The document discusses calculating volumes of revolution by rotating an area about the x-axis or y-axis. It provides the formulas for finding these volumes using integration, with examples of setting up the integrals to calculate specific volumes. It also covers cases where the curve needs to be rearranged in order to substitute it into the integral when rotating about the y-axis.

1. Volume of Revolution

2. We’ll first look at the area between the lines
y = x , . . .
• x = 1, . . .
• and the x-axis.
• Can you see what shape you will get if you rotate the area
through about the x-axis?

3. x = 1, . . .
hrV 2
3
1

We’ll first look at the area between the lines
y = x , . . .
and the x-axis.

4. The formula for the volume found by
rotating any area about the x-axis is
a b
x
)(xfy 
dxyV
b
a 2

where is the curve forming the upper edge of the
area being rotated.
)(xfy 
a and b are the x-coordinates at the left- and right-hand edges of the
area.
We leave the answers in terms of 

5. r
h
0 1
So, for our cone, using integration, we get
dxV 
1
0
 2
x
1
0
3
3







x







  0
3
1
 
3
1
xy 
We must substitute for y using before
we integrate.
dxyV
b
a 2


6. x
y
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 volume
hr 2
  2
y dx
The formula comes from adding an infinite number of
these elements.
dxyV
b
a 2


7. 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.
)1( xxy e.g. 1(a) The area formed by the curve
and the x-axis from x = 0 to x = 1 is
rotated through radians about the x-
axis. Find the volume of the solid formed.
2

8. )1( xxy 
dxyV
b
a 2

area rotate about
the x-axis
A common error in finding
a volume is to get
wrong. So beware!
2
y
)1( xxy 
222
)1( xxy 
)21( 222
xxxy 
4322
2 xxxy 
(a) rotate the area between
.10)1( tofromaxis-theand xxxy 

9. )1( xxy 
dxyV
b
a 2

4322
2 xxxy 
 dxxxxV  
1
0
432
2
a = 0, b = 1
(a) rotate the area between
.10)1( tofromaxis-theand xxxy 

10.  dxxxxV  
1
0
432
2
1
0
543
54
2
3







xxx

 











 0
5
1
2
1
3
1

30
1
 
30


2

11. Volumes of Revolution
To rotate an area about the y-axis we use the
same formula but with x and y swapped.
dxyV
b
a 2
 dyxV
d
c 2

The limits of integration are now values of y giving
the top and bottom of the area that is rotated.
Rotation about the y-axis
As we have to substitute for x from the equation of
the curve we will have to rearrange the equation.

12. Volumes of Revolution
xy 
dyxV
d
c 2

e.g. The area bounded by the curve , the
y-axis and the line y = 2 is rotated through
about the y-axis. Find the volume of the
solid formed.
xy 

360
2y
dyyV 
2
0
4

xy 
2
yx 
42
yx 

13. Volumes of Revolution
dyyV 
2
0
4

2
0
5
5







y

 








 








0
5
52

5
32


14. Exercise
2
xy  0x
the y-axis and the line y = 3 is rotated through
radians about the y-axis. Find the
volume of the solid formed.
2
1(a) The area formed by the curve for
x
y
1
(b) The area formed by the curve , the
y-axis and the lines y = 1 and y = 2 is rotated
2through radians about the y-axis. Find the
volume of the solid formed.

15. 2
xy 
Solutions:
2
xy (a) for , the y-axis and the line y = 3.0x

3
0
dyyV 
dyxV
d
c 2

3
0
2
2







y

2
9


16. Solution:

2
1 2
1
dy
y
V 
dyxV
d
c 2

2
1
1







y

 





 


  1
2
1

(b)
x
y
1
 , the y-axis and the lines y = 1 and y = 2.
2


y
x
x
y
11

2
2 1
y
x 