The document is a set of solved problems from Applied Mathematics II course at Adama Science and Technology University, focusing primarily on multiple integrals. It includes explanations of vertically and horizontally simple regions, as well as evaluations of various iterated double integrals. Additionally, it involves applications of concepts such as polar coordinates in calculating areas and volumes under curves.

1. Applied Mathematics II Math 1102 ASTU
Solved Problems On Multiple Integrals Page 1
y
x
R
2


x
5
 
0
,
2
1
2
1
2
1
y
x
x
y




-1
4
R
1
2
x
y
2
2





y
x
x
y
x
y
y
x



 2
Adama Science And Technology University ( ASTU )
School Of Applied Natural Science, Applied Mathematics Department.
Applied Mathematics II (Math 1102) Solved Problems On Multiple Integrals.
1. Let be the region between the graphs of , the x-axis and the line . Show that is
simple.
Solution
is a vertically simple region between the graphs of and for .
i) is a horizontally simple region between the graphs of and for .
ii) From i) and ii), we conclude that is simple.
2. Let be the plane region between the graphs of the equations and . Show that R is
simple.
Solution
i) is a horizontally simple region between the graphs of and for .
ii) is a vertically simple region between the graphs of √ and √ and
between and √ .
iii) From i) and ii), we conclude that is simple.
3. Evaluate each of the following iterated double integrals.
a) dx
dy
xy
x
x
 

1
0
1
Solution
 
 
 
  










 1
0
1
0
2
2
2
1
0
1
1
2
2
1
dx
x
x
x
dx
y
x
dx
dy
xy
x
x
x
x
 
 

1
0
1
2
2
dx
x
x
 







1
0
2
2
1
dx
x
x
1
0
4
3
2
3










x
x
12
7


2. Applied Mathematics II Math 1102 ASTU
Solved Problems On Multiple Integrals Page 2
0 2
R
x
Y
x
y 
 2
2
R
R
0
4
2

x
2
x
y
y
x



Y
x
b) dy
dx
x
  
3
1
3
0 2
9
2
Solution dy
x
dy
dx
x 
  














3
1
1
3
1
3
0 2
3
0
3
tan
3
1
2
9
2
dy
 


















 

3
1
1
1
3
0
tan
3
3
tan
3
2
   
 dy





3
1
1
1
0
tan
1
tan
3
2
 dy
 

3
1
0
4
3
2  dy


3
1
6

3


4. Evaluate  
 
R
xy
dA
e
x
x 1 if the triangular region bounded by the lines , and .
Solution
is a vertically simple region between the graphs of and
Thus,  
  




R
x
xy
xy
dx
dy
e
x
x
dA
e
x
x
2
0
2
0
)
1
(
1  
 






 
2
0
2
0
1 dx
e
x x
xy
 
 
 


 
2
0
2
)
1
(
1
2
dx
x
e
x x
x
 
  


 
2
0
2
0
2
)
1
(
1
2
dx
x
dx
e
x x
x
2
0
2
2
0
2
2
2
1 2

















 
x
x
e x
x
= 0
5.By reversing the order of integration, evaluate  
 
4
0
2
3
cos
y
dy
dx
x .
Solution:Note that it is impossible to evaluate the integral as it is. The plane region R is given as a
horizontally simple region between the graphs of √ and for .
As a vertically simple region, is a region between the graphs of and for .
Thus ,    
 
  
2
0 0
3
4
0
2
3
2
cos
cos
x
y
dx
dy
x
dy
dx
x   dx
y
x
x
 








2
0
0
3
2
)
(
cos
 


2
0
3
2
cos dx
x
x
  2
0
3
3
sin x

3
8
sin


3. Applied Mathematics II Math 1102 ASTU
Solved Problems On Multiple Integrals Page 3
z
x
Y
R
4
Y
x
R
2
4 x
y 


2
4 x
y 


6. Find the volume of the solid region bounded by the paraboloid and the plane z = 4.
Solution The intersection of the paraboloid and the plane is the circle and this determines the
region .
and R is a vertically simple region between the graphs of √ and
√ for .
Volume 

R
dA
y
x
f
V )
,
(  







2
2
4
4
2
2
2
2
)
(
x
x
dx
dy
y
x  







2
2
4
4
2
2
2
2
)
(
x
x
dx
dy
y
x
dx
y
y
x x
x




 

















2
2
4
4
3
2 2
2
3
         dx
x
x
x
x
x
x
x
x
 





























2
2
2
2
2
2
2
2
2
2
4
4
3
1
4
4
4
3
1
4
    dx
x
x
x
x












2
2
2
2
2
2
4
4
3
1
4
2 dx
x
x
 














2
2
2
2
4
3
2
3
4
2
Let (trigonometric substitution)
Also and






d
cos
2
cos
2
)
sin
4
(
3
2
3
4
2 2
2
2
 












 





d








 2
2
2
2
2
cos
sin
3
8
cos
3
4
8
  





d


 2
2
2
2
2
cos
sin
2
cos
3
32
  





d



 2
2
2
2
2
)
cos
(
)
cos
1
(
2
cos
3
32
  




d


 2
2
4
2
cos
2
cos
3
3
32
2
2
3
8
3
sin
cos
8
3
sin
cos
4
1
2
2
2
sin
2
3
3
32
































 
8

7. Find the area of the plane region bounded by the graphs of and .
Solution

4. Applied Mathematics II Math 1102 ASTU
Solved Problems On Multiple Integrals Page 4
1
R 2
R
y
x
0 1
-1
x
y 2

 x
y 2

2
3 x
y 

2
9 x
y 

3
-3
3
x
Y
y
x
2
1
R
is a vertically simple region between the graphs of
and for .
is also a vertically simple region between the
graphs of and for 0 .
is also a vertically simple region between the graphs of and for 0 .
Area of is 

 


2
1
1
1
1
R
R
R
dA
dA
dA
A    







0
1
3
2
0
1
3
2
2 2
x
x
x
x
dx
dy
dx
dy
 



 

















0
1
1
0
3
2
3
2
2
2
dx
y
dx
y
x
x
x
x
 
   
 
 








0
1
1
0
2
2
)
2
(
3
)
2
(
3 dx
x
x
dx
x
x
 
 









0
1
1
0
2
2
)
3
2
(
3
2 dx
x
x
dx
x
x
3
5
3
5


3
10

8.Change the integral dx
dy
y
x
x
 



3
3
9
0 2
2
2
1
to an iterated integral in polar coordinates and evaluate it
Solution: is a vertically simple region between the graphs of and √ for
√
In polar coordinates, is a region between the polar graphs of for
Thus, 

d
dr
r
r
dx
dy
y
x
x



 
 


0
3
0
3
3
9
0 2
2
1
1
2


d
dr
 

0
3
0

3

9. Let be the region bounded by the circles and for
Evaluate  
 
R
dA
y
x2
.
Solution      
  



d
dr
r
r
r
dA
y
x
R
   


2
0
2
1
2
2
sin
cos
  



d
dr
r
r
r
  

2
0
2
1
2
2
sin
cos

5. Applied Mathematics II Math 1102 ASTU
Solved Problems On Multiple Integrals Page 5


r
 

 

2
1
r
Y
x
R
z
x
y
R
1

2

  



d
dr
r
r
  

2
0
2
1
2
2
3
sin
cos  



















2
0
2
1
3
4
2
3
)
(sin
4
)
(cos d
r
r
 











2
0
2
sin
3
7
cos
4
15
d  











 





2
0
sin
3
7
2
2
cos
1
4
15
d
 
 












2
0
sin
3
7
2
cos
1
8
15
d





2
0
2
0
cos
3
7
2
2
sin
8
15








 
4
15

10. Find the volume of the solid region bounded by the paraboloid and the plane
Solution :Note that we have solved this problem by using rectangular coordinates and we have seen how
tedious the calculation is.
is a region between the polar graphs of and for
 

 


R
R
dA
y
x
dA
y
x
f
V 2
2
)
,
(   



2
0
2
0
2
d
dr
r
r  



2
0
2
0
3
d
dr
r 

d
r
 








2
0
2
0
4
4
1


d


2
0
4


2
0
4
 
8

11. Find the area of the shaded region shown below.
Solution
Area or is
8
1
3
0
)
(
2
1
0


 



  


d
dr
r
dA
A
R
12. Find the surface area of the portion of the sphere that lies inside the cylinder
.
Solution

6. Applied Mathematics II Math 1102 ASTU
Solved Problems On Multiple Integrals Page 6
x
x
Y
x
R
 
0
,
2  
0
,
4
The surface is the sum of the surface and which are of equal surface areas. is the graph of the
function √ on where is the cylinder .
In polar coordinates, R is a region between the polar graphs of and
for ⁄ ⁄ .
Thus, the surface area S of is given by  
   
 
 


R
dA
y
x
fy
y
x
fx
S 1
,
,
2
2
2
√
and
√
 
























R
dA
y
x
y
y
x
x
S 1
16
16
2
2
2
2
2
2
2
 






R
dA
y
x
y
y
x
x
1
16
16
2 2
2
2
2
2
2
 


R
dA
y
x 2
2
16
4
2




d
dr
r
r
 


 2
2
cos
4
0 2
16
1
8   



d
r
 







 2
2
cos
4
0
2
16
8
  



d
r
 






 2
2
0
cos
4
2
16
8   



d
 




 




 


 2
2
2
2
cos
4
16
0
16
8
  



d



 2
2
2
cos
1
4
4
8
  ,
1
sin
1
32 2
2




d


 since 1
cos
sin 2
2

 
     








 







d
d 2
0
0
2
sin
1
sin
1
32
   













2
0
0
2
cos
cos
32





  
2
32 
 

7. Applied Mathematics II Math 1102 ASTU
Solved Problems On Multiple Integrals Page 7
2
x
y 

2
1
R
Y
x
0
x
x
y 
 2
13. Let be the plane region between the graphs of and . Find the moments of about
the x and the y axes. Also find the centroid of .
Solution
is a vertically simple region between the graphs of and for .
dx
dy
y
dA
y
M
R
x
x
x
x   



 2
1
0
2
2
dx
y x
x
x
 










2
1
0
2 2
2
2
   
 dx
x
x
x
 


 2
1
0
2
2
2
2
2
1
 dx
x
x
 
 2
1
0
2
3
2
2
1 2
1
0
3
4
3
2
2
1










x
x




















8
1
3
1
16
1
2
1
2
1








3
1
4
1
16
1
192
1


dx
dy
x
dA
x
M
R
x
x
x
y   



 2
1
0
2
2
  dx
xy
x
x
x
 










2
1
0
2
2
   
 dx
x
x
x
x
 


 2
1
0
2
2
  dx
x
x
x
 


 2
1
0
2
3
3
  dx
x
x
 

 2
1
0
2
3
2
2
1
0
3
4
3
2 










x
x















8
1
3
1
16
1
2
1









3
1
4
1
8
1
96
1

dx
y
dx
dy
dA
A
x
x
x
R
x
x
x 
   














2
1
0
2
1
0
2
2
2
2
   
 dx
x
x
x
 


 2
1
0
2
2
  dx
x
x
 

 2
1
0
2
2
2
1
0
2
3
2
3
2 










x
x















4
1
2
1
8
1
3
2
24
1

Hence , ,
Therefore, the center of gravity of is at the point ( )

8. Applied Mathematics II Math 1102 ASTU
Solved Problems On Multiple Integrals Page 8
D
z
x
Y
1
2
2
2


 z
y
x
1
2
2

 y
x
1
14. Evaluate the iterated triple integral dx
dy
dz
xyz
y
x
   
1
0
1
0
2
2
2
Solution : dx
dy
z
xy
dx
dy
dz
xyz
y
x
y
x  
   










1
0
1
0
2
2
1
0
1
0
2
2
2
2
2
2
)
(
  dx
dy
y
x
xy
  




 


1
0
1
0
2
2
2
2
dx
dy
xy
y
x
xy
  










1
0
1
0
3
3
2
2
2 dx
xy
y
x
xy
 
















1
0
1
0
4
2
3
2
8
4
dx
x
x
x
 










1
0
3
8
4
dx
x
x
 









1
0
3
4
4
7
8
3

15. Evaluate the iterated triple integral dy
dz
dx
z
x
z
y z
   
2
0 0
3
0 2
2
Solution : dy
dz
z
x
dy
dz
dx
z
x
z y z
y z
 
   
















2
0 0
3
0
1
2
0 0
3
0 2
2
tan    
  dy
dz
y
 




2
0 0
1
1
0
tan
3
tan
dy
dz
y
 

2
0 0 3

dy
z
y
 








2
0
0
3

dy
y


2
0
3

3
2

16. Evaluate  
D
dV
x
y 2
1 where is the region shown in the following figure.
Solution
It is sometimes advantageous to interchange the roles of and .In this example, we interchange the role
of and . is a solid region between the graphs of √ and on , where is a
simple region between the graphs of √ and √ for .
Therefore,    



 





D
x
x z
x
dx
dz
dy
x
y
dV
x
y
1
1
1
1
1
1
2
2
2
2 2
2
1
1
45
28
 (check !)
17. Evaluate 
D
y
dV
e where is the solid region bounded by the planes
and .
Solution : is composed of two subregions and .

9. Applied Mathematics II Math 1102 ASTU
Solved Problems On Multiple Integrals Page 9
x
y 
1

Y
x
1
1
R 2
R
1

y
x
y 

z
x
Y
D
R
2
2
4 y
x
z 



x
R
2
4 x
y 

0 2
Y
is a solid region between the graphs of and on where is a plane region between the
graphs of and for , while is a solid region between the graphs of
and on , where is a plane region between the graphs of and for .
Thus,  
 

1 2
D D
y
y
D
y
dV
e
dV
e
dV
e dx
dy
dz
e
dx
dy
dz
e y
x
y
y
x
y
  
   

 
1
0
1
0
0
1
1
0
   
2
2 


 e
e 4
2 
 e
18. Find the volume of the solid region bounded above by the circular paraboloid and
below by the plane and on the sides by the parabolic sheet and .
Solution : is a solid region between the graphs of and on , where is a
vertically simple plane region between the graphs of and for .
Thus,
 
   




D
x
x
y
x
dx
dy
dz
dV
V
1
0
4
2
2
2
2
1    
 
  



1
0
2
2
2
2
4
x
x
dx
dy
y
x
 
 
  


1
0
2
2
2
2
4
x
x
dx
dy
y
x
105
71
 (check!)
19. Express the triple integral as an iterated integral in cylindrical coordinates and evaluate it: 
D
dV
xz ,
where D is the portion of the ball that is the first octant.
Solution

10. Applied Mathematics II Math 1102 ASTU
Solved Problems On Multiple Integrals Page 10
z
x
Y
D 
r
S
In polar coordinates, is the region between the polar graphs of and for . Thus, in
cylindrical coordinates, is the region between the graphs of and √ for in .
Therefore,  
   


D
r
d
dr
dz
r
z
r
dV
z
x 2
0
2
0
4
0
2
cos


  
  

 2
0
2
0
4
0
2
2
cos



r
d
dr
dz
z
r
  


d
dr
z
r
r2
4
0
2
2
0
2
0
2
2
cos

 
   


d
dr
r
r 4
2
2
0
2
0
4
cos
2
1

   15
32
 (check!)
20. Let D be the solid region in the first octant bounded by the sphere and the planes
√ and Evaluate 
D
dV
z .
Solution
We first determine the ranges of and .
Since D is a first octant region , . √
√
( ) (
√
)
( )
Hence , Now ,since we have
   

D
d
d
d
dV
z 









4
6
2
0
2
4
0
sin
cos 








d
d
d
  

4
6
2
0
4
0
2
5
cos
sin








d
d
  
















4
6
2
0
4
0
2
7
cos
sin
7
2







d
d
sin
cos
7
256 4
6
2
0
 

63
128
 (check!)
21. Find the volume of the solid region bounded above by the sphere and below by the
upper nape of the cone .
Solution : √
, ,

11. Applied Mathematics II Math 1102 ASTU
Solved Problems On Multiple Integrals Page 11
R
3
Y
x
 
z
y
x ,
,
z
x
Y
D
R
x

Volume 

D
dV
V Thus ,   

 





2
0
4
0
2
0
2
sin d
d
d
V




 
d
d
  














2
0
4
0
2
0
3
sin
3



 
d
d
 

2
0
4
0
sin
3
8




d
 









2
0
4
0
cos
3
8
 





2
0
3
2
2
4
d
  
3
2
2
8 

22. Find the center of gravity of the solid region bounded above by the sphere and below
by the plane, is equal to the distance form to the z-axis.
Solution
is the solid region, in spherical coordinates given by
√ √ .
Mass :  


D
dV
z
y
x
m ,
,
  

D
dV
y
x 2
2
 
  

 







2
0
2
0
3
0
2
sin
sin d
d
d
  

 





2
0
2
0
3
0
2
3
sin d
d
d  

 



2
0
2
0
2
sin
4
81
d
d   




 

 



2
0
2
0 2
2
cos
1
4
81
d
d
 
  

 



2
0
2
0
2
cos
1
8
81
d
d 




d
 















2
0
2
0
2
2
sin
8
81



d


2
0 2
8
81


2
2
8
81


 2
8
81


We evaluate the three moments:
 


D
xy dV
z
y
x
z
M ,
,
    
  

 









2
0
2
0
3
0
2
sin
sin
cos d
d
d
  

 






2
0
2
0
3
0
2
4
cos
sin d
d
d  

 



2
0
2
0
2
cos
sin
5
243
d 



d
 
















2
0
2
0
3
3
sin
5
243

12. Applied Mathematics II Math 1102 ASTU
Solved Problems On Multiple Integrals Page 12




2
0
5
81
d
5
162

 


D
xz dV
z
y
x
y
M ,
,
    
  

 










2
0
2
0
3
0
2
sin
sin
sin
sin d
d
d
 
  

 






2
0
2
0
3
0
3
4
sin
sin d
d
d  

 




2
0
2
0
3
sin
sin
5
243
d
d
 
  

 





2
0
2
0
2
sin
sin
cos
1
5
243
d
d 





d
 


















2
0
2
0
3
sin
3
cos
cos
5
243





2
0
sin
5
162
d  


2
0
cos
5
162

 0

 


D
yz dV
z
y
x
x
M ,
,
    
  

 











2
0
2
0
3
0
2
sin
sin
cos
sin d
d
d
  

 






2
0
2
0
3
0
3
4
cos
sin d
d
d = 0 (Check !)
Center of gravity ( ) ( ) ( )