18MAB102T Advanced Calculus and Complex Analysis Multiple Integrals

SRM Institute of Science and Technology

Ramapuram Campus
Department of Mathematics
Year / Sem: I / II
Branch: Common to ALL Branches of B.Tech. except B.Tech. (Business Systems)
Unit 1 – Multiple Integrals

Part – B (Each question carries 3 Marks)

3 2 1
1. Evaluate ∫2 ∫1 𝑑𝑥 𝑑𝑦.
𝑥𝑦

Solution
3 2
31 21
1
∫∫ 𝑑𝑥 𝑑𝑦 = [∫ 𝑑𝑦] [∫ 𝑑𝑥] = [log 𝑦]32[log 𝑥 ]12
𝑥𝑦 2 𝑦 1 𝑥
2 1
3
= (log 3 − log 2)(log 2 − log 1) = (log ) (log 2)
2
𝝅
sin 𝜃
2. Evaluate ∫𝟎 ∫0 𝟐 𝑟 𝑑𝑟 𝑑𝜃.

Solution
𝜋
sin 𝜃
∫0 ∫0
2 𝑟 𝑑𝑟 𝑑𝜃
𝜋 𝜋 𝜋
2 sin 𝜃 2 2
𝑟2 (sin 𝜃 )2 1 1 1 𝜋 𝜋
= ∫( ) 𝑑𝜃 = ∫ [ ] 𝑑𝜃 = ∫ sin2 θ dθ = ∗ ∗ =
2 0 2 2 2 2 2 8
0 0 0

2 2
3. Evaluate ∫0 ∫0 𝑑𝑥 𝑑𝑦 .

Solution
2 2 2 2

∫ ∫ 𝑑𝑥 𝑑𝑦 = ∫[𝑥 ]20𝑑𝑦 = ∫[2 − 0]𝑑𝑦 = [2 y]20 = (2)(2) − 0 = 4

0 0 0 0

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18MAB102T Advanced Calculus and Complex Analysis Multiple Integrals

3 2
4. Evaluate ∫0 ∫0 (𝑥 2 + 𝑦 2 )𝑑𝑥 𝑑𝑦 .

Solution
3 2 3 2 2
𝑥3 8
𝐼 = ∫ ∫ 𝑥 + 𝑦 𝑑𝑥𝑑𝑦 = ∫ [( ) + 𝑥𝑦 2 ] 𝑑𝑦 = ∫ [ + 2𝑦 2 ] 𝑑𝑦
( 2 2)
3 0
3
0 0 0 0
3 3 3
8y 2y 8∗3 2∗3
=[ + ] = + = 8 + 18 = 26
3 3 0 3 3

𝑎 𝑏 𝑐
5. Evaluate ∫0 ∫0 ∫0 𝑑𝑥 𝑑𝑦 𝑑𝑧.

Solution

𝑎 𝑏 𝑐 𝑎 𝑏 𝑎 𝑏
∫ ∫ ∫ 𝑑𝑥 𝑑𝑦 𝑑𝑧 = ∫ ∫ (𝑥 )𝑐0 𝑑𝑦 𝑑𝑧 = ∫ ∫ (𝑐 − 0)𝑑𝑦 𝑑𝑧
0 0 0 0 0 0 0
𝑎 𝑎 𝑎
= 𝑐∫ (𝑦)𝑏0 dz = 𝑐 ∫ (𝑏 − 0)𝑑𝑧 = 𝑏 𝑐 ∫ 𝑑𝑧 = 𝑏𝑐 (𝑧)𝑎0
0 0 0
= bc(a − 0) = abc

𝝅 𝑎
6. Evaluate ∫𝟎 ∫0 𝑟 𝑑𝑟 𝑑𝜃.

Solution

𝜋 𝑎 𝜋 𝑎 𝜋 𝜋
𝑟2 (𝑎) 2 𝑎2 𝑎2
∫ ∫ 𝑟 𝑑𝑟 𝑑𝜃 = ∫ ( ) 𝑑𝜃 = ∫ [ − 0] 𝑑𝜃 = ∫ 𝑑𝜃 = (𝜃 )𝜋0
2 0 2 2 2
0 0 0 0 0
2 2
𝑎 𝜋𝑎
= (𝜋 − 0) =
2 2
2 2
7. Evaluate ∫0 ∫0 𝑒 𝑥 + 𝑦 𝑑𝑥 𝑑𝑦 .

Solution
2 2 2 2

∫ ∫ 𝑒 𝑥+𝑦 𝑑𝑥𝑑𝑦 = ∫ 𝑒 𝑥 𝑑𝑥 ∫ 𝑒 𝑦 𝑑𝑦 = [𝑒 𝑥 ]20 [𝑒 𝑦 ]20

0 0 0 0

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18MAB102T Advanced Calculus and Complex Analysis Multiple Integrals

= (𝑒 2 − 𝑒 0)(𝑒 2 − 𝑒 0)=(𝑒 2 − 1)2

2 2−𝑦
8. Evaluate ∫1 ∫0 𝑥𝑦 𝑑𝑥 𝑑𝑦.

Solution
2−𝑦
2 2−𝑦 2 𝑥2 1 2
∫1 ∫0 𝑥𝑦 𝑑𝑥 𝑑𝑦 = ∫1 ( 𝑦 ( )) 𝑑𝑦 = ( ) ∫1 (𝑦(2 − 𝑦)2𝑑𝑦
2 2
0
2 2
1 1
= ∫ 𝑦(4 + 𝑦 2 − 4𝑦)𝑑𝑦 = ∫(4𝑦 + 𝑦 3 − 4𝑦 2 )𝑑𝑦
2 2
1 1
2
1 𝑦2 𝑦4 𝑦3
= [4 ( ) + ( ) − 4 ( )]
2 2 4 3 1
1 4 16 8 1 1 1
= {[4 ( ) + ( ) − 4 ( )] − [4 ( ) + ( ) − 4 ( )]}
2 2 4 3 2 4 3
1 5 5
= { }=
2 12 24

1 1 𝑥
9. Evaluate ∫0 ∫𝑦 𝑑𝑥 𝑑𝑦.
𝑥 2 +𝑦 2

Solution
1 1𝑥 1 𝑥 𝑥 1 𝑦 𝑦 =𝑥
−1
∫ ∫ 2 2 𝑑𝑥𝑑𝑦 = ∫ ∫ 2 2 𝑑𝑦𝑑𝑥 = ∫ (𝑡𝑎𝑛 ( ) ) 𝑑𝑥
0 𝑦 𝑥 +𝑦 0 0 𝑥 +𝑦 0 𝑥 𝑦 =0
1
= ∫ (𝑡𝑎𝑛−1 (1) − 𝑡𝑎𝑛−1 (0)) 𝑑𝑥
0

𝜋 1 𝜋 1 𝜋 𝜋 𝜋
= ∫ ( − 0) 𝑑𝑥 = ∫ 𝑑𝑥 = (𝑥)10 = (1 − 0) =
0 4 4 0 4 4 4

SRM IST, Ramapuram. 3 Department of Mathematics

18MAB102T Advanced Calculus and Complex Analysis Multiple Integrals

32
10. Evaluate   x y x  y  dy dx .
00

Solution

 
32 32

  x y x  y  dy dx    x y  x y dy dx
2 2

00 00

2
 x2 y2
3
y3 
    x  dx
0
2 3 0

3
 8 
   2 x 2  x  dx
0
3 

3
 x3 8 x 2 
  2    30

 3 3 2 0
1 1
11. Evaluate ∫0 ∫0 (𝑥 + 𝑦) 𝑑𝑥 𝑑𝑦 .

Solution
1 1 1 1
𝑥2
∫ ∫ (𝑥 + 𝑦) 𝑑𝑥 𝑑𝑦 = ∫ [( + 𝑥𝑦)] 𝑑𝑦
2 0
0 0 0

1 1
= ∫0 (2 + 𝑦) 𝑑𝑦
1
𝑦 𝑦2
= (2 + )
2 0

1 1
= (2 + 2) − (0 + 0)

=1

𝜋 1
12. Find the value of ∫0 ∫0 (𝑥 2 sin 𝑦) 𝑑𝑥 𝑑𝑦 .

Solution
𝜋 1 1 𝜋

∫ ∫ (𝑥 2 sin 𝑦) 𝑑𝑥 𝑑𝑦 = ∫ 𝑥 2 𝑑𝑥 ∫ sin 𝑦 𝑑𝑦
0 0 0 0

SRM IST, Ramapuram. 4 Department of Mathematics

18MAB102T Advanced Calculus and Complex Analysis Multiple Integrals

𝑥3 1
= ( ) (−cos 𝑦)𝜋0
3 0

1
= ( − 0) (−cos 𝜋 + cos 0)
3
1
= ( − 0) (1 + 1)
3
2
=
3
c b a
13. Evaluate    ( x  y  z ) dx dy dz .
0 0 0

Solution
a
c b a c b
 x2 
   ( x  y  z) dx dy dz      x y  x z  dy dz
 2


0
0 0 0 0 0

c b
 a2 
  
 2  a y  a z  dy dz

0 0  
c
 a2 b2 
   ba  a z b  dz
0  
2 2

b
c
 a2 y2 
 
 2 y  a  a z y  dz

0  2 0
c
 a2 b2 z2 
  b z  a z  a b 
 2 2 2 0

a b c (a  b  c)

2

4 x x y
14. Evaluate    z dx dy dz .
0 0 0

Solution
4 x xy

I=   
x=0 y=0 z=0
z dz dy dx

x y
4xz2 
=    dydx
0 0  2  0

SRM IST, Ramapuram. 5 Department of Mathematics

18MAB102T Advanced Calculus and Complex Analysis Multiple Integrals

1 4x
=   x  y dydx
2 00
x 4
1 4 y 2  1  2 x 
4 2
34 2 3  x 3 
=   xy  dx =   x   dx =  x dx   16
2 0  2  2 0 2  40 4 3 
0  0

1 1 y 2
dx dy
15. Evaluate   1 x2  y2
.
0 0

Solution

1 1 y 2
dx dy
I  
0 0 1 x2  y2

1 y 2
1  1  
  tan 1  x  dy
   
 1 y  1 y
2 2
0  0

 
  dy
1
  1
tan 1 (1)  tan 1 (0)

 1 y
2
0 

 
1
dy
  log (1  2 )
0
4 1 y2 4

a ay
16. Evaluate   x y dx dy .
0 0

Solution
a2  x2 ay
a
 x2  a

  y dy dx   y  2  dy
0 0 0 0

a
1 a4
2 0
 y a y dy 
6

SRM IST, Ramapuram. 6 Department of Mathematics

18MAB102T Advanced Calculus and Complex Analysis Multiple Integrals

a a2  x2 a2  x2  y2
dz dy dx
17. Evaluate  
0 0
 0 a  x2  y2  z2
2
.

Solution

a a2  x2 a2  x2  y 2
dz dy dx
Let I  
x 0

y 0

z 0 a  x2  y 2  z 2
2

a2  x2  y 2
a a2  x2   z 
  sin 1   dy dx
0 0   a  x  y   0
2 2 2

 
a2  x2 a2  x2

a a a

  1  sin  0 dy dx   2 0

 
a2  x2
sin 1 1
2  0  dy dx  y dx
 
0
0 0 0 0

    2 a2
a
 a
a2  x   x
a2  
 
2 0
 a  x dx  
2
a  x  sin 1      0 
2
  0 
2
0 
2

2 2 2  a  0 2  2 2  8

 (x  y 2 )dy dx over the region R for which x, y  0, x  y  1.

2
18. Evaluate
R

Solution

The region of integration is the triangle bounded by the

lines x  0, y  0, x  y  1

Limits of y : 0 to 1 – x ; Limits of x : 0 to 1
1 1 x

 ( x 2  y 2 )dy dx =   x  y 2 dydx
2

0 0
R

1 x
 1
y3


= x 2 y  
3 
dx
0 0
1
 2 (1  x)3 
 0  x (1  x ) 
3 
dx
1
 x3 x 4 (1  x)4 
   
3 4 12  0

1
 2 (1  x)3 
 0  x (1  x ) 
3 
dx
1
 x3 x 4 (1  x)4 
   
3 4 12  0

SRM IST, Ramapuram. 7 Department of Mathematics

18MAB102T Advanced Calculus and Complex Analysis Multiple Integrals

1 1 1
  
3 4 12
1

6

19. Find the area bounded by the lines x  0 , y  1 and y  x using double integration.

Solution

𝐺𝑖𝑣𝑒𝑛 x  0 , y  1 𝑎𝑛𝑑 y  x .

𝐻𝑒𝑛𝑐𝑒 𝑥 𝑣𝑎𝑟𝑖𝑒𝑠 𝑓𝑟𝑜𝑚 0 𝑡𝑜 1 𝑎𝑛𝑑 𝑦 𝑣𝑎𝑟𝑖𝑒𝑠 𝑓𝑟𝑜𝑚 𝑥 𝑡𝑜 1.

1 1 1 1 1
x2 1 1
𝐼 = ∫ ∫ 𝑑𝑦𝑑𝑥 = ∫[𝑦]1𝑥 𝑑𝑥 ( )
= ∫ 1 − 𝑥 𝑑𝑥 = [x − ] = 1 − =
2 0 2 2
0 𝑥 0 0

20. Find by double integration the area between the parabolas y 2 = 4ax and x2 = 4ay .

Solution
4a 4ax 4a 4a
 x2 
 Area =   dydx =   y  x2 =   4ax -  dx
4ax
dx
0 
0 x2 0 4a
4a 
4a
4a
 x 2 1 x3 
3
 1 2
4a
=   2 a x - x  dx =  2 a
1
2
- 
 4a   3 4a 3 
0
 2 0
4 a 3 1
= (4a) 2 - (4a)3
3 12a
5
4 a 32 32 1 4 2 42 1
= (4) (a) - 64a 3 = a - 64a 3
3 12a 3 12a
5
(22 ) 2 2 16 2 32 2 16 2
= a - a = a - a
3 3 3 3
16 2
= a
3

SRM IST, Ramapuram. 8 Department of Mathematics

18MAB102T Advanced Calculus and Complex Analysis Multiple Integrals

21. Find the area of the circle x2 + y2 = a2 using double integration.

Solution
Area of circle = 4  Area in first quadrant
a a2  y2
 4  dx dy
0 0
a
 4   x 0
a2  y2
dy
0
a
 4 a 2  y 2 dy
0
a
y 2 a2 1 y 
 4 a  y2  sin  
 2 2  a  0
a2  
 4   a
2

 2 2

22. Find the area of the circle x2 + y2 = a2 using polar coordinates.

Solution
x  r cos , y  r sin 
x2  y2  r 2
r 2  a2

  2 r  a
Area =   r dr d
 0 r 0
  2
a2
=  2
d
 0

=  a2

SRM IST, Ramapuram. 9 Department of Mathematics

18MAB102T Advanced Calculus and Complex Analysis Multiple Integrals

23. Find the area of the cardioid r = a(1 + cosθ) by using double integration.

Solution

Given the curve in polar co ordinates r = a(1 + cosθ)

 Area of the cardioid = 2(Area above the initial line)
θ varies from 0 to π
r varies from 0 to r = a(1 + cosθ)
π a(1+cosθ)
Area = 2  r drdθ
0 0
a(1+cosθ)
r  π 2
= 2   dθ
0 
2 0
π
=  a 2 (1 + cosθ)2 dθ
0
π
= a 2  (1 + 2cosθ + cos 2θ)dθ
0

 1  cos 
π π
  3 1 
= a 2  1 + 2cosθ +    dθ = a 2   + 2cosθ + cos2θ  dθ
0   2  0 
2 2 
π
3 1 sin2θ 
= a  θ + 2sinθ +
2
sinnπ = 0, n
2 2 2  0
3  3πa2
= a2  π  =
2  2

*****

SRM IST, Ramapuram. 10 Department of Mathematics