Volumes of Revolution
We’ll first look at the area between the lines
y=x, ...
x = 1, . . .
and the x-axis.
yx
0
1
Can you see what shape you will get if you rotate the
area through 360  about the x-axis?
Ans: A cone ( lying on its side )
Volumes of Revolution
We’ll first look at the area between the lines
y=x, ...
x = 1, . . .
and the x-axis.
yx
r
0
h
1
V  13  r 2 h
For this cone,
r  1, h  1

V  13 
Volumes of Revolution
The formula for the volume found by rotating any
area about the x-axis is
y  f ( x)
V 

b
y 2 dx
a
x
a
b
where y  f ( x ) is the curve forming the upper edge
of the area being rotated.
a and b are the x-coordinates at the left- and righthand edges of the area.
We leave the answers in terms of

Volumes of Revolution
V 

b
y 2 dx
a
So, for our cone, using integration, we get
We must substitute
for y using y  f ( x ) before
1 2
we integrate.
V   x dx
0
x
 
 3
 1
 
 3
 1
3
3
yx
1


0

 0

r
0
h
I’ll outline the proof of the formula for you.
1
Volumes of Revolution
The formula can be proved by
splitting the area into narrow strips
. . . which are rotated about the
x-axis.
Each tiny piece is approximately a
cylinder ( think of a penny on its
side ).
Each piece, or element, has a
2
volume
2
 r h   y
y
x
Volumes of Revolution
The formula can be proved by
splitting the area into narrow strips
. . . which are rotated about the
x-axis.
Each tiny piece is approximately a
cylinder ( think of a penny on its
side ).
Each piece, or element, has a
2
volume
2
  r h   y dx
y
x
dx
Volumes of Revolution
The formula can be proved by
splitting the area into narrow strips
. . . which are rotated about the
x-axis.
Each tiny piece is approximately a
cylinder ( think of a penny on its
side ).
Each piece, or element, has a
2
volume
2
y
x
dx
  r h   y dx
The formula comes from adding an infinite number
of these elements.
V 

b
a
y 2 dx
Volumes of Revolution
e.g. 1(a) The area formed by the curve y  x(1  x )
and the x-axis from x = 0 to x = 1 is
rotated through 2 radians about the xaxis. Find the volume of the solid formed.
(b) The area formed by the curve y  e x , the
x-axis and the lines x = 0 and x = 2 is
rotated through 2 radians about the xaxis. Find the volume of the solid formed.
Solution: To find a volume we don’t need a sketch
unless we are not sure what limits of integration
we need. However, a sketch is often helpful.
As these are the first examples I’ll sketch the curves.
Volumes of Revolution
(a) rotate the area between
y  x(1  x ) and the x - axis from 0 to 1.
y  x(1  x )
area
V 

b
y 2 dx
a
A common error in finding
a volume is to get y 2
wrong. So beware!
rotate about
the x-axis
y  x(1  x )
2
2
2
 y  x (1  x)
y 2  x 2 (1  2 x  x 2 )
2
2
3
4
y  x  2x  x
Volumes of Revolution
(a) rotate the area between
y  x(1  x ) and the x - axis from 0 to 1.
y  x(1  x )
V 

b
y2  x2  2x3  x4
2
y dx
a
1

a = 0, b = 1

 V    x 2  2 x 3  x 4 dx
0
Volumes of Revolution
1


 V    x 2  2 x 3  x 4 dx
0
1
x
2x
x 
 



42
5 0
 3
 1 1 1

        0 
 3 2 5

1

30


30
3
4
5
Volumes of Revolution
(b) Rotate the area between
y  e x and the lines x = 0 and x = 2.
y  ex
x2
V 

b
a
y 2 dx
 
y  e
2
e 
x 2
x 2

 e x  e x  e2x
Volumes of Revolution
(b) Rotate the area between
y  e x and the lines x = 0 and x = 2.
y  ex
x2
V 

b
a
y 2 dx
 
y  e
2
2
 V    e 2 x dx
0
x 2
e
2x
Volumes of Revolution
2
V   e
0
2x
dx
2
e 
 

 2 0
2x
0 
 e4
e

  


2
2


Remember that
4
 e4


1
e 1

  
    

2
2
2




e0  1
Volumes of Revolution
Exercise
1
1(a) The area formed by the curve y 
the x-axis
x
and the lines x = 1 to x = 2 is rotated through 2
radians about the x-axis. Find the volume of
the solid formed.
(b) The area formed by the curve y  x x ,
the x-axis and the lines x = 0 and x = 2 is
rotated through 2 radians about the xaxis. Find the volume of the solid formed.
Volumes of Revolution
Solutions:
1 , the x-axis and the lines x = 1 and x = 2.
1. (a) y 
x
V 

b
a
y dx  V  
2

 V  
2
1
2
1
1
x
2
dx
x 2 dx
2
x 
 V  

 1  1
2
 1
 V   
 x 1
1
Volumes of Revolution
Solutions:
2
 1
V   
 x 1



V   

V 

2

 1






1

 2



Volumes of Revolution
(b)
y  x x , the x-axis and the lines x = 0
and x = 2.
Solution:
V 
yx x

b
2
y dx
y2  x x  x x  x3
a
 V  
2
x
0
3
2
x 
 V   
 4 0
 V  4
4
dx
Volumes of Revolution
Rotation about the y-axis
To rotate an area about the y-axis we use the
same formula but with x and y swapped.
b
V    y dx
a
2
d
V    x 2 dy
c
Tip: dx for rotating about the x-axis;
dy for rotating about the y-axis.
The limits of integration are now values of y giving
the top and bottom of the area that is rotated.
As we have to substitute for x from the equation of
the curve we will have to rearrange the equation.
Volumes of Revolution
e.g. The area bounded by the curve y  x , the
y-axis and the line y = 2 is rotated through 360 
about the y-axis. Find the volume of the
solid formed.
y2
y
V  
d
c
y x
 x  y2
 x2  y4
x
2
4

V


y
x dy
0 dy
2
Volumes of Revolution
2
V    y 4 dy
0
2
 y 
 

 5 0
 5

 2

  
   0 
 5 




5

32
5
Volumes of Revolution
Exercise
1(a) The area formed by the curve y  x 2 for x  0
the y-axis and the line y = 3 is rotated through
2 radians about the y-axis. Find the
volume of the solid formed.
1
(b) The area formed by the curve y 
, the
x
y-axis and the lines y = 1 and y = 2 is rotated
through 2 radians about the y-axis. Find the
volume of the solid formed.
Volumes of Revolution
Solutions:
(a)
y  x 2 for x  0 , the y-axis and the line y = 3.
yx
 V 
d
V    x dy
2
c
3
0 y dy
3
 y 
9
 
 
2
 2 0
2
2
Volumes of Revolution
1
(b) y  , the y-axis and the lines y = 1 and y = 2.
x
Solution:
d
V    x 2 dy
c
 V 
2
1
1
2
dy
y
2
 1
  
 y 1

  

1
1
y
 x
x
y
1
2
 x  2
y
  1    1   



 2 
2

Volumes of Revolution