Prepared by Mrityunjoy Dutta Unit III – Multiple Integral (May-Jun-2006) 1. Evaluate by changing the order of integration. Ans 0 0 2 x x y xe dydx 0 0 2 x x y xe dydx by changing the order of the equation we will have 0 2 y x y xe dydx 0 0 2 2 y y x x y y xe dydx xe dx dy 0 0 2 2 2 2 2 2 2 2 2 2 2 y y t y y x x y y y xe dydx e dy y x x y xe dx taking t y xdx ydt x x y y y y y xdx dt xe dx e dt e y 00 / . 2 x y x dydx xe
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II sem (csvtu) Mathematics Unit 3 (Multiple Integral )Solustions
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Prepared by Mrityunjoy Dutta
Unit III – Multiple Integral
(May-Jun-2006)
1. Evaluate by changing the order of integration.
Ans 0 0
2xx yxe dydx
0 0
2xx yxe dydx
by changing the order of the equation we will have 0
2
y
xyxe dydx
0 0
2 2
y y
x xy yxe dydx xe dx dy
0 0
2
2 2
2
2
2
22
2 2 2
y
y
t
y y
x xyy yxe dydx e dy
y
xxyxe dx taking ty
xdx ydt
xx yy y yyxdx dt xe dx e dt e
y
0 0
/ .2
xyx dydxxe
Prepared by Mrityunjoy Dutta
0 0 0 0
2 20
2 2 2y
x xy y yy y y yxe dydx e dy e dy e dy
yy
0 0
21 1. 1/ 22 1y
x yey yxe dydx y e
(Ans).
2. Find the volume common to the cylinder 2 2 2 2 2 2,x y a x z a .
Ans:
2 2 2 2 2 2 2 2
2 2 2 2 0 0 0
Volume 8a a x a x a a x a x
a a x a x
dxdydz dxdydz
2 2 2 2
2 22 2
00 0 0 0
Volume 8 8a a x a a x
a xz dxdy a x dxdy
2 2
2 2 2 2 2 2 2 20
0 0 0
Volume 8 . 8 . 8 ( )a a a
a xa x y dx a x a x dx a x dx
3 3 3 3
2 2
0
2 16Volume 8 8 83 3 3 3
ax a a aa x a
(Ans).
3. Prove that . 1,0,)1(
!.)1().(log 1
1
0
mnm
ndxxx n
nnm
Prepared by Mrityunjoy Dutta
(Nov –Dec 2006)
4. Evaluate the following integral by changing of order of integration 2
4 2 20
a a
ax
y dxdyy a x
.
Ans:
Here region of integration is x varies from 0 to a, y varies from ax to a.
After changing order y varies from 0 to a, x varies from 0 to2y
a.
2 /2 2
4 2 2 4 2 20 0 0
y aa a a
ax
y dxdy y dydxy a x y a x
putting 2a as common we get
2
2
4 2 2 220 0 0 2
2a a a
ax
y y dydxay dxdy ay a x y x
a
2
2 21
24 2 2 220 0 0 02
0
22
sina a a a
ax
yay y dxay dxdy y xa dy dy
a yy a x y x aa
2
2 2 21 1 1
2 24 2 20 0 0
0
2
sin sin sin 0a a a a
ax
ya y
y dxdy y x y ady dya ay yy a x
a a
Prepared by Mrityunjoy Dutta
2 2 2
4 2 20 0 0
02 2
a a a a
ax
y dxdy y ydy dya ay a x
2 2 2
3
04 2 20 02 6 6
a a aa
ax
y dxdy y ady ya ay a x
(Ans)
5. Find the volume common to gas cylinder 2 2 2 2 2 2,x y a x z a .
Ans:
2 2 2 2 2 2 2 2
2 2 2 2 0 0 0
Volume 8a a x a x a a x a x
a a x a x
dxdydz dxdydz
2 2 2 2
2 22 2
00 0 0 0
Volume 8 8a a x a a x
a xz dxdy a x dxdy
2 2
2 2 2 2 2 2 2 20
0 0 0
Volume 8 . 8 . 8 ( )a a a
a xa x y dx a x a x dx a x dx
3 3 3 3
2 2
0
2 16Volume 8 8 83 3 3 3
ax a a aa x a
(Ans).
6. Define Beta function and show that ( ) ( )( , )( )m nm nm n
.
Prepared by Mrityunjoy Dutta
Ans: The Beta function is defined as 1 1
1 1
0 0
( , ) (1 ) , , 0(1 )
nm n
m nxB m n x x dx dx m nx
.
We know that 1
0
( ) .x nn e x dx
By putting x az dx adz we get
1
0
( ) .( )az nn e az adz
1
0
( ) .n az nn a e z dz
Putting z = x we get
1
0
( ) .n ax nn a e x dx
Now, taking a = z we get
1
0
( ) .zx n nn e x z dx
By multiplying 1z me z both side we get
1 (1 ) 1 1
0
( ) .z m z x n m nn e z e x z dx
Now taking integration with respect to z both side from 0z we get
1 1 (1 ) 1
0 0 0
( ) .z m n z x m nn e z dz x e z dzdx
Now, let (1 )1 1
y dyz x y z dzx x
we get
1
1
0 0
.( ) ( )(1 )
y m nn
m n
e y dyn m x dxx
1
1
0 0
( ) ( ) .(1 )
ny m n
m n
xn m e y dy dxx
1
0
( ) ( ) ( )(1 )
n
m nxm n m n dxx
Prepared by Mrityunjoy Dutta
1
0
( ) ( ) ( ) ( ). ( , )(1 )
n
m nxm n m n dx m n B m nx
( ) ( )( , ) .( )m nB m nm n
(proved)
(May-Jun-2007)
7. Change the order of integration in 2
1 2
0
x
x
I xydxdy
and hence evaluate the same.
Ans: From 2
1 2
0
x
x
I xydxdy
, region of integration is 2 , 2 , 0, 1y x y x x x
By changing of order of integration we get two regions
In one region limit of x is 0,x x y and limit of y is 0, 1y y
In another region limit of x is 0, 2x x y and limit of y is 1, 2y y
So, 2
21 2 1 2
0 0 0 1 0
y yx
x
I xydxdy xydydx xydydx
21 2
0 0 1 0
y y
I xydx dy xydx dy
21 22 2
0 10 02 2
y yx y x yI dy dy
1 22 2
0 1
0 (2 ) 02 2
y y yI dy dy
1 22 3
2
0 1
2 22 2y yI dy y y dy
Prepared by Mrityunjoy Dutta
1 23 3 4
2
0 1
26 3 8y y yI y
1 16 16 2 10 4 16 3 8 3 8
I
4 96 128 48 24 16 3
24I
164 153 11
24 24I
(Ans)
8. Find the volume bounded by the cylinder 2 2 4x y and the planes 4, 0y z z
Ans:
From the figure 4z y is to be integrated over the circle 2 2 4x y in the xy plane.
To cover the shaded half of the circle, x varies from 0 to 24 y
So, 2 24 42 2
2 0 2 0
Volume 2 2 (4 )y y
zdxdy y dxdy
2
24
02
Volume 2 (4 ) yy x dy
2
2
2
Volume 2 (4 ) 4y y dy
2 2
2 2
2 2
Volume 8 4 2 4y dy y y dy
22
1
2
4 4Volume 8 sin 02 2 2
y y y
Prepared by Mrityunjoy Dutta
{second term is zero because of odd function}
1 1Volume 8 0 0 2sin 1 2sin ( 1)
1Volume 8 4sin 1 8 4 162 (Ans).
9. Solve 2 2 /2( )nxy x y dxdy over the positive quadrant of 2 2 4x y supposing n+3>0. Ans:
2 2
2 2 /2 2 2 /2
0 0
( ) ( )a a x
n nxy x y dxdy xy x y dxdy
2 2
32 2 2
2 2 /2
0
0
( )( ) 322
a xn
an
y
x x yxy x y dxdy dxn
3
2 2 /2 2 2 2 2
0
1( ) ( 0)3
a nnxy x y dxdy x x a x dx
n
3 322 2
2 2 /2
0 0
( )3 3 2
n n aan a a xxy x y dxdy xdx
n n
73
222 2 /2( )
3 2 2( 3)
nn
naa axy x y dxdy
n n
(Ans)
(Nov –Dec 2007)
10. Define Beta Function.
Prepared by Mrityunjoy Dutta
Ans. The Beta function ,m n is defined as
1
0
1 1, 1m nm n x x dx for m>0 and n>0 It is also known as first Eularian Integral
11. Evaluate .
Ans:
2 2 2c b a
x y z dxdydzc ab
3 32 2 2 2 22 2
3 3
ac b c bz ax z y z dxdy ax ay dxdyc cab b
3 3 3 3
2 2 2 2 22 2 4 42 43 3 3 3
bc b a c cay a y ab a bx y z dxdydz ax y dx abx dxc a c cbb
3 3 3 3 3 3
2 2 2 4 4 4 8 8 83 3 3 3 3 3
cc b a abx ab x a bx abc ab c a bcx y z dxdydzc a cb
2 2 2 2 2 283
c b a abcx y z dxdydz a b cc ab
(Ans)
12. Prove that .
Sol.
1
10
1 !log
1
nn
n
nmx x dxm
1
0
log nmx x dx ……………………………………………….1
Let log tx t x e tdx e dt
1 becomes 0
0
1 1n nm t m te t dt e t dt
……………………………….2
1
1dpm t p dt
m
2 becomes 0
11 1
nppe dpm m
c
c
b
b
a
a
dxdydzzyx )( 222
1,0,)1(
!.)1().(log 1
1
0
mn
mndxxx n
nnm
Prepared by Mrityunjoy Dutta
0
11 11
nn
p ne p dpm
=
0
1 1 11 11
nn npe p dp
m
0
1 11 11 11 1 11 1
n nn nnpe p dp n
m m
1 1 !1 1 1 11 1
nn
n nn nm m
Hence Proved
13. Find the area enclosed by the parabolas 푦 = 4푥 − 푥 , 푦 = 푥 .
Ans:
23 4
0
Areax x
x
dxdy
2
3 3 34 2 2
0 0 0
Area (4 ) (3 )x x
xy dx x x x dx x x dx
32 3
0
27 27 27 9Area 3 0 02 3 2 3 6 2x x
(Ans)
(May-Jun-2008)
14. Define Gamma Function.
Ans. Gamma function is defined as 1
0
, 0x nn e x dx n
15. Evaluate the integral: 1
loglog
1 1
xye ezdxdydz
Prepared by Mrityunjoy Dutta
Sol. 1
loglog
1 1
xye ezdxdydz
1 1 1
log log loglog log 0 11
1 1 1 1
xx
x xy y ye e e eezdz dxdy z z z dxdy e x e dx dy
1 1 1
log log loglog 1 1
1 1 1
xx x x x x
y ye e e e yzdxdydz e x e dx dy e x e e x dy
1
1 1
1 1
loglog log 2 log 2 1
1 1log
log log 2 log 11 1
x
x
ye e ezdxdydz y y y y e e dy
ye e ezdxdydz y y y y e dy
2
2 2
1
loglog log log 1
4 11 1
xy
eye e yzdxdydz y y y y y e
2 2
1
log 5log 1 0 5 / 4 0 12 41 1
xye e e ezdxdydz e e e e
2
1
loglog 2 13 / 4
41 1
xye e ezdxdydz e
Ans.
16. Given 1
0 1 sin( )
nx dxx n
Show that 1
sin( )n n
n
Hence evaluate 40 1
dyy
Sol. Given that
1
0 1 sin( )
nx dxx n
We know that ( , ) m nm nm n
1
0 1
n
m nx m ndx
m nx
1 1let m n m n
1
0
1 111
n
m nx n ndx n nx
1
0
1sin( )1
nx dx n nnx
Prepared by Mrityunjoy Dutta
Therefore 1sin( )
n nn
Now for
40 1
dyy
……………………………………….1
Let 4y =x
Y= 1/4x
Y= 3/414
dy x dx
3/4 1/4 1
40 0 0
1 1 1 1 111 4 1 4 1 4 sin 4 4 2 2sin
4 2
dy x dx x dxy x x
(Ans)
17. Find the area enclosed by the parabolas 푦 = 4푎푥,푥 = 4푎푦 .
Area is given by 4
0
2
2/4
a axdxdy
x a
4 4
0 0
2 22/42/4
a aax axdxdy y dxx a
x a
4 42
0 0
22 / 4
2/4
a aaxdxdy ax x a dx
x a
44 3 2 2 2
0 0
2 3/2 1 32 16 1623 / 2 4 3 3 3 32/4
aa ax x x a a adxdy a
ax a
(Ans)
(Nov –Dec 2008)
Prepared by Mrityunjoy Dutta
18. Write only the value of .
Ans By using Legendre duplication formula
1 22 12 2
m m mm
14
1 1 1 1 12 214 4 2 4 22 122 2
19. Evaluate .
Ans:
2 2 2c b a
x y z dxdydzc ab
3 32 2 2 2 22 2
3 3
ac b c bz ax z y z dxdy ax ay dxdyc cab b
3 3 3 3
2 2 2 2 22 2 4 42 43 3 3 3
bc b a c cay a y ab a bx y z dxdydz ax y dx abx dxc a c cbb
3 3 3 3 3 3
2 2 2 4 4 4 8 8 83 3 3 3 3 3
cc b a abx ab x a bx abc ab c a bcx y z dxdydzc a cb
2 2 2 2 2 283
c b a abcx y z dxdydz a b cc ab
(Ans)
20. Change the order of integration 0
y
x
e dxdyy
and hence evaluate it.
Ans 0
y
x
e dxdyy
43.
41
c
c
b
b
a
a
dxdydzzyx )( 222
Prepared by Mrityunjoy Dutta
0
y
x
e dxdyy
by changing the order of the equation we will have 0 0
y ye dxdyy
0 0 0 0
y yy ye edxdy dx dyy y
0 0 0 0
yy y ye edxdy x dyy y
0 0 0 0
y y yye edxdy y dy e dy
y y
0
0 0 0 0
1y y y
y ye edxdy y dy e dy ey y
(Ans).
21. Find the area of the loop of the curve 푎푦 = 푥 (푎 − 푥) . Ans.
Prepared by Mrityunjoy Dutta
Area of the loop of the curve 푎푦 = 푥 (푎 − 푥) .will be given by 0 0
a xx aadxdy
0 0 0 0
a x a xx xa aa adxdy dy dx
……………………………..1
0 0 0 0 0 0
a x a x a xx x xa aa a a adxdy dy dx y dx
0 0 0 0 0 00 0
a x a x a xx x xa aa a a aa a xdxdy dy dx y dx x dxa
0 0 0 0
a xx aa a a xdxdy x dxa
…………………………..2
Take 2sinx a
Prepared by Mrityunjoy Dutta
Becomes /2 /2
2 2 2 4 2
0 0 0 0
sin cos 2 sin cos 2 sin cos
a xx aadxdy a a d a a d
/2 22 4 2 2
0 0 0
3 1 12 sin cos 2
6 4 2 2 16
a xx aa adxdy a a d a
(Ans).
(May-Jun-2009)
22. Write the relation between Beta and Gamma function.
Ans: ( ). ( )( , )( )m nB m nm n
23. Evaluate the following integral by changing of order of integration 2
4 2 20
a a
ax
y dxdyy a x
.
Ans:
Here region of integration is x varies from 0 to a, y varies from ax to a.
After changing order y varies from 0 to a, x varies from 0 to2y
a.
2 /2 2
4 2 2 4 2 20 0 0
y aa a a
ax
y dxdy y dydxy a x y a x
putting 2a as common we get
2
2
4 2 2 220 0 0 2
2a a a
ax
y y dydxay dxdy ay a x y x
a
Prepared by Mrityunjoy Dutta
2
2 21
24 2 2 220 0 0 02
0
22
sina a a a
ax
yay y dxay dxdy y xa dy dy
a yy a x y x aa
2
2 2 21 1 1
2 24 2 20 0 0
0
2
sin sin sin 0a a a a
ax
ya y
y dxdy y x y ady dya ay yy a x
a a
2 2 2
4 2 20 0 0
02 2
a a a a
ax
y dxdy y ydy dya ay a x
2 2 2
3
04 2 20 02 6 6
a a aa
ax
y dxdy y ady ya ay a x
(Ans)
24. Evaluate 2 2( )
0 0
x ye dxdy
by changing to polar coordinates. Hence show that 2
0 2xe dx
.
Ans: 2 2 2
/2( )
0 0 0 0
x y r
r
e dxdy e rdrd
2 2 2
/2( )
0 0 0 0
1 22
x y r
r
e dxdy re dr d
2 2 2
/2( )
00 0 0
12
x y re dxdy e d
2 2
/2( )
0 0 0
1 (0 1)2
x ye dxdy d
2 2/2
/2( )0
0 0 0
1 1 12 2 2 2 4
x ye dxdy d
(Ans)
Now, 2 2( )
0 0 4x ye dxdy
2 2( ) ( )
0 0 4x ye dx e dy
Prepared by Mrityunjoy Dutta
2 2( ) ( )
0 0 4x xe dx e dx
2
2
( )
0 4xe dx
2
0 2xe dx
(Proved)
25. Find by double integration, the area lying between the parabolas 24y x x and the line y x .
Ans:
23 4
0
Areax x
x
dxdy
2
3 3 34 2 2
0 0 0
Area (4 ) (3 )x x
xy dx x x x dx x x dx
32 3
0
27 27 27 9Area 3 0 02 3 2 3 6 2x x
(Ans)
(Nov –Dec 2009)
26. Find the value of 2
0
axe x dx
.
Ans: - ∫ 푒 푥 푑푥∞ Let 푎푥 = 푡 ⇒ 푎푑푥 = 푑푡 ⇒ 푑푥 = 푑푡
= ∫ 푒 푡 푑푥∞ = Γ(3) = × 2 = (Ans).
27. Change the order of integration and evaluate 43