to change the order of integration, the region is split into

three parts BCOH, COFD, DFEG.

In the region BHOC, y varies from y = + k ) 2 toy= k and x varies from x = - k to x = 0.

In the region COFD, which is a square of side k, y varies from 0 to k and x varies from 0 to k.


In the region DFEG, y varies from y = 0 to y = k 2- ( x- k) 2 and x varies from x = k to x = 2k. Hence,

:.I= J0 2 -k k -(x+k) o 2 2 J 2kJ k -(

x-k ) f ( d dx + k 0 x,y) y ·

J a J 2a-x E.X. 10: Change the order of integration 0 -x2 f (x,y) dy dx. Sol. : The region of integration is given by y = a2 - i.e. x2 + i = a2, a

. I 2a . 2a . h 1· ctrc e, y = -x t.e. x + y = , a strrug t me, x = 0, they-axis and x = a, a line parallel to the 28) y-axis. Thus, the region of integration is ABCD.

When we change the order of integration,

the region is split into two parts DAB and DBC.

In the region DAB, x varies from x = i tox=aandyvaries fromO toa.

x=a In the region DBC, x varies from x = 0 to

x = 2a -y andy varies from y = a to y = 2a. Fig. (9.15)

J a J a J 2a J 2a-y :.1 = f (x,y)dxdy+ f (x,y)dx dy. 0 a 0

E.X. 11 : Change the order of integration J� J f (x, y) dy dx.

Sol. : The region of integration is given by y = x, i.e. a straight line; y = 1 + 1-x 2 i.e. (y-1)2= 1-xZ i.e.xZ+(y-1)2= 1 i.e. a circle with centre(O, 1)andradius

y 1; x = 0, the y-axis and x = 1, a line parallel to the y-axis.

The line y = x will intersect the circle

y= 1 + wherex= 1 + i.e. (x- 1 i = 1 -x2 i.e. xZ - 2x + 1 = 1 - xZ :. a2- 2x = 0 :. 2x (X-1) = 0 :. X= 0, X= 1.

X

Fig. (9.16) From y = x, when x = 0, y = 0 and when x =

1, y = 1. Hence, the points of intersection are 0 (0, 0) and A (1, 1 ). When we change the order of integration the region is split into two parts, OAC and CAB.


In the region OAC, x varies from x = 0 to x = y and then y varies from y=Otoy= 1.

In the region G'AB, x varies from x = 0 to x given by x2 + (y- 1)2 = 1

i.e. x2 + l � 2y = 0 i.e. x = y 2 . Then y varies from y = 1 toy= 2.

J1Jy

:.!= 0 0 f(x,y)dxdy+ 1 0 · f(x,y)dxdy.

h f. . J2 J(6-x)/2

f d Ex. 12 : Change t e order o mtegl'6ltiOn 2 (x, y) y dx. 0 (x +4)/4 2

Sol. :The region is bounded by y = x + 4

i.e. 4y- 4 = x2 i.e. x2 = 4(y- 1), 4

a parabola with vertex at (0, 1) and opening upwards, y = 6 x i.e. x + 2y = 6 2

is a straight line, x = 0 is they-axis and x = 2 is a line parallel to they-axis. To change the order of integration,

the region is divided into two parts ABC and CBD.

In the region ABC, consider a strip parallel to the x-axis. On this strip x varies

from x = 0 to x = y- 1 . Then y varies from y = 1 to y = 2.

In the region CBD, consider a strip parallel to the x-axis. On this strip x varies from x = 0 to x = 6- 2y. Then y varies from y = 1 toy= 3.

J2J2F-i

J3J6-2y

:.

I= 1 0 f (x, y) dx dy + 2 0 f (x, y) dx dy. Ex. 13 : Change the order of integration

J4s9-

y 0 y/2 f (x, y) dx dy. Sol. : The region of integration is given by

x = y/2 i.e. y = 2x, a straight line through the origin, x = 9 -y i.e. x + y = 9 a straight line;

y = 0, the x-axis and y= 4, a straight line parallel

to the x-axis.

The two lines y = 2x and x + y = 9 intersect at 9 -x = 2x i.e. x = 3, y = 2x =6. The line y = 4 intersects y = 2x in C (2, 4) and the line x + y = 9 in (5, 4). Thus, the required region is OABC.

Engineering Mathematics • II (9-25) Multiple Integrals

If we change the order, the region is split into there parts, ODC,.DCBE, EBA.

In the region ODC, y varies from y = 0 toy= x and x-varies from x = 0 tox = 2.

In the region DEBC, y-varies from y = 0 toy= 4 and x-varies from x = 2 to x = 5.

In the region EAB, y-varies from y = 0 to y = 9 - x and x-varies from x= 5 tox=9.

f2J2x

j5J4 J9J9-x :.1= 0 0 f(x,y)dydx+ 2 0f(x,y)dydx+ 5 0 f(x,y)dydx.

fa/2 Jx-(x2 Ia)

Ex. 14: Change the order of integration of · 2 f (x, y) dy dx. 0 x Ia Sol. : Here, the region of integration is bounded

by y = �Ia; i.e. � = ay, a parabola; y =x-(�/a) i.e. ay = ax-x2 . 2 z.e.x -ax=-ay i.e. [x-(a/2)]2 =-a [y-(a/4 )] , a parabola

with vertex at [(a/2), (a/4)] and openin g

downwards; x = 0, the y-axis and x = a/2, a line parallel to they-axis. The two parabolas intersect

at�=ay=ax-x2 i.e. 2x2-x=O. :. x(2x-a)=O :. x = 0 or x = a/2 . When x = 0, y = 0.

a a2 a When x= 2 ,y= 4a =4.

Fig. (9.19)

Hence, the two parabola intersect at (0, 0) and [(a/2), (a/4)]. Now, the parabolay =.t-(�Ia) can be written as ay =ax-� i.e. �-ax=- ay i.e. [x- (a/2)]2 = (a2/4)-ay.

The parabola y = �Ia can the be written as � = ay i.e. x = ± ..[;.

Thus, on the strip parallel to the x-axis, x varies from x = !!. - a 2 - ay . 2 4

Engineering Mathematics - II (9·26) Multiple Integrals

(Note this carefully from the graph) to x = ,J;Y and then y varies from y = 0 to y =a/4.

Ja/4J,Ja; :.1 " (x, y) dx dy. 0 14)-ay

Ja Ex.15 :.Change the order of integration in 0 12! (x, y) dy dx. Sol. : The region of integration is bounded

-x2 by

2 . 42 2 2. 42 2 1.e. y = a -x 1.e. + y =a

2 2 i.e. � + = 1 which is an ellipse; by

a 2 a2 I 4 y = 2 -x 2 i.e. y2 = a2 x2 i.e. � + l = a2 which is a circle; x = 0 y-axis and x = a, the line parallel to the y-axis.

Fig. (9.20)

If we change the order of integration, the region is split into two parts CABandCBD.

In CAB, consider a strip parallel to the x-axis. On this strip x varies from

x = 2 -4 y 2 to x = 2 - y 2 and then y varies from y = 0 to y = a/2. In CBD, consider a strip parallel to the x-axis. On this strip :c varies from

x = 0 to x = 2 - y 2 and then y varies from y = a/2 to y = a.

a/2 a :.1= J J f(x,y)dxdy+ J J f(x,y)dxdy. 0 a/2 0

Ex. 16 : Change the order of integration I J2

(I+Ji="Y) f (x, y) dx dy. 2y

Sol. : The region of integration is given by x = 2y, a line through the origin;

Fig. (9.21) x = 2 (1 + y ) i.e. (x- 2) = 2 y


i.e. (x- 2)2 = 4 (1- y) i.e.� -4x=- 4y i.e. y=x - (�/4) i.e. y =- i (� - 4x) :. y- I = - .!. (x2- 4x + 4) i.e. y - 1 = - .!. (x- 2)2 which is a parabola with

4 4 vertex at (2, 1) and opening downwards; y = 0, the x-axis andy= l, a line parallel to the y-axis.

When we change the order of integration, the region of integration is

divided into two parts, OAB and ABC. In the region OAB, consider a strip parallel to they-axis. On this strip, y

varies from y = 0 to y = x/2. Then x varies from x = 0 to x = 2. In the region ABC, y varies from y = 0 to y= x- (x2/4) and then x varies

from x = 2 to x = 4. Hence,

J4Jx (x2/4)

I= f(x,y)dydx+ 2 0 f(x,y)dydx.

Exercise - IV

Change the order of following integrals

1. ( ( f(x, y) dy dx [ Ans.: ( ( f(x, y) dx dy] 0 y 0 0

(S�e fig. 7.3 page (7.2)) 4a 2..

2.J J f(x,y)dydx 0 X (See fig. 7.25 page (7.6))

4a y [ Ans.: J J 2!4a f(x, y) dx dy] 0 y

I I+� 3.J J f(x,y)dy dx

I I [Ans. :J f f(x, y) dx dy]

0 2y-y2 0 0 (See fig. 7.11 page (7 .3), b = 1)

2 4 4. 0 f (x,y) dx dy [ Ans. :J J f(x, y) dy dx] Y 0 x2-4y

(Fig. to fig. 9.21 page (9.26)) I llx ·

5. J J f (x, y) dy dx 0 X

(See fig. 9.39 page (9.42))

oo 1/y I y [Ans.:J J f(x,y)dy dx+J J f(x,y)dy dx]

.,[i 6. f f(x, y) dy d;<

x21a ·

I 0 0 0 a .yay

[ Ans. :J J . f(x, y) dx dy] · 0 y21a

(See fig. 10.43 page (10.27))


2 4-x 7. I f f(x, y) dy dx (See fig. 9.15 page (9.23), a= 2)

0 2 2

4 4-y

[ Ans.: I I f(x, y) dx dy +I I f(x, y) dx dy] 0 2 0

3 l+x s.I I t

.... l:l•ucc• JIIY tYinema"tiCS - 11 (9-29) Multiple Integrals

Type II

f i J4 x2 Ex. 1 :Change the order of integration and evaluate 0 4Y e dx dy.

Sol. : The region of integration is x = 4y, a line through the origin; x = 4, a line parallel to the y-axis; y = 0, the x-axis; andy= 1, the line parallel to the x-axis. Thus, the region is the triangle OAB.

(S.U. 2007)

To change the order of integration consider a strip parallel to the y-axis. On this strip y varies (4,0) from y = 0 to y = x/4. Then, x varies from x = 0 to Fig. (9.22) x=4.

3

J4 Jx/4 x2 :. I = 0 0 e dy dx

f4 x2 [ ]x/4 dx J4 x2 X dx == e y = e ·-o 0 0 4 = �[ I= He" -1]

(For integration put x2 = t)

Ex. 2 : Change the order of integration and evaluate.

I 5 I 2+x 0 2 x dy dx

Sol. : The region of integration is ABC. When the order of integration is changed the region

splits into two.

In the region ABD, x varies from 2 - y to 5 and y varies from - 3 to 2. In the region ADC, x varies from y - 2 to 5 and y varies from 2 to 7. Hence,

2 5 7 5 :. I= I I dxdy + I I dxdy

-3 2-y 2 y-2 2 5 7 5

= J [x]2 ydy+ I [x]y-2dy -3 2

(S.U. 19.85)

B(S-3) Fig. (9.23)


2 7 I= I (3 + y) dy +I (7- y) dy 3 2 [ y2 ] 2 [ y2 ] 7 = 3y+2 _/ 7y-2 2

= [ 17- ] + [ 37 - �] = 25 Ex. 3 : Change the order of integration and evaluate.

JaJx dydx o Sol. : The given region of integration is the triangle OAB bounded by y = 0 i. e. the x-axis, the line y = x and the line x = a. When the order of integration is reversed, first x varies from y to a and then y varies fromO to a.

Multiple Integrals

(S.U. 1985, 90) y B(a,a)

Fig. (9.24) a a dxdy :. I=Jo JY [ Complete the square on x ] a a dxdy =Ioiy j[ [ a-y ] 2 [ a+y ] 2 ] (y+a) - x-r a+ y ·1a a 1 . 1 = fo (y+ a) sm

Y

dy

= ( (y a) [ sin I ( 1) - sin I ( - 1 )] dy = ( a) [ � + �] dy

a dy = 1t I 0 (y +a) = 1t [log (y +a)]� = 1t [ log 2a -log a ]

= 1t log 2 Ex. 4 : Change the order of integration and evaluate

eY o o (eY -y2 (S.U. 1985, 87, 2002)


Sol. : The limits of y are 0 and � and those of x are 0 and 1 . We, therefore, draw the curve

y = � i.e. the upper-half of the circle x2 + i = 1. The region of integration is OAB.

X

To change the order of integration we

consider a strip parallel to x-axis. Now x varies from Fig. (9.25) 0 to � andy varies from 0 to 1.

1 � eY · J - J J . dxdy · · - o o ( e Y + 1) (1 - y 2)- x 2

l y [ ]� - sin 1 x

d - Jo (eY+ 1) � 0 Y

Jl eY 1t 1t 1 = -- ·- · dy = - [ log (e Y + I) ] o eY+ 1 2 2 o 1t [e+1)

JaJ.[; x

Ex. 5 : Change the order of integration and evaluate 0 y 2 2 dx dy.

X +y

Sol.: The region of integration is given by x = y, a y

line through the origin; x = .[; i.e. 2 = ay, a parabola through the origin and opening upwards,

y = 0, the x-axis and y = a, a line parallel to the y-axis.

In the region of integration, consider a strip Fig. (9.26)

parallel to the y-axis. On this stripy varies from y = 2ta toy = x and then x varies from x = 0 to x = a.

:. I= r J\ X 2 d y dx = r[�tan -l L]x

dx 0 x /ax2+y 0 x x x2/a = s:[ tan-11-tan -l ; ]dx = s:( �-tan -1;) dx

1 1 1r · a -1 � _ X· ·- dx =4- [xJ.- [x·tan ( .) J l+(x'Ja') a o


= na [xtan 1�-�log(x2 +az)] a 4 a 2 0 =na -[an -�log(2a2)-0+�loga2]=�log2. 4 4 2 2 2

3 Ex. 6 : Change the order of integration and evaluate J J 2 dx dy

0 y 19

(S.U.1984, 86)

Sol.: The limits for x are il9 and -y2and those for y are 0 and 3. We,

../10.

y therefore , draw the curves x = il9 i.e. the parabolai=9xandx = -y2 i.e. the upper half of the circle XZ. + i = 10. The region of integration is thus OACD. Solving the equations

C( ..fiO.O) x i = 9x and XZ. + y2 = 10 we get the points of intersection A (1, 3) and B (1,- 3).

Now, to change the order, if we consider a strip parallel to the y-axis, the region has to be divided into two parts ODA and ADC.

In region ODA, y varies from 0 to 3fi and x varies from 0 to 1.

In the region ADC, y varies from 0 to x varies from 1 to

:. 1= J�s:.rx dydx+JlJiO dydx 1 3Vx 1

Now,/1=f0 [Y]0 dx=f0 3-Vxdx

= 3 [ 1x312 ] :=2 {10

l =f 2 1 0 1


= [ i 10 sin -1 ..).[10 2 .f I =5·�

3 2 -2 -5sin -1 I

110

· · I= 1 + ["

1

1 I2 - a= n .

2 {3,-2 -

sm - 1 I ] = s· -I 3 3

= 1 + 5 · [� -sin - lko) = + 5 sin -I [ }o) Ex. 7 : Change the order of integration and evaluate

I X I dx I 2 2 dy (S.D. 1987, 88, 97, 2004) 0 X X +y

Sol. : The limits for y are x and -x 2 , and those for x are 0 and I. We, therefore, draw the curves y = x which is a straight y line andy= -x 2 which is the upper half of the circle� + J = 2. The region of integration is OACD. Solving the equationsy=x andx2 + J = 2 we get the points of intersection A(1,1) and •• x B (-1,-1).

If we consider the strip parallel to the x-axis, the region has to be divided into two parts OAD and ADC. Fig. (9.28)

In the region ODA, x varies from 0 toy andy varies from 0 to 1. andyvariesfrom 1 to.fi.

11

1y X J,Ji X .. I= 0dy 0

dx+ 1 dy 0

dx

x2+y2 x2+y2

I [ ]y I Now I1 = J dy x 2 + y 2 = I ('v2 · y- y) dy 0 0 0 [y 2] I 1 1 1 = (-fi- I) 2 o = 2 (-fi- 1)= -¥2 - 2

Vi And h = J dy [ x 2 + y 2

]

I 0 Vi

= J (-¥2 -y)dy I


[ y2 ]Yz 3 = ..y-2 I =2-../2

:. I l = I1 + !2 = l - ..

Ex. 8 : Change the order of integration and evaluate

faJx eY o o -x)(x- y)

dy dx.

Sol. : Since integration with respect to y is complicated

Multiple Integrals

we change the order of integration. x = a The limits for y are y = 0 toy= x and for x are x = 0

to x =a. The region of integration is the triangle OAB. Fig. (9.29) Now, consider a strip parallel to the x-axis. On this strip x varies from x

= y to x =a and for the stripy varies from y = 0 toy= a.

faJa eY :. I= d.x dy 0

fa J a eY = dxdy 0 Y -(a+ y)x]

= J: J; ( )'

)' dx

a- y a+ y -- - x-2 2

fa y [ . -1 I = e �n 'Y o (a-y)/2 y = J: eY [ �-( -�)] dy = nJ; eYdy =n[eY J: =n(ea - 1) .

Aliter : The integral can also be evaluated by putting x -y = ? i.e. X= y + r :. dx = 2t dt.

Whenx=y,t=O; whenx=a,t=


. _ Ja J � e y • 1 21 dt dy .. !- o o a 2dt = J eYdyJ 2 o 0 (a y) t

dy = 2 J; e Y · [ � - 0] dy = 1C J; e Y dy =1C [ea-1] .

Ex. 9 : Change the order of integration and hence evaluate

o dx dy. (S.U. 1986, 87) Sol. : The limits for x are 2 - 4 y 2 and 2 + 4 -y 2 and those for y are

y 0 and 2. We, therefor e, draw the curves x = 2 - 4 y 2 and x = 2 + 4 - y 2 i.e. the circle (x -2)2 + i = 4 which has the centre at (2, 0) and radius 2.

X Now, x = 2- is the arc OA and

x = 2 + 4 y 2 is the arc AB. Since y = 0 is the x-axis and y = 2 is the line AD the region of integration is the semi-circle OAB.

8(4.0)

Fig. (9.30)

To change the order of integration we consider a strip parallel to the

y-axis. On this strip y varies from 0 to 4 ( x 2)2. This strip sweeps the region when x varies from 0 to 4.

J4 J4 :.!= dyd'C= [y] dx 0 0 0 0 4

= J - (x 2) 2 · dx 0 =

[ (x 2) 4 - (x - 2) 2 + sin - 1 ( x 2 ) ] :


= [ 2 . ) - [- 2 . ) = 2 1t. Ex. 10: Change the order of integration and evaluate

r r ydxdy

o (S.U. 1991, 2005) Sol. : In the given region x varies from ila (i. e. i =ax) to x = y andy varies from 0 to a. Thus, the region is OAB bounded by the line x = y and the arc of the parabola i = ax. When the order is reversed y varies from x to and .,faX and x varies from 0 to a.

. I = ( I..[tU Fig. (9.31)

· • o x a I .,fOX =- Io (a-x) dx

=I: Now, put X= a cos2 e :. dx =- 2a cos e sin e de

ITCJ2 cos e .

:. I = 0 sine . 2a Stn e cos e de TCJ2

= a I 2 cos2 e de 0 Tt/Z [ sin 2 e ] 1t!Z 1t =ai0 (I +cos26)d6=a 8+ 0 =za


Il cos-lx dxd o o

y

Sol. : The limits for yare 0 and I and for x are 0 and x = � i. e. � + i =I. Hence, the region of integration is the first quadrant of the circle x2+/= I.

Now, if we change the order of integration, y varies from 0 to � and x varies from 0 to I. Hence,

(S.U. 2003)


:. I = I I Iv'i7 cos-lx dydx o o � -x2-y2

II cos-1x [ . ]

v'i7

= sm -I dx � o n 1 cos 1x dx = 2 I 1 2 [ Put cos 1 x = t, 2 = - dt ) 0 -X I -X

1t 0 1[ rr/2 = - - I t dt = - I t dt 2 rtt2 2 o

1t [ t2 ] n/2 1[3 =2 2 0 =16

Ex. 12 : Change the order of integration and evaluate faJ2a-x

2 xydydx. 0 x Ia (S.U.1985,86,88,92,94,2005) Sol. : The limits for yare �Ia and 2a- x and those for x are 0 and a. We, therefore, draw the curves

y = .ila i.e. the parabola � = ay and y = 2a- x i.e. the straight line x + y = 2a. The region of integration is OACB. Solving the equations x2 = ay and x + y = 2a we get points of intersection as A (a, a) and D (- 2a, 4a).

X Fig. (9.33)

Now, to change the order if we consider a strip parallel to the x-axis, the region has to be divided into two parts OAB and BCA.

In the region OBA, x varies from 0 to .. andy varies from 0 to a. In the region BCA, x varies from 0 to 2a-y andy varies from a to 2a.

a ..[;iY 2a 2a -y :. I = I I xydxdy+ I I xydxdy 0 0 a 0

Ia [x2 ]..fciY

Ia a 2 a [v3 ] a a4 Now, I = y -· dx = - y dy = - � = -l o 2 o o 2 2 3 o 6

2a [ 2 ] 2a-y 2a 1 1z = Ia y x2 0 dy = t 2 ( 4a 2 y 4ay 2+ y 3 ) dy

1 [ 4a 4 ] 2a 5a 4 _ _ 2a 2 y 2 __ y 3+ X. = _ - 2 3 4 a 24

. I - a 4 + 5a 4 = l. a 4 .. - 6 24 8

Mathematics - II (9-38) Multiple Integrals

Ex.13 :Evaluate ( dy xy log (x+a) dx o o (x-a)2

(S.U. 1985, 87, 98, 2006) Sol. : In the given form, the integral is to be evaluated w. r. t. x first, which y obviously is complicated. We, therefore, change B the order of integration.

Fig. (9.34)

The limits for x are 0 and a- a 2-y 2and for y are 0 and a. We draw the curves x = 0 i. e.

x

they-axis and x = a - a 2 -y 2 i. e. the left half of the circle (x- a)2 + i = a2.

It is clear that A is (a, a) andB is (0, a). The region of integration is OAB. Now, to change the order of integration if we consider a strip parallel to

they-axis, y varies from a 2 - ( x- a) 2 i. e . 2ax- x 2 to a and x varies from 0 to a.

. _ a dx a xy log ( x + a) d .. I -I (x-a)2 Y 0 _ ( [�]a dx -

o 2 a X log ( x +a) 1 I a = I 2( )2 [ a 2-2ax + x 2 ] dx = 2 x log ( x +a) dx 0 x-a o 1 [ x 2 I x 2 ]a = 2 log ( x +a) . 2 - 2( x +a) dx o 1 [ x 2 I I x 2 -a + a 2 ] a = 2 2 ·log(x +a)-2 (x+a) ·dx o l [ x2 I a 2 dx]a = 2 2 · log ( x +a)-2 I ( x-a ) dx -2 I x +a 0 l [ x 2 1 [ x 2 ) a2 ]a =2 2·Iog(x +a) 2 T ax Tlog(x+a) 0 l [a2 a2 a2 a2 ] = 2 2 log 2a + 4 -2 log 2a + 2 log a 1 [ a 2 a 2 ] a 2 = 2 4 + 2 log a = 8 [ 1 + 2 log a]

Ex. 14 : Change the order of integration and evaluate 2 f2f2 y d dx 2 Y · y -4x


Sol. : Here, the region of is bounded by y = .[2; i.e. / = 2x a parabola; y = 2, a line parallel to the x-axis; x = 0, the y-axis and x = 2, a line parallel to they-axis.

If we consider a strip parallel to the x-axis, on

this strip x varies x = 0 to x = //2 and then y varies from y=O toy=2.

J2 Ji /2 idx dy . I-·· - o o y - X 1 t Ji l2 i = 2 o � I 2)2 -x2 dx dy

l 12 � H:" I = � J� i [sin-11-sin-1 0 ]dy

= Ex. 15: Change the order of integration and evaluate

Jl/2 1+x2 dxdy. o o

Fig. (9.35)

Sol. : Here, the region of integration is bounded by x = 0 i.e. they-axis.

:.x2 =l-4y2

2 :. x2 = l.

Fig. (9.36)

X (1/4) It is an ellipse with semi-major axis 1 and semi

minor axis l/2. The line y = 0 is the x-axis and the line y = 112 is a tangent of the ellipse parallel to the x-axis.

Thus, the region of integration is the first quadrant of the above ellipse.

To change the order of integration, consider a strip parallel to they-axis.

On this stripy varies from y = 0 to y = � . Then x varies from 0 to 1. 2

Engineering Mathematics - II (9-40) :.I =

l+x2. dy dx 0 o y2

=J� l+x2 y 0 J1 l+x2 [. _1 1 . _10Jdx = · sm --sm o 2

= !E. f l+x2 dx 6 o

To find the integral, put X= sin e, dx = cos e de. :. I =

!E.Jn/2 l+sin 2 0 ·cosO dO 6 o cosO = n Jn12 (1 +sin 2 0) dO = :=.[{o}n12 +.!. :=.] 6 0 6 0 2 2 =�[%+�] = '/C82.

Multiple Integrals

Ex. 16 : Change the order of integration and evaluate JJ x 2 dx dy where R is the region in the first quadrant bounded by xy = 16, x � 8, y = 0 andy = x. Sol. : The region of integration is bounded by xy = 16, a rectangular hyperbola; x = 8, a line parallel to the yaxis; y = 0, the x-axis andy= x, a line passing through the origin. The region is OBCDO.

If we change the order of integration, the region is split into two parts, OAB and ABCD.

Now, consider a strip in the region OAB, parallel Fig. (9.37) to the y-axis. On this strip y-varies from y = 0 to y = x. Then x varies from x= 0 tox=4.

Also consider a strip in the region ABCD, parallel to they-axis. On this strip, y varies from y = 0 toy = 16/x. Then x varies from x = 4 to x = 8.

2 2 :.1= dydx+ x dydx

.. 11!:f111'1'CJI UI!: IWIGLJICIIIdLu.;;:t - II \l:I�41J Multiple Integrals

= J: x2 [Y ]� dx+ J: x2 [Y ]�6/x dx 4 8 x4 · 8

[ ]

4 = Jox3dx+ J416xdx= 4 o + [

8x2 ]4 = [64- 0] + [512 -128] = 448.


Jl dxfco e-YyX logy dy. x=O y=l Sol. : Since all the limits of integration are constants, the order can be changed without taking the help of a diagram.

:. I= J= e-YJogydy· J 1 yxdx y=l x=O =J e-Y logy -- dy 00 y x l y=l logy

= Jco e-Y(y-1) dy = Jco (y e-y- e-Y) dy y=l I

= [ -ye-Y -e-y +e-Y ]� = [ -ye-Y ]� =e-1 = �.

Ex. 18 : Express as a single integral and evaluate

IIJ.{y J 3 J l I = d y dx + dy dx. 0 I -1 Sol. : Let I= h + h·

Now, for h. the limits are x =- .{Y and x = .[Y i.e. x2 = y, a parabola with vertex at the origin and opening upwards. The limits for y are y = 0 toy= 1. The region is DAB.

For I2, x varies from x =- 1 to x = 1 and

=3

y variesfromy= 1 toy=3. The region isABCD. We have, to sweep both the regions i.e. the

region OADCBO. Fig. (9.38)


Now; consider a strip parallel to they-axis extending from the parabola

to the line CD.

On this stripy varies from y = 2- toy= 3. To sweep the whole area the strip has to move from x =- 1 to x = 1.

:.I=JI J\ dydx=JI [y J\ dx I X I X = fi[3-x2]dx=2J�(3-x2)dx [ ·: Even function]

Ex. 19: Change the order of integration and evaluate

JI/x y x (1+xy)2(1+l> dydx. (S.U. 2003)

Sol. : The curve y =x is a s�ight line through the origin andy = i. e. xy = 1 Y B is a rectangular hyperbola, x = 0 is the y-axis and

x = 1 is a line parallel to they-axis. Thus, the region of integration is OAB.

To change the order of integration we consider

a strip parallel to the x-axis. Now, we have to consider

the regions separately. In the region OAC, consider a

Fig. (9.39) strip parallel to the x-axis.

:. II = J J f(x, y) dx Rl y=I x=y

=�=o t=o (l +y2) dxdy - Y dx

1 _.1_ [ 1 ])I = Jo1+y2dy - l+xy o

r.uymee.-mg JVJal:nemal:JCS - 11 (9-43)

To find the first integral, put y = tan 8 . 1 - rr14 sec2 e d e . . Jo (1 + Y 2)2 Jo sec4 e

rrJ4 de rrJ4 = J �e = J cos2 e de 0 sec 0

= t4 ( 1 + cos 2 e ) d 8 = 1_ [ 8 + sin 2 e ] 1t1

4 0 2 2 2 0

_.L[�+.L]= �+1_ -2 4 2 8 4

1 dy I 1t J =[tan-ly] =-0 ""f+.Y2 0 4 1t 1 1t 1t 1 :. h =- 8 - 4 + 4 = 8- 4

Multiple Integrals

In the region CAB, consider a strip parallel to they-axis. :. /2 = J J f(x, y) dx dy R2

=r y=l

=r I

x= lly J

y x=O (1 +xy)2(1 +y2) dxdy

1/y ydx I+y2 J o (l+.xy)2

00 dy [ 1 ] 1/y =fl 1+y2 l+.xy 0 00 dy [ 1 ] =f1 1+y2 1 oo d y 1 _ 1 co = 2 Jl 1 y2 = 2 [tan y] I 1[1t 1t] 1t

=2 2-4 =g :. Required integral

1t 1 1t =h+/2=g-4+8 1t I 7t-l =

Ex. 20 : Change the order of integration and evaluate rr12 2a cos 6

J I r sin e dr de 0 0

Engineering Mathematics - II (9-44) Multiple Integrals Sol. : From the given limits we find that we have first to integrate w. r. t. r

y from r = 0 to r = 2a cos e and then w. r. t. e r=2acos9 from 0 to 1t/2. The curve r = 2a cos 8 i.e. = 2a r cos e i.e. :l + l = 2a.x i.e. (x- a)2 + l = a2 is a circle with centre (a, 0) and radius a. To change the order of integration we have

a X to integrate w.r.t. e first. For this we have to Fig. (9.40) consider a radial strip on which e varies from

e = 0 toe= cos -l(r/2a).Then we integratew.r.t. rfrom r= 0 to r=2a. 1FI2 2a cos a 2a cos 1{r/2a)

:.I I r sine dr de= I I r sin e dr d8 0 0 r=O 9=0 2a cos 1{rf2a) 2a

=I r [-cos eJ0 d e= I r [-cos cos I


·24

6. r y dy dx (S.U. 1997) [ Ans.: � log 2 1 0 0 (1- _ y2 (See fig. 9.25 page (9.31)) a!..fi [ 1 ] 7. I0 Jx i dy dx (S.U. 1990) Ans. : 32 a 4(n + 2) (See fig. 9.28 page (9.33)) I X

[ 1 ] 8. Io II e-Y y log X dy dx (S.U. 2000) Ans. : e (Fig. similar to fig. 7.3 page (7.2))

I 1t/2 1t/2 cosy

9. 0 Ix -y- dx dy (S.U. 1997, 2004) [ Ans. : 1 ] (Fig. similar to fig. 7.3 page (7 .2))

1 1 +V'I"=Y [ 4 ] 10. I J dx dy Ans.: -3 0 (See fig. 9.14 page (9.22), k = 1) 2 x+2

11. L Jx2 dy dx (See fig. 7.140 page (7.36))

IaJ

y

12• o o 2 - x 2)(a - y)(y- x) (See fig. 9.24 page (9.30))

(S;U. 1996) [ Ans. : � ]

[Ans.: 1ta]

[Ans·1-1!.] 13. 0 1 X 2)(1- X 2 y 2) (S.U. 1989, 94) .. 4 (See fig. 7.7 page (7.3))

( ( 14• o o 2 + x 2)(a - y)(y- x) (See fig. 9.24 page (9.30))

[ Ans. : 1t log (1 +.fi) ]

15. r: I� tflY dy dx (S.U. 1995, 99, 2003, 04)[ Ans. : 1 (Fig. similar to fig. 9.35 page (9.39)) 1C/2 y

16. /0 f0 cos2y 1- a sin x · dx dy (S.U. 1988, 90, 2003) (Fig. similar to fig. 7.3 page (7 .2)) [ Ans. : [ (1 -a 2) 3/2- 1]]

a 17. J0 (x2 + l) dy dx (S.U.l988, 95, 2000)


(See fig. 9.31 page (9.36))

( ( x2 18. 0 Y 2 + y 2 dx dy

(See fig. 9.29 page (9.34)) I 2-x

19. J0 Ix2 xy dy dx (See fig. 9.61 page (9.59))

20. ( Jox x (a 2_ y 2)(x 2 - y 2) dy dx

(See fig. 7.3 page (7.2))

21. ( (' e -x'l(l + ,2) x dx dt (First quadrant.)

a 2Vi 22. I0 /0 � dy dx

(See fig. 9.35 page (9.39))

Ioo I co e Y 23. 0 x ydwdy (See fig. 7.3 page (7 .2))

J3 j


Hence, in polar from x/4 a/cos&

I = J0 J0 �cos 6drd6

Multiple Integrals

= (14 cos a [2-]a/cos 0d e = !!.!_ J'IC14 sec2 ad e 0 3 0 3 0 - a 3 [ tan e ]7C/4 = !!.!_ - 3 0 3

a a x2 Ex. 2 : Change to polar co-ordinates and evaluate J J dx dy. 0 y x2+y2

Sol. : The region of integration is the same as in the above example. Hence, putting X = r COS a, y = r sin 6, d.t dy = r dr d 6, We get

I7r/4 Ja/cosfJ r 2 cos 2 (J d I= ·r r d(J o o r J7r/4Ja/cosfJ 2 2 (J d =0 0 r cos dr 6 [ 3 ]alcosfJ = J:/4 r3 o ·cos2 (J d(J

J1rl4a3 1 2 a3J"'4 = -·--·cos 6d6=- sec6d6 o 3 cos3 6 3 o

a3 [ ]"� = J

log (sec 6 + tan 6) 0

[ log(./2 +1)-log 1]

=

a3

3 log ( l + -/2).

I4a Jy Ex. 3: Evaluate 0 Y 214a dx dy by changing to polar co-ordinates. (S.U. 1995) Sol. : The region of integration is bounded by the parabola x = y214a i. e. l = 4ax and the line x = y. By putting X = r cos e. y = r sin a the parabola be comes � sin2 e = 4ar cos e i. e. r = 4a cos 6/ sin29

y

and the line becomes r cos a = r sin 6 i. e. 6 = 7t14. Fig. (9.42) Hence, r varies from 0 to 4a cos atsin26 and 6 varies from 1t/4 to 1Cf2.

"'


:. I 1t/2 4a cos at sin" a I rt/2 [ r 2 ] 4a cos atsin2 a

= I I rd 0 dr = 14 2 d 0 rt/4 0 1t 0 = .l Irt/2 16a 2 8 d 8 2 n/4 sm4 8

rt/2 [ 3 ] rt/2 = 8a 2 J cot 2 8 cosec 2 8 d 8 = Sa 2 - cot 8 rt/4 3 rt/4 = [0-1]=

Ex. 4 : Express the following integral in polar coordinates and evaluate. a dy dx

I 2 2 2 (S. u. 1984, 86, 87, 90, 92) 0 a -x -y Sol. : The limis of y are ax x 2 and 2 -x 2 i.e. the upper half of the circles

(i) x2 +l-ax= 0; (x- a/2)2 + y2 = a2/4 and y (ii) i2 + y2 = a2 To change the given integral to polar

coordinates we put x = r cos 8, y = r sin 8 and dxdy = rd 8 dr. The equations of the circles now become

(i) ? -ar cos 8 = 0 i.e. r = a cos e

(ii) ? = a2 i.e. r= a.

a/2 x=a Fig. (9.43)

Hence, r varies from a cos 8 to a and 8 varies from 0 to rc/2.

:. I -('2 r - 0 acose r2

dO 0 acosa rt/2 rt/2 = I0 a sine de =

[ - a cos e 10 = a

Ex. 5 : Change into polar coordinates and evaluate. a

J f e-


:. I n/2 a 2 rt/2 [ 1 2]

a = J J e-r rdrde= J - - e -r de 0 0 0 2 0

l Jrt/2 [ · 2) 1 [ 2]

rt/2 = - 2 o e- a

- I de = - 2 e - I [ e] o

= [ I- e -a2]

J I .T 2 2 2 Ex. 6: Evaluate by changing to polar coordinates e- (x + Y > dy dx. 0 0

Sol. : The region of integration is bounded by y = 0, Y N

the x-axis; y = i.e.� - 2x + i = 0 i.e. � Cl)

(x- 1)2 + i =I i.e. a circle with centre at (1, 0) and

radius I. x = 0, they-axis and x = I, a line parallel to they-axis.

If we put X= r cos e. y = r sine, y = X 2 Fig:-(9.45) i.e. x2 + i = 2x becomes ? cos2 e + ? sin2 e = 2r cos 9 i.e. r = 2 cos e.

Now, consider a radial strip in the region OAB. On this strip r varies from r = 0 to r = 2 cos e. Then e varies from 0 to rr./2.

Jn/2 J 2cos8 2 Jn/2

J2cos8 3

:. I= r r dr d8 = r dr d8 0 0 0 0

n/2 r 4 I n/2 [ ] 2cos8

= fo 4 o d8= 4 fo 2

4cos48d8

= 4[� . .!_. !:.] -= 3n 4 2 2 4 .

Ex7·E l

• . va uate J - dx d

0 y

Sol. : The curve x = a + 2-y2 is a circle (x- a}2 + i = a2 • We evaluate this integral by changing the co-ordinate:. to polar. In polar form the

circle i. e. x2 + i = 2ax becomes r = 2a cos e. The

line X = y in polar form becomes 8 = . The region of integration is OAB. Hence,

(S.U. 1989, 2003)

8 X

Fig. (9.46)


_ Jlt/4J2acos9 rdOdr = j_Jit/4[ 1 )2acosad O 1- o o (4a2+r2)2 2 o 4a2+r2 o

=-_!_ r'4[ 1 - I lde 2 ° 4a 2 + 4a 2 cos 2 0 4a 2 Sa 2 1 + cos 2 0 J r'4[1- sec 2 O ]de [Putt= tanO, sec28 dO= dt, sec28 = 1 + r] Sa2 0 l+sec20

= =

Ex. 8 : Express in polar coordinates and evaluate. 4a y2 (x2-y2 ]

(S.U. 1984, 88, 91)

Jo Jy2t4a x2 + y2 dx dy y a=rr2 Sol.: The limits ofxare }=4axandx=y. Putting

X = r COS 8, y = r sin 8, ; = 4ax gives ? sin2 8 = 4ar COS 8 i. e. r = 4a COS 8/sin2 8 andy = X gives r sin 8 = r cos 8 i.e. 8 = n/4. Further, (� - })!(� + l> = cos2 e -sin2 e and dx dy changes to r de dr. Fig. (9.47)

In the given region bounded by the parabola and the line, r varies from 0 to 4a cos 8/sin2 8 and 8 varies from 1t/4 to 1t/2. (See Ex. 3 page 9.47)

'

Jn/2 J4acos6/sin 2 6( 2 . 2 O dOd :. 1= cos 0-sm )r r 'IC/4 0 [ 2 ]4acos6/sin 2 6

= Jn12 (cos 2 8- sin 2 0) .!:_ dO n/4 2 0

8 2fn12( 20 . 2 0)cos20 dO = a COS -SID --'lr/4 sin 4 0 =Sa 2fnl2 (cot4 8- cot 2 0) dO n/4 =Sa 2J

"12 (cot2 8-cosec28-2cosec 20+ 2) dO n/4

[ 3 ]"'2 [ ] 2 cote 2 n 5 =Sa =Sa 2-3 n/4

Engineering Mathematics - II (9-51) Multiple

a x dy dx Ex. 9 : Evaluate by changing to polar coordinates J J 2 + 0 X X y Sol. : The region of integration is y =xi. e. a straight line, y = 2- x 2 i.e. x2 + i = a2, the circle with centre at the origin and radius a, x = 0 i.e. the y-axis and x = a i. e. a straight line parallel to the y-axis. Thus, the region of integration is OAB.

If we put x= r cos O,y = r sin 8 the linex = y becomes cos e = sine i.e. tan e = 1 i.e. e = n/4. The circle x2 + i = a2 becomes r = a. The line X= 0 i.e. r cos e = 0 becomes 8 = n/2. The line

y 8

a=!! 2

x x=a

Fig. (9.48) X =a, the circle x2 + i = a2 and the line X = y intersect at A for which e = n/4.

Hence, r varies from 0 to a and 8 varies from n/4 to n/2.

:. I rt/2 a 8

= J J r cos r d 8 dr rt/4 o r rt/2 a rt/2 [r2 ]a

= J J r cos 8 d 8 dr = J cos 8 2 d e rt/4 0 rt/4 0 a 2 Jrt/2 a 2 . rt/2 = 2 rt/4 COS 8 d 0 = 2 [Sin 0 )rt/4

a 2 [ �] =2 l-v'2 Ex. 10: Change to polar coordinates and evaluate

a� xy J J - . . e - ( x 2 + Y 2 ) dy dx

0 (x2 y2)

Sol.: The curvey = i. e. x2 ax+ i = 0 i.e. (x- a/2)2 + i = (a/2)2 is a circle with centre (a/2, 0) and radius a/2. The curve y = 2- x 2 i. e._..;.+ i = a2 is a circle with centre at (0,0) and radius a. (See Ex. 4).

X (a/2/0) (a,O) Fig. (9.49)

To change to polar coordinates we put x = r cos 8, y = r sin e. The equation� + i - ax = 0 changes to ? - ar cos 8 = 0 i. e r = a cos 8 and the equation x2 + i = a2 changes to r = a. Hence, r varies from a cos 8 to a and e varies from 0 to n/2.

rtl2a 2·8 2 . I = f I r Sin cos • e r r dr de · · o acosa r2


7t/2 -r2 a = I sin e cos e [- �2 ] de 0 a cos a

1 I1t/2 [ 2 2 2 ] = -2 0 e -a sin e cos e - e a cos 9 sin e cos e d e 1 7t/Z 2 1 I1t12 2 2 = - 2 I e a sin e cos e d e + 2 e -a cos s sin e cos e d e 0 0

-a2 7t/2 --a2 [ . 2e ] 1t'2 I = - l sine cos 8 d 8 =- .§! I 2 0 2 2 0

e -a 2 =---

4 For f.;,_. put cos2 8 = t, - 2 sin 8 cos 8 = dt 1 Io 2 ( dt ) 1 Jo 2 . I = - e a t - - = - e -a t dt ··2 21 2 41

:. I

1 [ e-a 2t ] 0 1 2 =4 -a2 !=-4a2[e-a -1]

e-a 2 1 2 1 =- 4a2 e-a + 4a2

1 2 =4a2[1-(a2+l)e a]

Ex. 11 : Change to polar coordinates and evaluate JJ dx dy over R (l+x2 + y2)2

one loop of the lemniscate (� + /)2 = � -/. Sol. : If we put X= r cos e, y = r sine, (x2 + /)2 = �-i becomes r4 =? (cos2 e- sin2 8) i.e.?= cos 2 e.

Y e = n/4

(1+x2 + y2)2 (1+r2)2 Now, on the loop rvaries from 0 to

and e varies from -1t/4 to n/4. 6=-n/4

:. I = Jn/4 r d r dO Fig. (9.50)

-n/4 0 (l+r2)2

=- -1 dO= 1- dO fn/4[ 1 ] Jn/4[ 1 ] 0 1 + cos 20 0 1 + cos 20

\Engineering Mathematics - II (9-53} Multiple Integrals = I:"[ I- =

2'9 ] do=[o-

1r I 1r-2 =---=--4 2 4

Ex. 12 : Change to polar coordinate and evaluate J� J: (x + y) dy dx. Sol. : The region of integration is bounded by the line y = 0 i.e. the x-axis, the line y = x, the line x = 0 i.e. the y-axis and the line x = I.

To change the coordinate system, we put X= r cos e, y = r Sin e. Then the line y =X becomes r COS e = r sine i.e. tan e = I, i.e. e = 7tl4 and the line X = I becomes r cos e = I i.e. r =sec e.

x

\Now, consider a radial strip as shown in the figure. On this strip r varies from r = 0 to r = sec e and then e varies from e = 0 to e = 7tl4.

1tr/4Jsec8 . :.1= 0 0 (rcosO+rstnO)rdrdO Jtr/4Jsec8 2 = 0 0 (cosfJ+sinO)r drd(J n/4 . r 3 [ ]

sec8 = J0 (cosO+smO) J 0 dO

I'Jtr/4 . 3 = J O (COS(} + SID (J) sec (} d (}

=- sec (}dO+ -- ·sinO d(J I [Jtr/4 2 Jtr/4 I ] 3 o 0 cos38 =.!.[{tane} �/4 + { I 2 }"'4] · 3 2cos (} 0 = .!. [1+.!.(2-l)]= .!.( t+l) = .!..

3 2 3 2 2

[Put cos e = t.]

ll

Ex. 13 : Change to polar coordinates and evaluate fJ dx dy where R is the region of integration bounded by� + l - x = 0 :0d y;: 0. Sol. :T he curve� + ; - x = 0 i.e. [x- (1/2)]2 + ; = (112i is a circle with

Engineering Mathematics - II (9-54) Multiple Integrals centre [(1/2), 0] and radius l/2. The line y = 0 is the x-axis. The region of integration is the semi-circle OAB.

To change to polar put X= r cos e, y = r sin e. Then� +1-x=O changes to? cos2 e +? sin2H =

r cos e i.e.?= r cos e i.e. r =cos e.

Now, consider a radial strip. On this strip r Fig. (9.52) varies from r = 0 to r = cos e. Then e varies from e = 0 to 9 = 1C/2.

frr/2 Jcos9 1 d d :. I= r r 8 0 0 = r /2 ros8 d r d () = Jrr/2 l [r ]cosO d () 0 0 . 0 °

frr /2 1 = ·cosO d ) 0

(1,0) X

=J:12 sin l/2() cos 112 e d () [See Beta, Gamma functions.]

= ..!.B(..!. �) = L 2 4 • 4 2 It

= � · v'i ·n= Ex. 14: Change to polar coordinates and evaluate

. fa!./2 0 JY log(x

2+y2)dxdy. (S.U. 1997, 2003)

Sol. : The region of integration is bounded by x = y, a line through the origin; x = 2 - y 2 i.e. � +I = a2, a circle with centre at the origin and radius a; y = 0, the x-axis andy = a/ ".J2, a line paprallel to the x-axis.

When x = y, x2 + I = a2 gives � = a2 i.e. x =a/ J2 Y and then y = a/ J2 . Thus, the line and the circle intersect inA (alii, alv'2 ) .

To change to polar we, put X= r cos e,y = r sin e. Then� +I changes to ? and x = y changes to

r cos 9 =· r sin 9 i.e. 9 = 1t/4. Fig. (9.53)

Engineering Mathematics - II (9-55) Multiple Integrals Now, consider a radial strip in the region of integration OAB. On the

strip r varies from r = 0 to r =a. Then 9 varies from 9 = 0 to 9 = n/4.

Jn/4Ja 2 Jn/4Ja :. I= 0 0 log r ·rdrdO = 0 0 2logr·rdrd0 n/4 ,.2 ,.2 l [ = 2J0 logr· T - J 2.; dr 0 dO

]a n/4 r2 r2 n/4 a2 a2 =2J logr·--- d0=2f -loga-- dO 0 2 4 0 0 2 2 2 a2 ] [ Jn/4 = a loga-2 0 dO= a loga-2 0 0 2 ( ' ) 1r =a log a -2 .4.

Ex. 15 : Change to polar and evaluate.

d dx 0 2� y y Sol. : The curve y = 2 ..[ax i. e. i = 4ax is a parabola andy= x 2 i.e. 2 2 x2 + i- Sax= 0 i.e. [x- + y 2= [ is a circle with cente at{Sa/2, 0) and radius Sa/2.

y A(a,2a) To change to polar coordinates we put X = r cos e, y = r sin 8. The equation i = 4ax

. 2 4acos0 changes sm 8 = 4ar cos 0 i.e. r = sin2 0

x and the circle x2 + i = Sax ctmnges to

Fig. (9.54)

? = Sar cos 8 i.e. r =Sa cos 9. The circle and the parabola intersect at 0

(0,0) and at A (a, 2a). Now for A, tan 9 = 2a/a = 2 and for 0, 8 = n/2. Hence, 8 varies from

tan- 1 2 to 1t/2. And r varies from 4a cos 9/ sin2 8 to Sa cos 8.

I= r r Jn/2 J5acos0 ( r ) d dO tan -I 2 4acos0/sin 2 0 r 2 sin 2 0 =

('2 5acos6 tan-12 �acos9/sin2e sm e


In/2 Sa cos a I = tan-1/r ]4a cos a/sin2 a 2 6 d 6 In/2 [ 4a cos 6] d 6

= 1 sa cos 6 . 2 6 -:-6 tan 2 sm sm 7tl2

= I 1 [ Sa cosec 6 cot 6 -4a cosec2 6 cosec 6 cot 6 ] d 6 tan- 2 7tl2

= [- Sa cosec 6 + cosec3 6] tan_12 =[-sa+ Sa

+ =]" (S..t5-ll) Ex. 16 : Evaluate by changing to polar coordinates

2 I Io (x 2 + y 2)2

Sol. : The curve y = I ± 2x- x 2 i. e. (y - 1 )2 = 2x- �i.e. (x- 1)2 + (y- 1)2 = 12 is a circle with centre at (I, I) and radius = 1. By putting x = r cos 6, y=rsin 6, we get, � +r =? and (x-I)2 + (y - 1 )2 = 1 changes to (r cos 6 -1)2 + (r sin 6-I)2 = 1 i.e. ? -2r (sin 6 +cos 6) + I = 0.

2 (sin 6 +cos 6) 6 +cos 6)2-4 :. r = 2

=(sin 6 +cos 6) ± sin 2 6

1 Fig. (9.SS)

Fig. (9.56)

2

Thus,rvariesfrom 2 6 to(sin6+cos6)+ 2 6 and 6 varies from 0 to 1t/2.

Iw2

I (sin a+ cos a)+ 2 a rd 6 dr

.·.].= � 0 (sin a+ cos 0) 2 a (r ) 7tl2 A + B dr = I I :;-!' d 6 where A = sin 6 + cos 6, B = 0 A-B r

f'Itl2 [ I ]A+ B - I I'ltl2 [ 1 1 ] = 0 -

2r 2 A -B d 6 -

2 o (A - B)2 - (A + B)Z

d 6

= ! I:Z (A 2)2 d 6 = 2 I07tl2 (sin 6 +cos 6) sin 2 6 d 6 n/2

= 2..fi I (sin3'2 6 cos112 6 + sin112 6 cos312 6) d 6 0

Engineering Mathematics - II (9·57) Multiple Integrals ·: J:12 sin 312 Scos 112 8d8 = f:12 sin 112 8cos312 8d8 = �(5/4,3/4)

= 2.n · 2[ -t ��374] =2F2·{ It li = 2.J2 ·� . .J21t = 1t.[-: G = = Ji ·1t J

Exercise - VI Change to polar coordinates and evaluate

a 1. I I y 2 + y 2> dy dx 0 0

(See fig. 9.53 page (9.54))

Ji. 2. J J log(x2 + y2) dx dy 0 0 '

(See fig. 9.53 page (9.54), a= 2)

3. I4 0 y

[ Ans. : 210 1ta 5 ]

(S.U. 1997, 2003)

[Ans. : n ( log 2 - �) ] (S.\J.

(See fig. 9.47 page (9.50)) [Ans. : [ *-J tan -IJ ] ] Iaia xdydx 4. 0 Y (See fig. 9.41 page (9.46))

I -x 2) 5. f f (x 2 + Y 2) dy dx 0 0 (See fig. 9.45 page (9 .49))

f2a xdydx 6. o o + y 2)

(See fig. 9.45 page (9.49)) ll a x x3dydx

7. Io Io x 2 + Y (See fig. 9.41 page (9.46))

I 4xy 2 2 sf I e-ddx • o o y

(S.U. 1997) [ Ans. :

3n [Ans.: --1] 8

(S.U. 1996, 2000)

[Ans.:

(S.U.1999)

[Ans.: a44 log(l+v'2)]

(S.U. 1994, 98)

Engineering Mathematics - If (9-58)

(See fig. 9.46 page (9.49))

9· 0 2 2 - y

(See fig. 9.43 page (9.48), a = I)

I I dx dy

10. -oo -oo (a 2 +X 2 + (Entire x-y plane)

a Ja x2dxdy

11. f0 Y (x 2 + y (See fig. 9.41 page (9.46))

Multiple Integrals

(Ans.:l]

(S.U.l990)

[Ans. :

(S.U.l995)

[Ans.:i]

12. r r e- (x 2 + y 2) dx dy (S.U. 1985) (First quadrant) [ Ans. : ]

13. s:a (x2 + y 2) dy dx ·(See fig. 9.46 page (9.49))

7. Evaluation of Integral over a given region

(S.U.1987)

Sometimes we are given the region not in terms oflimits for x and y but in terms of a curve or area bounded by curves. To evaluate such an integral, it may be noted, that the choice of order of integration or changing to polar coordinates is sometimes more convenient.

Type IV

Ex. 1 : Evaluat� J J (:Xl- y2) x dx dy over the positive quadrant of the circle :Xl +

y2 = a2. (S.U. 1986)

Sol. : Over the region X varies from 0 to 2 -y 2 andy varies from to 0 to a. a y

:.1 =f J (x2-y2)xdxdy 0 0

fa [x4 x2 =o 4-TY o dy _

J" [ (a 2 _ y 2)2 (a 2 _ y 2]

-0 4 2 dy

Fig. (9.57)

Mathematics - ll (9-59) Multiple Integrals

=[.l[a4y-2a2y3 _ys ] ]a

4 3 5 3 5 0 2 . 1 a5

= l5 a 5 15 a 5 = TI ·

.Ex. 2 : J J dx dy over the region bouned by the x-axi� ordinate at x = 2a and the parabola x2 = 4ay. (S.U. 1985, 87) Sol. : In the region y varies from 0 to �/4a and x varies from 0 to 2a.

2a x214a Y :. I = l0 J0 xy dy dx 2a �]x2/4a

:. l = JO [ x 2 0 dx 2a x x4 1 J2a = Jo 2 · I6a

2 dx = 32a 2 o x 5 dx

I [x6]20 1 64a

2 a4

= 32a2 6 0 = 32a2

= T·

Fig. (9.58)

Ex. 3 : Evaluate Jf 4 I 2 dx dy where R is the region x � 1, y � �. R X

+ y (S.U. 2005)

Sol. : The boundaries of the region are y = x2, a parabola with vertex at the origin and opening upwards. The line x = 1 is a line parallel to the y-axis. The region of integration is

the region between the line x = 1 and the branch of the parabola in the first quadrant.

In this region consider a strip parallel to the

y-axis. On this stripy varies from y = � toy = oo. Then x varies from x = 1 to x = oo.

foo foo dy :. I= 2 2 2 dx . I X y +(X )

= - tan - dx foo[ } 1 Y ]00

I x2 x2

x2

=foo-l [�-�]dx=

�foodx I x2 2 4 4 I x2

=�[-;]� =�[-0+1]=�.

y

y = x2

A x=1

Fig. (9.59)


Ex. 4 : Evaluate f f xy dx dy over the area bounded by the parabolas y = XZ. and x = -l. Sol. : The two parabo las are as shown in the figure. They intersect in 0 (0, 0) and A (-1, 1).

In the region OAB, consider a strip

parallel to the x-axis. On this strip x varies from

x = -.JY to x = -l and then y varies from y=O to L JlJ y2 :.1= 0 � y·xdxdy

�H�

25


Sol. : We shall first integrate w. r. t. y. Now y varies 0 to 1 -x. We shall first find Y

1-x :. h = I0 Let 1-x=a

= (a y) dy Puty =at I

= I a l 12 tll2 a112 (1 _ t)l/2 a dt 0 I - -

= a 2 I tll2 (1- t)l/2 dt =a 2 13/213/2 0 2 iTi2 . iTi2 -� =a 2 -87t

I 1 a 21t 7t

I 1 . I - x l/2 -dx=- x 112 (1 -x)2 dx . . - 0 8 8 0

1t 13/2 i3 - 1t i172. 2! = 8 - 8 21t = 105 [See also Ex. 6 page (9.4)]

Fig. (9.62)

[·:a= 1-x]

Ex. 7 : Evaluate J J (� + l> dx dy over the area of the triangle whose vertices are (0, 1), (1, l), (1, 2). (S.U.1985) Sol. : The equation of the line AC is

y-2 x-1 = i. e. y = x + I

I X+ I

y C(1,2)

:. I = J J (x2 +y2)dyd.x 0 I 1 [ 3 ]x + I = I 0 x2 y +

y3 dx Fig. (9.63) I

=I� [ { x 2 (x + I) + (x +3 1) 3 } -{ x 2 + t } ) dx Jl 1[ 3x2 } 1 --I (4 x3+3x2+3x )dx =- x4+x3+-- 3 0 3 2 0 [ t+I+t ] =t

Ex. 8 : Evaluate fJ x-y dx dy over the area of the triangle formed by R

X = 0, y = 0, X + y = I.


Sol. : The region of integration is shown in the figure. Now, consider a strip parallel to the x-axis. On this

strip x varies from x = 0 to x = 1 - y. Then y varies from y = 0 to y = 1.

:.1= J�J�-y Now, put x = (1 -y) t :. dx = (1 - y) dt When x = 0, t = 0; when x = 1 - y, t = 1.

y

:. I= f� f� y(1- y) · y) (1 y) · t · (1- y) dj dy = f�J�y(l- y)2 · t y � ·dt dy

= f � f � y (1- y) 512 • [ t(l t) 112 J. dt dy = f� y (l:_ y)512 ·B( 2,%) dy = s( 2,%) J� y � (l y)sndy

= s(2·�) s(2,7.) = �1312 .�17 12 = j312 2 2 j7 I 2 j111 2 j111 2 _,3/2 - fil 16 -,11/2-(9/2)(7 /2)(5/2)(3/2),3/2 = 945

Ex. 9 : Prove that fJ e ax+by dx dy = 2R where R is area of the triangle whose R

boundaries are x = 0, y = 0 and ax + by = 1. Sol. : Consider a strip parallel to the x-axis. On this strip x varies from x = 0 to x = (1 - by)la. Then y varies from y = 0 toy= lib.

. - J l/b J (I-by)/a ax+by dx d .. 1- o o e y [ ](1-by)/a f llb eax d = - y o a 0

y

B (0,1/b)

II ><

X

y= (1/a,O) Fig. (9.45)


fllbehy [ 1 b J 1 Jllb b = 0 --; e - Y - 1 dy = -;_ 0 ( e- e Y) dy

= = 2[ �(; )( �)] = 2R where R is the area of the triangle OAB.

Multiple Integrals

Ex. 10 : Evaluate fJ Y dy 2

where R is the region bounded by

i =ax andy=x. Sol. : We shall first integrate with respect to y and then with respect to x.

The curve i = ax is a parabola with vertex at the origin and opening on the right. y = x is a line through the origin.

In the region of integration consider a strip parallel

y

Fig. (9.66)

to they-axis. On this stripy varies from y = x toy= ..[;. Then x varies from x= 0 tox=a .

... I= ydy o (a-x) dx ax-y2

= J: y2 t dx = Jdx o (a-x) =r .r; dx 0

Now, putx= a sin2 8 :. dx= 2a sin 8 cos 8 d8

Jn/2 fa . :.1= ·2asm0cos0d0 0

Engineering Mathematics - II

fn/2 2 = 2a 0 sin (} d(} 1 n na =2a ·-·- = -. 2 2 2

(9-64) Multiple Integrals

Ex. 11 : Evaluate fJ x ( x -y) dx dy where R is the triangle with vertices (0, 0), R (1, 2), (0, 4).

Sol.: Let 0 (0, 0), A (1, 2), B (0, 4) be the vertices of the triangle OAB. The equation of the line OA is y-O X-O . 2x Th . f h 1" -- = z.e. y = . e equatton o t e me 0- 2 0-0

. y-2 x-I . AB1s -- =-- z.e. y=-2x+4. 2-4 1 - 0

Fig. (9.67)

Now, consider a strip parallel to they-axis. On this stripy varies from y

= 2x toy=.. 2x + 4 and thenx varies fromx = 0 to x = 1.

J I J-2x+4 2 :.1= 0 2x ( x -xy)dydx

[ 2 ]-2x+4 = J� x 2y-

2x dx

= J�[x2(-2x-4)-�(-2x+4)2 - x2 ·2x+�(2x)2 ]dx

= J�[ -2x3 +4x2 -2x3 +8x2 - 8x-2x3 +2x3]dx

= J�(-4x3+12x2-8x)dx

[ 4 2 2 ] 1 = - x +4x -4x 0 =-1. Ex.12: Evaluate JJ xm- lyn-1 dxdyover region bounded by x + y = h, x = 0, y = 0. Sol. : The region is bounded by the x-axis, y-axis and the line x + y = h.

(h, 0) X On the strip y varies from 0 to h- x and then the strip moves from x = 0 to x = h. Fig. (9.68)


fhfh-x ' '

:. I == 0 0 x m- y n- dy dx

Let I,== f�l x yn-1 dy=

0 1 n =-(h-x) 1l

Now, l=fhxm-l . .!.(h x)11 dx. o n

Put X= ht

== f h h m 1 . t m-1 .!. h n (1-t) n . h dt o n

- hm+n J'tm-1(1-t)n dt -n

o

�zm+n lmln+l = --. = n lm+n+l (m+n)lm+n

Multiple Integrals

Ex. 13 : Evaluate fJ xy-y 2 dx dy where R is a triangle whose vertices are R

(0, 0), (10, 1) and (1, 1). Sol.: The triangle OAB is as shown in the figure. The equation of the line OA

. y-I x�I . Th

. f IS --= -- 1.e. y = x. e equatton o e 1-0 1-0

y-1 x-10 line OB is --= -- i.e. x = 1 Oy. 1-0 10-0

y

B (10, 1)

Now, consider a strip parallel to the x-axis.

On this strip x varies from x = y to x = I Oy. Then y varies fromy = 0 to y = 1.

Fig. (9.69)

:.I= f�f�Oy y2 dx dy

d =J'(9y2-0)

3/2 o (3 I 2)y y o dy y


2fl [ y3 ] 1 = 3 0 27 . y 2 dy = 18 3 0 = 6. Ex. 14: Evaluate J J (x + y)2 dx dy over the area bounded by the ellipse

2 2 L+L =1 a 2 b 2 . Sol. : We have

a I = J_a J_; (x2 + 2xy + y2) dy dx

(S.U. 1986, 92)

= 2 (12 { a3 b sin 2 e cos 2 e + 1. ab 3 cos 4 e } d e -rc/2 3

= 4 J ;'2 { a3 b sin 2 e cos 2 e + ab 3 cos 4 e } d e = 4 [ a3 b . 1. . 13T2 13T2 + 1. ab 3 . 1. . I il 1572 ) 2 13 3 2 13 = 4 [a3 b 2 + 1. ab 3 .1.. lili · 3/2 · l/2 · lili 2 2 3 2 2 = 4 [ a 3 b · 1t + ab3 ] = 1tab [a 2 + b 2] 16 16 1t 4

Ex. 15: Evaluate J J i dx dy over the area outside x2 + i -ax = 0 and inside � + i - 2ax = 0.

(S.U. 1986,89,93, 2006) Sol. : First we note that x2 + i -ax = 0 i.e. (x- a/2)2 + i = (a/2)2 and x2 + i- 2ax = 0 i.e. (x- a )2 + i = a2 are circles as shown in the figure.

We change the given integral into polar co

ordinates. Putting X = r cos e, y = r sin e, in Fig. (9.70)


:l- + l -ax= 0 we get?-- ar cos 9 = 0 i.e. r =a cos 9 and in x2 + l- 2ax = 0 we get;.- 2 ar cos e = 0 i.e. r = 2a cos 9.

Replacing dXdy byrd 9 dr, we get '/t/2 2a cos e

I = 2 f f r 2 sin 2 9 · r d 9 dr 0 acose

1tl2 [ 4 ] 2a cos e 1t/2

=2 f0 sin29 r4 d9= tf0 sin29[16a 4 cos4e-a 4cos4S]d9 a cose

_.12. a 4f1t12

cos 4 9 sin 2 9 d9 = 15a 4 . 1. . s(.i 1.) = 151ta 4 - 2 0 2 2 2' 2 64 .

(x2 +y2) 2 Ex. 16 : Evaluate f f x 2 Y 2 dxdy over the area common to the curves x2 + l =ax and x2 + l = bx. (a, b > 0) . Sol. : First we note that :l- + l =ax i.e. (x- a )2 + l = a2 and :l- + l =by i.e. :l- + (y- b/2)2 = b2 I 4 are the circles as shown in the figure.

If we change to polar coordinates by putting x = r cos 9; y = r sin 9, the equations of the circles become r = a cos 9, r = b sin 9. At the point of intersection a cos 9 = b sin 9

Also, :. tan 9 =alb i.e. 9 =tan -I (a/b)= a. say.

r4 (x 2 + y2) 2 x 2 y2 r 4 sin 2 9 cos 2 9

Now, integral over the region DBA. fafbsine r4

I = r d9 dr I o o r 4 sin 2 9 cos 2 9 fafbsine 1

= r d9 dr o o sin2Scos29 a 1 [ 2]bsine

=f

L d9 o sin2Scos29 2 0

a 2 [ )a = 1. b2 f sec29d(:) = lL tanS 2 0 2 0 1. b2 tan a. = 1. b2 1! = .! - 2 2 b 2

And integral over the region DCA. f1t/2facose 1

f. = r d9 dr 2 a 0 sin 2 9 cos 2 9

X

Fig. (9.71)

Engineering Mathematics • II (9-68)

TCI2 1 [ 21a cos 6

=I r_ de IX sin2ecos2e 2 o

Multiple Integrals

2 7tl2 2 [ ]n/2 -a 2 =g_I cosec2ede=fL -cote = -2 [0-cota.] 2 IX 2 IX =1 a2 !=1 a b

:. Required integral = ab x 2y2 Ex. 17 : Evaluate I I x 2 + y 2 over the annular region Y

b�tween circles x2 + l = a2 and x2 + l = b2; a> b (>0). Sol. : Changing to polar coordinates we see that

I1t'2I

a r4sm2ecos2e I = 4 0 b r 2 r de dr

= 4 I; sin 2 e cos 2 e [ [ de Fig. (9.72)

J1t/2 • 1 = (b 4- a 4) sm 2 8 cos 2 8 dO = (b 4- a 4) · -0 2 13

= (b4-a 4). l ( 1/2)1172 · ( 1/2) 1172 = (b4-a 4) . 7t 2 2! 16

Ex. 18 : Evaluate I I dx dy over the area of the positive quadrant of the circle�+ I = I by changing to polar co-ordinates.

(S.U. 1985, 90, 97) Sol. : Since�+ 1 = 1 �a circle with centre at the Y (0,1) origin and radius 1, in polar co-ordinates r varies from 0 to 1 and e varies from 0 to 7t/2.

7tl2 1j{ 1-r2 } I= I 0 I0 1 + r 2 · r dr de To obtain the first integral we put r 2 = t.

:. 2r dr = dt. When r = 0, t = 0; when r = 1, t = 1. 1J{ 1-r2 } 1 1 ){ 1-t } · f --2 · r dr = -2 f - ·dt ··o l+r o 1+t

1 1 I - t I [ 1 dt 1 t dt ] dt=2

(1,0) X Fig. (9.73)


= [

{sin- I t }� + - t 2 }�] = ( - 1) :. I

_ J

1t12 l (� - 1) de = l (� - 1] [ e 1 1t12 - 0 2 2 2 2 o = ( - 1] = -

Multiple Integrals

Ex. 19 : Calculate I I ? dr de over the area between the circles r = 2 sin e and r = 4 sin e. (S.U. 1986) Sol. : The circle r = 2 sin e i.e. ? = 2r sin e becomes in cartesian system 2- + l = 2y i.e. 2- + (y - 1)2 = 1. Similarly the circler= 4 sin e i. e. ? = 4r sin 8 becomes in cartesian system 2- + l = 4y i.e. x2 + (y- 2)2 = 4. In the given region r varies from 2 sin e to 4 sin e and e varies from 0 to n.

J 7t J

4 sin a J

7t [ r4 ] 4 sin a :.I = . r3drd8= - d e 0 2 sm a 0 4 2 sin a

y

Fig. (9.74)

1 7t 7t = - f [44 sin4 e- 2 4 sin4 e 1 de= 6o I sin4 e de 4 0 0

7t/2 3 1 1t 45 = 120 f sin4 e de= 120. 4 . 2 . 2 = 2 . 1t 0

Ex. 20 : Evaluate I I re -,2 /a2 cos e sine dedr over the upper half of the circle r = 2a cos e. (S.U. 1985, 88) Sol. : The circle r = 2a cos 8 i.e. r2 = 2a r cos 8 y i.e. x2 + l = 2ax i.e. (x- a) 2 + l = a2 has centre at (a, 0) and radius a.

In the given region r varies from 0 to x 2a cos e and e varies from 0 to 7tl2.

:. I rt/2 2a cos a

= J f re-?-la2 sin e cos e de dr 0 0 7t/2 [ 2 2 ] 2a cos 0 .

= IO -a2 e-?-la O

Sil 8 COS 8 d 8

= _ a2

2 (12 [ e- 4 cos2 a-- I) sin e cos e d 8

Fig. (9.75)

]rt/2 a 2 [ l e- 4 cos2 0 _l sin2 e =- 2 8 2 0 [Put 4 cos2 e = t]


Ex. 21 : Evaluate Jf x2 over the semicircle x2 + l =ax in the positive quadrant. (S.U. 2006) Sol. : Putting X = r cos e. y = r sin e the circle x2 + l = ax becomes r2 = ar cos e i.e. r = a cos e.

As in the above example

Jrc/2Jacos6 2 2 I= a r rdrdO 0 0 rc/2[ (a2- r2 )3/2 ]acosO l

= -f o 3/2 . 2 . dO 0 = -� J; 12 [ (a2 - a2 cos2 0)312 - a3] dO

a3 Jrc/2 . 3 = - (l- sm 0) dO 3 0 a3 [Jrc/2 Jrc/2 3 J = 3 0 dO 0 sin 0 dO a3 [1t 2 J a3 = 3 2- 3 ·1 = 1s (31t - 4)

J I rde dr Ex. 22 : Evaluate 2 +a 2 over one loop of the lemniscate?= a2 cos 20.

(S.U. 1986) Sol. : In the given region r varies from 0 to a 2 e and e varies from - to .

:. I J rr/4

J 2 0 rd e dr

= 1tl4 o 2 + r 2 rr/4 [ 1 ] 2 0 = J (r 2 + a 2)2 . d e -rr/4 0 = r�:4 [ a 2 + a 2 cos 2 8 - a] d e

Fig. (9.76)

Fig. (9.77)

rr/4 ( ] [ ] rr/4 = a J_ rr/4 {2 . cos e - 1 d 8 = 2a {2 . sin e -e 0


= 2a [ 1 * J = }: (4- n) Ex. 23 : Evaluate J J

2 over one loop of the lemniscate (1+x +y)

(-� + i)2 = x2- i. (S.U. 2004) Sol. : Changing to polar coordinates by putting x = r cos 8, y = r sin 8, dxdy = rd 8dr the equation of the lemniscate becomes r4 = ? (cos 2 8-sin28) i.e. ? =cos 2 8. Then as above

J rr/4 2 e rd 8 dr

J rr/4 1 [ 1 2 e 1 = = de

- rr/4 0 (I + r 2)2 -- rr/4 2 (1 + r 2 ) 0 =--. 2 -1 d8 I rr/4 [ 1 ] 2 Jo 1 +cos

7tf4 [ 1 ] =- J0 2 sec 2 8- I d8 [ ] rr/4 =- t tan S-8

0 = - [± - *] = t JJ

dxdy Ex. 24 : Evaluate + 2 + 2 over one loop of lemniscate (x

2 + i ) X y

= 4 (x2 - i>. (S.U. 1985) Sol. : In Ex. 22, put a = 2

:.1=(4-n)

Exercise - VII

1. Evaluate fJ (x2 - dx dy over the area of the triangle whose vertices are at the points (0, 1), (1, (1, 2) (See fig. 9.63 page (9.61)) [ Ans.:- t]

2. Evaluate fJ r sin 8 d A over the cardiode r =a (1 +cos 8) above initial line. (See fig. 7.88 page (7.1 9)) (S.U. 2006) [ Ans.: t ·a 3]

3. Evaluate fJ xl i ym i dx dy over the triangle given by x 2:0, y 2:0, x [ im·IT ]

+ y:: 1. (See fig. 9.66 page (9.63)) (S.U. 1985) Ans. : lm + 1 + 1 4. Evaluate fJ dx dy throughout the area bounded by y = 2, x + y = 2.

(See fig. 9.61 page (9.59)) r Ans.: l

Engineering Mathematics - II (9-72) Multiple Integrals 5. Evaluate fJ (x2 -l) x dx dy over the positive quadrant of the circle

2 2 2 [A · �] x + y =a . (See fig. 7.7 page (7.3)) ns .. 5

ff 2 x 2 X.: 6. Evaluate (x2 - y ) dx dy over the area of the ellipse ;;z + b 2 = 1. (See fig. 7.29 page (7.7)) [ Ans.: 7t (a 2 + b 2)]

7. Evaluate Jf x (x2 - y2) dx dy over the positive quadrant of the ellipse + � = 1. (See fig. 7.29 page (7.7)) [ Ans.: a 2b (2a 2 + b 2)]

8. Evaluate Jf xm-l yn-l dxdy over the positivequadrant of the ellipse =l.(Seefig. 7.29 page(7.7))

ff x2 l 9. Evaluate 2 7 dx dy over area between the circles x2 + l = 1 X +y-and x2 + i = 4. (See fig. 9.71 page (9.68)) [Ans. : 1

51t ] 16

10. Evaluate Jf x2 l dx dy over the area of the ellipse + � = 1. [ 1ta 3 b 3] (See Fig. 7.29 Page (7.7)) Ans.: 24

11. Evaluate H R r4 cos3 e dr de where R is the region of the curve r = 2a cos 8 (See fig. 9.75 page (9.69)) (S.U. 2000)[ Ans.: 7a5/2]

12. Evaluate Jf xy (x + y) dx dy over the area of y = x2, y = x. (See fig. 9.26 page (9.31)) (S.U. 1997) [ Ans.: 3/56]

13. Evaluate fJ x2 y2 dx dy over the area of the circle x2 + l = 1. (See fig. 7.7 page (7.3)) (S.U.1988, 91) [ Ans.: n/24)

14. Evaluate fJ xy dxdy over the area nounded by the x-axis, the ordinate at x = 2a and the curve x2 = 4ay. (See fig. 9.10 page (9.20)) [ Ans. : a4!3 )

15. Evaluate fJ x3 dx dy over the circle x2 + i = 2ax = 0. 7a 5 (See fig. 9.75 page (9.69)) (S.U. 2003) [ Ans. : 4 ]

16. Evaluate Jf y dx dy through out the area bounded by y = x2 and x + y = 2. (See fig. 9.61 page (9.59)) (S.U. 1986) [Ans. : 356 ]

17. Evaluate fJ e3x+4 Y dx dy over the triangle x = 0, y = 0, x + y = 1. (S.U.1986,88,98,2000)


(See fig. 9.62 page (9.60)) [ Ans. : (3e 4 -4e 3 .+ 1)) 18. Evaluate fJ dx dy over the positive quadrant of the circle

1-i x2 + i =I. (See fig. 7.7 page (7.3)) (S.U. 1985,95, 99) [ Ans.: i]

19. Evaluate Jf xy (x2 + i)312 dx dy over the positive quadrant of the circle�+ l = I. (See fig. 7.7 page (7.3)) (S.U. 1985, 2005) [ Ans.:

20. J J x2 l dx dy over the circle�+ i = a2• (See fig. 7.7 page (7.3))

[ Ans. : · a 6]

21. J J dx dy over the region bounded by � + l- x = 0, y � 0. (S.U. 1988, 2003)

(See fig. 7.9 page (7.3), a= 112) [Ans . l jl . j1 = �] ""24 4 .fi 22. J J sin (� + l) dx dy over the circle x2 + l = a2.

(See fig. 7.7 page (7.3)) [ Ans.: n(l- cos a2)] dxdy .

23. J J over the regiOn bounded by y = 0, x = y, x = I.

(See fig. 7.3 page (7.2)) [ 24. fJ (x2 + i) dx dy over the first quadrant of the circle x2 + l = a2.

7ta4 (See fig. 7.7 page (7.3)) (S.U. 1987, 97) [Ans.: ]

(x2 +l)2 25. Jf 2 dx dy over the area bounded by

X y (i) x2 + l =ax, ax= by (See pig. 7.141 page (7.36)) (ii) � + l =by, ax= by (See pig. 7.142 page (7.36)) (iii) x2 + y2 =ax, x2 + l =by (See pig. 9.71 page (9.66))

8. Change of variables

ab ab [Ans. : (i) T (ii) T (iii) ab]

We know that the set of equationsx = (u, v) andy= lJI (u, v) defines, in general, a transformation or mapping which establishes a correspondence between points in u - v plane and the points in x - y plane. Under these


transformations in general, a closed region R in x - y plane is mapped onto a closed region R' in u - v plane. Further, it can be shown that

dx dy = I I du dv Using this result, it is possible to convert a double integral in x, y through

suitable transformation to a simpler integral in u, v.

TypeV

Ex. 1 : Given that x + y = u, y = uv, change the variables x, y to u, v in the integral f f (1 -x-Y)dx dy taken over the area enclosed by the lines x = 0, y = 0, x + y = 1. Show hat the value of the integral is 27t/105.

(S.U.1985,89,98,2001)

Sol. : The region of the integration in x- y plane is shown on I. h. s. and that in the u - v plane is shown on r. h. s.

From x + y = u, y = uv we get x + uv = u i. e. x = u(1 - v). When x = 0, u = 0 or v = I. When y = 0, u = 0 or v = 0.

When x + y = 1 , u = I. Thus, the region R in xy plane is mapped onto rgion R' in uv plane.

Fig. (9.78) ox ox

Further Fij = I - v, av = -u oy =v oy =u ou ' OV ox ox

= Fi av = 11-v u l �� =u v u ou OV

I = ff (1 x y) dx dy

R

I I = f f (1- v) · uv (1- u) · u · du dv u=O v=O


I I = f f u2(1 - u)l/2. vll2(1 - v) I I2 dvdu u=O v=O

Multiple Integrals

f I 2(1 )112 d f I 1/2(1 )1/2 d i313/2. j3/2 i3f2 - u -u u v -v v- -- 0 0 19/2 13

(I 3/2)3 (I /2)3 (f172)3 2n = 19/2 = (7/2)(5/2)(3/2)(1/2)!112 = 105 (See also Ex. 6 page 9.60)

Ex. 2 : By using the transformation x + y = u, y = uv show that

Jl Jl-x 1

0 0 eYI(x+ y) dy dx = 2 (e- 1) (S.U. 1985,87,88, 91, 93)

Sol. : The regions of integration on the two planes are as shown above in Ex. ( 1) and also J = u as before.

I I I I :. I = f f euvlu u du dv = f f ev udu dv. u=O v=O u=O v=IJ

f I I f I = [ u e v ) du = u ( e - 1) du

0 0 0

[u 2] 1 1 = (e- 1) -2 = 2 ( e- 1)

0

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Multiple Integrals - Weeblydigirit.weebly.com/.../16653588/em-ii_multiple_integrals.pdf · 2019. 8. 14. · Multiple Integrals 1. Introduction b You know that an integral I f(x) dx