-
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)
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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
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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
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Engineering Mathematics - II (9-4) Multiple Integrals
= · =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
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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 )
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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-.
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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)
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Engineering Mathematics - II (9-8)
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.
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Engineering Mathematics - II (9-9) Multiple Integrals
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)
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Engineering Mathematics - II (9-10) Multiple Integrals
= [ 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)
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Engineering Mathematics - II (9-11)
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
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Engineering Mathematics - II (9-12) Multiple Integrals
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
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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)
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Engineering Mathematics - II (9-15) Multiple Integrals
= 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)
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Engineering Mathematics - II (9-18) Multiple Integrals
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 «
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Engineering Mathematics - II (9-19) Multiple Integrals
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.
-
Engineering Mathematics - II (9-20) Multiple Integrals
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.
-
Engineering Mathematics - II (9-21)
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
Ex. 7 : Change the order of integration
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
-
Engineering Mathematics - II (9-22)
Ex. 8 : Change the order of integration
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.
-
Engineering Mathematics - II (9-23) Multiple Integrals
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.
-
Engineering Mathematics - II (9-24) Multiple Integrals
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
-
Engineering Mathematics - II (9-27) Multiple Integrals
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))
-
Engineering Mathematics - II (9-28) Multiple Integrals
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)
-
Engineering Mathematics - II (9-30)
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)
-
Engineering Mathematics - II (9-31) Multiple Integrals
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
-
Engineering Mathematics - II (9-32) Multiple Integrals
= 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
-
Engineering Mathematics - II (9-33) Multiple Integrals
= [ 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
-
Engineering Mathematics - II (9·34)
[ 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=
-
Engineering Mathematics - II (9-35) Multiple Integrals
. _ 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 ) ] :
-
Engineering Mathematics - II (9-36) Multiple Integrals
= [ 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
Ex. 11 : Change the order of integration and evaluate
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)
-
Engineering Mathematics - II (9-37) Multiple Integrals
:. 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
-
Engineering Mathematics - II (9-39) Multiple Integrals
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.
Ex. 17 : Change the order of integration and evaluate
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)
-
Engineering Mathematics - II (9-42) Multiple Integrals
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
-
Engineering Mathematics - II (9-45) Multiple Integrals
·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)
-
Engineering Mathematics - II (9-46)
(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
-
Engineering Mathematics - II (9·47)
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.
"'
-
Engineering Mathematics - II (9-48) Multiple Integrals
:. 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-
-
Engineering Mathematics - II (9-49) Multiple Integrals
:. 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)
-
Engineering Mathematics - II (9-50) Multiple Integrals
_ 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
-
Engineering Mathematics - II (9-52) Multiple Integrals
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
-
Engineering Mathematics - II (9-56) Multiple Integrals
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)
-
Engineering Mathematics - II (9-60)
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
Engineering Mathematics - II (9-61) Multiple Integrals
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.
-
Engineering Mathematics - II (9-62) Multiple Integrals
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)
-
Engineering Mathematics - II (9-63)
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)
-
Engineering Mathematics - II (9-65)
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
-
Engineering Mathematics - II (9-66) Multiple Integrals
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)
-
Engineering Mathematics - II (9-67) Multiple Integrals
: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)
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Engineering Mathematics - II (9-69)
= [
{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]
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Engineering Mathematics - II (9-70) Multiple Integrals
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
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Engineering Mathematics - II (9-71) Multiple Integrals
= 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
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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)
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Engineering Mathematics - II (9-73) Multiple Integrals
(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
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Engineering Mathematics - II (9-74) Multiple Integrals
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
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Engineering Mathematics - II (9-75)
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