ma2149a6
1.
Multiple Integrals
Evaluate the following double integrals:
 xy dxdy , where S is the region bounded by the lines x = 0, x = 1, y = x 2 and y = 4.
(a)
S
 x
(b)
2
dxdy , where S is the region bounded by y  2 x and x 2  y  8 .
S
Sol:
(a)
4

1 47
xy
dxdy

xy
dy

S
0  2 dx  ...  4  12  12
x

1
(b)
y  2x
y  2x
y  2x
 y  2x



 x  4  x  2
 2



or 
 2
x  y  8
x  2x  8  0
( x  4)( x  2)  0
 x  4 or x  2
 y  8  y  4
8  x 2 
S x dxdy  4  2x x dy dx  ...


2
2
2
 192  211.2  120  100.8
2.
Evaluate
 xy dxdy where S is the region enclosed by the 4 parabolas
y 2  x , y 2  2x ,
S
x 2  y , x 2  2 y using the change of variable u =
x2
y2
.
,v 
y
x
Sol:
x2
y2
, the boundaries y 2  x goes to v  1 ,
,v 
y
x
2
2
y  2 x to v  2 , x  y to u  1 and x 2  2 y to u  2 . And the region S in x-y plane which are
Under the transformations u =
enclosed by the 4 parabolas y 2  x , y 2  2 x , x 2  y , x 2  2 y goes to the region S in u-v plane
which are enclosed by the lines u  1, u  2, v  1, v  2
 ( x, y )
1
1

 ... 
 (u , v)  (u , v)
3
 ( x, y )
Therefore,
J
3
 xy dxdy   uv J dudv  ...  4
S
S
1
3.
Evaluate E z =
 0z
4 0
 ( x
S
1
2
y z )
2
2
3
dxdy where S is the disc x 2  y 2  a 2 , which
2
represents the z-component of the electric field at the point (0, 0, z) due to a uniformly charged
circular disc lying in x 2  y 2  a 2 , z = 0.
Sol:
x 2  y 2  r 2 , dxdy  rdrd
 0z
Ez=
4  0
2 0 z

4  0 0

S
a
3
2 2
(x  y  z )
2
2
rdr
3
(r 2  z 2 ) 2
 z
 0
20
 2

dr
3

 ( r 2  z 2 ) 2 
 z
dxdy  0   
4  0 0  0
a
1
1


1
 
1
z
2
2 2
 (a  z )

 z
 0 
0
20 
a
1
(r 2  z 2 ) 2
rd
1
Let R be the region bounded by x  y  1 , x = 0, y = 0. Show that
4.
x y
 cos x  y  dx dy =
R
sin 1
, using the substitution x - y = u , x + y = v .
2
Sol:
x  y  1  v  1, x  0  u  v  0, y  0  u  v  0
x
uv
x y u
2 , Jacobian= u

vu
y
x y v
y
2
u
x
x y
1
u
x
v  1 / 2
y
v
 cos x  y  dx dy =  2 cos v  du dv  ... 
R
R'
5.
Evaluate
sin 1
2
 xe dxdy , where R denote the region inside x
y
2
 y 2  1 but outside
R
x 2  y 2  2 y with x  0, y  0 , using the change of variable u  x 2  y 2 , v  x 2  y 2  2 y .
Sol:
1
 e2 
3
2
2
6.
Evaluate
 e
xy
dxdy where S is the region enclosed by xy  1 , xy  2 , y  x , y  4 x using
S
the change of variable xy  u ,
y
 v.
x
Sol:
xy
2
 e dS  ...  e  e ln 4


S
Use the change of variables x  y  u , x  y  v to evaluate
7.
 e
( x y )
dxdy .
x  y 1
Sol:
x  y  1 is equivalent to the region bounded by x  y  1, x  y  1, x  y  1, x  y  1 .
Under the transformations: x  y  u , x  y  v ,
x  y  1 goes to u  1,  x  y  1 goes to u  1 ,  x  y  1 goes to v  1 and x  y  1 goes to
v  1.
 ( x, y )
1
1
1
The Jacobian J 


 .
 (u , v)  (u , v)
2
1 1 

det 
 ( x, y )
 1  1
So
 e
( x y )
x  y 1
1 v
 1 ev 
e
dxdy   e J dudv     dvdu  
2 
2
1 x1
1  1
1
1
v
1
1
1


1
du   e  e 1 du  e  e 1 .
2 1
1 y 1
8.
Evaluate
dxdydz
 (1  x  y  z)
3
, where V is the region bounded by the planes:
V
x  y  z  1, x  0, y  0, z  0 .
Sol:
1 x  y

dxdydz
1

dz

 dxdy
3
3



(
1

x

y

z
)
(
1

x

y

z
)


V
S  0
where S
is
the region bounded by
x  y  1, x  0, y  0 .
Then
1 1 x 1 x  y
1 x  y


 
dxdydz
1
1

dz
dxdy

dz

 

3

0  0  0 (1  x  y  z )3 dy dx
(1  x  y  z )3 
(
1

x

y

z
)

V
S 
0


 
 
1 1 x
1 x
 
1 x  y 
1
1
1 
  
dy
dx





0  0  2(1  x  y )2 8 dy dx
0
2(1  x  y  z ) 2
0 0

1
3
1
1

 1 1 x
1
y 1  x
1 
 
 
dx   


dx
0
2
(
1

x

y
)
8
4
8
2
(
1

x
)




0
0


x2


x
x
1
2  ln( 1  x )  1    1  1  1 ln 2    5  1 ln 2
  
8
2
16 2
 4
 0  4 16 2





Find the volume of the region bounded by the surfaces z  x  2 , z  0 , y  x 2 and
9.
y  2 x  3 . Notice the relationship between the double and triple integrals which give this
volume.
Sol:
32
10.
A solid tyre of outer radius a + b has a circular cross-section of radius a and consists of
material of constant density . Evaluate the mass of the tyre. Use cylindrical polar coordinates
with the z-axis along the axis of revolution (i.e. the axle) of the tyre. Evaluate the moment of
inertia about the z-axis and express the result in terms of the mass.
Sol:
2 b

mass   dxdydz      
V
a  0 
 b 
a
a 2
 
 1 2

r
dr
d

dz


2 2  
a  0 2 r
a z
 
a2  z 2
b a 2  z 2
b a  z
2
2

d dz

a 2
a


 2b    a 2  z 2 d dz  4b  a 2  z 2 dz
a  0
a




2
2
2
 4b  a(cos  )a cos d  4a 2b  cos 2 d  4a 2b
z  a sin


2



2
 2 a b
2

2
4
2
1  cos 2
d
2
2 b

moment of inertia   r dxdydz      
V
a  0 
 b 
a
2

a 2

1
      b  a 2  z 2
4 a  0 
  b 
4

a 2
 
 1 4

r
dr
d

dz



 
a  0 4 r


a2  z2
 
a2  z2
3
4

a 2  z 2 d dz
 
4
4

a 
1 4 
a 2 z 2  
a 2 z 2  
 b   1 

 1

dz

2
b 2 b 2  
b 2 b 2  
a 



1 4

b 
z  a sin 2
4
4
 a
  a
 
1

cos


1

cos





a cos d
  b
b




 
2
2
3 

 2 2 a 2b  b 2  a 2 
4 

-End-
5
b a 2  z 2
b a  z
2
2

d dz
