ii sem (csvtu) mathematics unit 3 (multiple integral )solustions
DESCRIPTION
Previous Years Solutions from 2006-2012 for both summer and winterTRANSCRIPT
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
0 1
( )y
x y dxdy
Prepared by Mrityunjoy Dutta
Ans: - ∫ ∫ (푥 + 푦)푑푥 푑푦 = ∫ ∫ (푥 + 푦)푑푦 푑푥 = ∫ 푥푦 + 푑푥
= ∫ 푥(4 − 푥 ) + ( ) 푑푥
= ∫ 4푥 − 푥 + 8− 4푥 + 푑푥
= 2푥 − + 8푥 − +
= 2(4 − 1) − (16 − 1) + 8(2 − 1) − (8 − 1) + (32 − 1)
= 6− + 8− + = = (Ans).
28. Find the volume bounded by the plane xy, the cylinder 2 2 1x y and plane 3x y z .
Ans: - Volume = ∭푑푥푑푦푑푧
= ∫ ∫ ∫ 푑푥푑푦푑푧√√ = ∫ ∫ 푧] 푑푥푑푦√
√
= ∫ ∫ (3 − 푥 − 푦)푑푥푑푦 = ∫ 3푦 − 푥푦 −√
√푑푥√
√
= ∫ (3− 푥)(√1− 푥 + √1− 푥 ) − ( )푑푥
= ∫ 2(3− 푥)√1− 푥 푑푥 = ∫ 6√1 − 푥 − 푥√1− 푥 푑푥
= 6 √ + 6 푠푖푛 푥 + ( ) /
= 6(0 − 0) + 3 + + (0 − 0) = 3휋 (Ans).
29. Prove that: 1 12
4 40 0 4 21 1
x dx dxx x
.
Ans: - ∫ √ Let 푥 = 푠푖푛푡 ⇒ 2푥푑푥 = 푐표푠푡푑푡 ⇒ 푑푥 = =
√푑푡
⇒ ∫ √= ∫ /
√푑푡 = ∫ √푠푖푛푡푑푡/ =
Γ Γ
Γ=
Γ
Γ√
⇒ ∫ √=
Γ
Γ√ ----------------------------------(1)
∫ √ Let 푥 = 푡푎푛푡 ⇒ 2푥푑푥 = 푠푒푐 푡푑푡 ⇒ 푑푥 = =
√푑푡
⇒ ∫ √= ∫ /
√푑푡 = ∫
√푑푡/ = ∫ √
푑푡/
⇒ ∫ √=
√∫ √
푑푡/ Let 2푡 = 휃 ⇒ 2푑푡 = 푑휃 ⇒ 푑푡 =
Prepared by Mrityunjoy Dutta
⇒ ∫ √=
√∫ √
/ =√
Γ Γ
Γ=
√
Γ
Γ√ --------------------(2)
Now, ∫ √× ∫ √
=Γ
Γ√ ×
√
Γ
Γ√
⇒ ∫ √× ∫ √
=√
Γ
Γ휋 =
√
Γ
Γ휋 =
√ (Ans).
(May-Jun-2010) 30. Compute, 훽 , .
Ans: -
9 7 7 5 3 1 5 3 1. . . . . . . .9 7 52 2 2 2 2 2 2 2 2,9 72 2 7.6.5.4.3.2.1 20482 2
(Ans).
31. Show that : ∫ √= √ ∫ √
=.
.
Ans: -
1
061 x
dx putting ddxxx cossin
31sinsin 3/23/13
2/
0
1)2/1(21)6/1(22/
0
3/22/
0
3/21
06
cossin31sin
31
cos
cossin31
1
dd
d
xdx
3/2
.6/161
3/222/1.6/1
31
1
1
06
xdx
--------------------(1)
By Legendre’s Duplication method nnn n 222
112
---------(2)
Now, putting 61
n in eqn (2) we get 3/12
3/26/1 3/2
3/2
3/126/13/2
So, equation (1) becomes
3/23/23/12
61
3/23/12
3/261
1
3/23/21
06
xdx
Prepared by Mrityunjoy Dutta
233/2
2
23/21
06 )3/sin(/
3/1.261
3/23/13/13/12
61
1
x
dx as
nrn
rnr
sin1.
3/7
333/21
06 2
3/1.4.
3.3/1.261
1
xdx
(Proved)--------(I)
Again
1
0312
3
xdx
putting ddxx cossin32sin 3/13/2
2/
0
1)2/1(21)3/1(22/
0
3/12/
0
3/11
03
cossin3
1cossin3
1cos
cossin32
23
123
dd
d
xdx
6/5
.3/132
16/52
2/1.3/13
1
123 1
03
xdx
-----------------(3)
By Legendre’s Duplication method nnn n 222
112
---------(4)
Now, putting 31
n in eqn (4) we get 3/22
6/53/1 3/2
3/1
3/226/53/2
So, equation (3) becomes
3/2
3/132
1
3/13/22
3/132
112
3 2
3/53/2
1
03
xdx
)3/sin(/
3/132
13/23/13/13/1
321
123 3
3/5
2
3/5
1
03
xdx
as
nrn
rnr
sin1.
3/7
33
3/5
1
03 2
3/1233/1
321
123
xdx
(Proved)--------------(II)
From (I) and (II)
Prepared by Mrityunjoy Dutta
3/7
3
1
03
1
06 2
31
123
1
xdx
xdx
(Proved)
32. Change the order of following integration: ∫ ∫ 푥푦 푑푥 푑푦 and evaluate.
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
1 23 3 4
2
0 1
26 3 8y y yI y
Prepared by Mrityunjoy Dutta
1 16 16 2 10 4 16 3 8 3 8
I
83
249
24155164
243162448128964
I (Ans).
33. Find the volume enclosed by the cylinders
x2 + y2 = 2ax and z2 = 2ax.
Ans: - 2 2 2 2 2 2 22 2 0 ( )x y ax x y ax x a y a
At 2 20, ( ) 0,2y x a a x a
Volume = 2
2
2 2 2
0 22
a ax x ax
axax x
dxdydz
Volume = 2
2
2 22
20 2
a ax xax
axax x
z dxdy
Volume = 2
2
2
2
2 2 22
20 02
2 2 2 2a ax x a
ax x
ax xax x
axdxdy a x y dx
Volume = 2
2
0
2 2 .2 2a
a x ax x dx
Volume = 2
2
0
4 2 . 2a
a x ax x dx Let 22 sin 4 sin cosx a dx a d
Volume = /2
0
4 2 2 sin .2 sin cos .4 sin cosa a a a d
Volume = /2 3
3 3 2 3
0
2 12864 sin .cos 64 .5 3 15
aa d a
(Ans).
(Nov –Dec 2010)
34. Write the relation between beta and gamma function. ( ) ( )( , ) .( )m nB m nm n
35. Change the order of integration and evaluate : 2
4 2
0 /4
a ax
x a
I dydx
Prepared by Mrityunjoy Dutta
2
4 2
0 /4
a ax
x a
I dydx By changing the order of the integration we will have figure below and the
solution will be 2 2
24 2 4
0 0/4 /4
aya ax a
x a y a
dydx dydx
=
24
0 2/4
ayadydx
y a =
4
0
24 22/40 2/4
aaya aydydx x dyy ay a
4 42
0 0
22 / 4
2/4
a aaydxdy ay y a dy
y a
44 3 2 2 2
0 0
2 3/2 1 32 16 1623 / 2 4 3 3 3 32/4
aa ay
y y a a adxdy aa
y a
(Ans)
36. Evaluate :
2 22 11 1
0 0 0
x yx
I xyzdydxdz
Sol.
2 22 22 2 111 1 1 1 2
0 0 0 0 0 02
x yx yx x xyzI xyzdydxdz dydx
Prepared by Mrityunjoy Dutta
2 22 2 2 211 1 1 1
0 0 0 0 0
12
x yx x xy x yI xyzdydxdz dydx
2 22 2 3 311 1 1 1
0 0 0 0 0 2
x yx x xy x y y xI xyzdydxdz dydx
2
2 22
12 3 2 4
11 1 1
0 0 0 0
0
2 2 42
x
x yxxy x y y x
I xyzdydxdz dx
2 222 2 4
2 3 2 211 1 1
0 0 0 0
1 1 1
4 4 8
x yx x x x x x xI xyzdydxdz dx
2 22 22 3 2 211 1 1
0 0 0 0
1 1 14 4 8
x yx x x x x x xI xyzdydxdz dx
2 22 3 3 5 5 311 1 1
0 0 0 0
24 4 8
x yx x x x x x x xI xyzdydxdz dx
2 22
12 4 4 6 2 6 4
11 1
0 0 0
0
22 4 4 6 2 6 4
4 4 8
x yxx x x x x x x
I xyzdydxdz
2 22 11 1
0 0 0
1 1 1 1 1 1 18 16 16 24 16 48 16
x yx
I xyzdydxdz
2 22 11 1
0 0 0
1 1 1 1 1 1 1 18 16 16 24 16 48 16 48
x yx
I xyzdydxdz
(Ans)
37. 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
Prepared by Mrityunjoy Dutta
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
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)
(May-Jun-2011)
38. Evaluate 3
2
Ans. We know that
( 1) ( )n n n
( 1)( ) nnn
3 13 2
322
Prepared by Mrityunjoy Dutta
13 2
322
………………………..1
As we know that
( 1)( ) nnn
Therefore we will have
1 111 2 2
1 122 2
…………………..2
From 1 and 2 we will have
121 1
3 2 23 3 12
2 2 2
143 2
2 3
Ans
39. 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.
Prepared by Mrityunjoy Dutta
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
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)
40. State and prove relation between beta and gamma function. Ans: The Beta function is defined as
1 11 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
Prepared by Mrityunjoy Dutta
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
1
0
( ) ( ) ( ) ( ). ( , )(1 )
n
m nxm n m n dx m n B m nx
( ) ( )( , ) .( )m nB m nm n
(proved)
Prepared by Mrityunjoy Dutta
41. 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).
(Nov –Dec 2011)
42. Write the relation between Beta and Gamma function
Ans: - The relation between beta & gamma function is ( ) ( )( , )( )m nm nm n
.
43. Change the order of integration 0
y
x
e dxdyy
and hence evaluate it.
Ans 0
y
x
e dxdyy
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).
44. Find the volume common to the cylinder 2 2 2 2 2 2,x y a x z a .
Ans:
Prepared by Mrityunjoy Dutta
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).
45. Prove that 2
1 2 12 2 1n
nnn
Ans.
2 11 2 1 12 2 2
nnn
2 1 2 1 2 1 2 1 2 3 2 31 1 ..........2 2 2 2 2 2 2
n n n n n nn
2 1 2 1 2 3 2 5 2 71 1 11 ............2 2 2 2 2 2 2 2
n n n n nn
2 1 . 2 3 . 2 5 . 2 7 ................112 2
n n n nn n
2 . 2 1 . 2 2 . 2 3 . 2 4 . 2 5 . 2 6 . 2 7 ................112 2 .2 . 2 2 . 2 4 . 2 6 ......................2
n n n n n n n nn n n n n n
1 2 !2 2 .2 . 1 . 2 . 3 ......................1
nn n n n n n
1 2 !22 2 . !
nn n n
1 2 122 2 . 1
nn n n
hence Proved
(May-Jun-2012)
46. Evaluate the following Integral:
Prepared by Mrityunjoy Dutta
2 3 2
0 1xy dxdy
3 23 22 3 2 22
0 1 0 01 0
26 26 52 / 33 3 6
xy x xxy dy dx dx dx
47. Evaluate the following integral by changing the order of integration:
0
y
x
e dydxy
Ans 0
y
x
e dxdyy
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).
48. Given 1
0 1 sin
nx dxx n
, show that ( ) (1 )sin
n nn
Prepared by Mrityunjoy Dutta
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
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)
49. Find the area included between the parabola: 24y x x and the line y x Ans:
Prepared by Mrityunjoy Dutta
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)
.