Exercises
Section 1.4 – Area of Surfaces of Revolution
1.
Find the lateral (side) surface area of the cone generated by revolving the line segment
y  x , 0  x  4 , about the x-axis. Check your answer with the geometry formula
2
Lateral surface area  1  ba se circumference  slant height
2
2.
Find the lateral surface area of the cone generated by revolving the line segment y  x , 0  x  4 ,
2
about the y-axis. Check your answer with the geometry formula
Lateral surface area  1  ba se circumference  slant height
2
3.
Find the lateral surface area of the cone frustum generated by revolving the line segment
y  x  1 , 1  x  3 , about the x-axis. Check your answer with the geometry formula
2 2
1
Frustum surface area   r  r
4.
2
  slant height
Find the lateral surface area of the cone frustum generated by revolving the line segment
y  x  1 , 1  x  3 , about the y-axis. Check your answer with the geometry formula
2 2
1
Frustum surface area   r  r
2
  slant height
5.
3
Find the area of the surface generated by y  x , 0  x  2, x  axis
9
6.
Find the area of the surface generated by y  2 x  x2 , 0.5  x  1.5, x  axis
7.
Find the area of the surface generated by y  x  1, 1  x  5, x  axis
8.
Find the area of the surface generated by x  2 4  y
32
0  y  15 ,
4
y  axis
9.
Find the area of the surface generated by x  2 y  1 5  y  1,
8
10.
y  1 x2  2
3


3/2
, 0  x  2;
evaluate the integral S 
11.

y  axis
y  axis (Hint: Express ds  dx 2  dy 2 in terms of dy, and
2 y ds with appropriate limits.)
Did you know that if you can cut a spherical loaf of bread into slices of equal width, each slice will
have the same amount of crust? To see why, suppose the semicircle y  r 2  x 2 shown here is
revolved about the x-axis to generate a sphere. Let AB be an arc of the semicircle that lies above an
interval of length h on the x-axis. Show that the area swept out by AB does not depend on the
location of the interval. (It does depend on the length of the interval.)
33
Section 1.4 – Area of Surfaces of Revolution
Solution
Exercise
Find the lateral (side) surface area of the cone generated by revolving the line segment y  x , 0  x  4 ,
2
about the x-axis. Check your answer with the geometry formula
Lateral surface area  1  ba se circumference  slant height
2
Solution
dy 1

dx 2
2
 dy 
5
1    1 1 
4
2
 dx 
b
S  2


a
2
dy
y 1    dx
 dx 
4
 2
0
 5

2
 x  5 dx
2 2

4
xdx
0
4
  5 1 x2
2 2
0

  5 42  0
4

 4 5
ba se circumference  2 r  2  2   4
slant height  42  22  20  2 5
Lateral surface area  1  ba se circumference  slant height
2
 1   4   2 5
2


 4 5
32
Exercise
Find the lateral surface area of the cone generated by revolving the line segment y  x , 0  x  4 , about
2
the y-axis. Check your answer with the geometry formula
Lateral surface area  1  ba se circumference  slant height
2
Solution
x  0  y  0
y  x  x  2y  
2
x  4  y  2
dx  2
dy
2
 dy 
1    1 4  5
 dx 
S  2


d
2
x 1   dx  dy
 dy 
c
2
 2
2 y 5dy
0
 4 5

2
ydy
0
2
 4 5 1 y 2
2
0
 2 5  4  0 
 8 5
ba se circumference  2  4   8
slant height  42  22  20  2 5
Lateral surface area  1  ba se circumference  slant height
2
 1  8   2 5
2


 8 5
33
Exercise
Find the lateral surface area of the cone frustum generated by revolving the line segment
y  x  1 , 1  x  3 , about the x-axis. Check your answer with the geometry formula
2 2
1
Frustum surface area   r  r
2
  slant height
Solution
dy 1

dx 2
2
 dy 
5
1    1 1 
dx
4
2
 
b
S  2



a
3
 2
2
dy
y 1    dx
 dx 

x  1  5  dx
2 2  2 
1
3
 5
2
x  1 dx
1
3
  5  1 x 2  x 
2 2
1



5 1 3 2  3  1 1 2  1 
    2    
2  2


5 9  3  3
2  2
2 

5 6
 
2
 3 5
r  1  1 1 r  3  1  2
1
2
2
slant height 
2
2
2
 2  12  3  12
1
Frustum surface area   r  r
2
 5
  slant height
  1  2  5
 3 5
34
Exercise
Find the lateral surface area of the cone frustum generated by revolving the line segment
y  x  1 , 1  x  3 , about the y-axis. Check your answer with the geometry formula
2 2
1
Frustum surface area   r  r
2
  slant height
Solution
y  x  1  2y  x 1  x  2y 1
2 2
dx  2
dy
2
 
1   dx   1  4  5
 dy 
2
S  2


2 y  1
 5  dy
1
2
 2 5
2 y  1 dy
1
2
 2 5  y 2  y 

1


2
 2 5  2   2  12  1 


 4 5
r 1 r  3
1
2
slant height 
 2  12  3  12
1
Frustum surface area   r  r
2
 5
  slant height
  1  3 5
 4 5
35
Exercise
3
Find the area of the surface generated by y  x , 0  x  2, x  axis
9
Solution
dy 1 2
 x
dx 3
2
 dy 
1     1  1 x4  1 9  x4
9
3
 dx 

S  2
2
0
 2
27
 2
27

54
x 3 1 9  x 4 dx
9 3

2

25

25
x 3 9  x 4 dx
u  9  x4
 du  4 x3dx  1 du  x3dx
0
4 
u1/2 1 du
9
 x  2  u  25

x  0  u  9
u1/2 du
9
25
  2 u3/2
54 3
9

  253/2  93/2
81

 98
81
36
4
Exercise
Find the area of the surface generated by y  x  1, 1  x  5, x  axis
Solution
y  x  1   x  1
1/2
dy 1
1/2
  x  1
2
dx
2
 dy 
1
1     1  1  x  1
4
 dx 
 1
1
4 x1
 4x  4  1
4  x  1
 1 4x  5
2 x 1
S  2

5
1


5

25
x  1 1 4 x  5 dx
2 x 1
4 x  5 dx
1

9
 x  5  u  25
u  4 x  5  du  4dx 
x 1  u  9
1 u1/2 du
4
25
  2 u3/2
43
9

  253/2  93/2
6

   98
6
 49
3
37
Exercise
Find the area of the surface generated by x  2 4  y
0  y  15 ,
4
Solution
dy
1/2
 2 1 4  y 
 1  1
2
dx
4 y
2
 dy 
1    1 1
4 y
 dx 
S  2

4  y 1
4 y

5 y
4 y

15/4

15/4

15/4
2 4 y
0
 4
5 y
4 y
dy
5  y dy
d 5  y   dy
0
 4
5  y 1/2  d 5  y  
0
3/2 15/4
 4 2  5  y 
3
0


3/2

3/2 
  8  5  15
 5  0 
4
3 


 3/2 3/2 
  8  5
5 
3  4



  8  5 5  5 5 
3  8

 
  8 5 5   7 
8
3
  8 5 5 1  1
8
3
 35 5
3
38
y  axis
Exercise
Find the area of the surface generated by x  2 y  1 5  y  1,
8
Solution
dy 1
1/2
  2 y  1
2  1
2
dx
2 y 1
2
 dy 
1    1 1
2y 1
 dx 

S  2

1

1

1
2y
2y 1
2y
2 y 1
2y 1
5/8
 2
dy
u  2 y  du  2dy
2 y dy
5/8
 2
2 
u1/2 1 du
5/8


1
u1/2 du
5/8
 2  2 y 
3
3/2 1
5/8

 3/2 5 3/2 
 2   2 


4
3 



 2  2 2  5 5 
8 
3 


 2  16 2 5 5 
8
3 


  16 2  5 5
12

39
y  axis
Exercise

y  1 x2  2
3

3/2
, 0  x  2;

the integral S 
y  axis (Hint: Express ds  dx 2  dy 2 in terms of dx, and evaluate
2 x ds with appropriate limits.)
Solution

dy  1 3 x 2  2
32

1/2
 2 x  dx  x


ds  dx   x x 2  2 dx 


2

x 2  2 dx
2

 dx 2  x 2 x 2  2 dx 2
 1  x 4  2 x 2 dx

1  x2



 dx
2
 1  x 2 dx
S

 2
2 x ds


2
0
 2


x 1  x 2 dx
 x  2  u  3
u  1  x 2  du  2 xdx  
 x  0  u  1
3
1
1 u du
2
3
  1 u2
2

1
  32  12
2

  8
2
 4
40
Exercise
Did you know that if you can cut a spherical loaf of bread into slices of equal width, each slice will have
the same amount of crust? To see why, suppose the semicircle y  r 2  x 2 shown here is revolved
about the x-axis to generate a sphere. Let AB be an arc of the semicircle that lies above an interval of
length h on the x-axis. Show that the area swept out by AB does not depend on the location of the interval.
(It does depend on the length of the interval.)
Solution
y  r 2  x2
dy
2 x
x


dx 2 r 2  x 2
2
r  x2
2
2
 dy 
1    1 x
 dx 
r2  x2
r2
r2  x2
r

r2  x2

S  2

a h
r2  x2
a
 2 r

r
r2  x2
dx
a h
dx
a
 2 rx
a h
a
 2 r  a  h  a 
 2 rh
41