Volumes
using Disks
Limerick Nuclear Generating Station, Pottstown, Pennsylvania
Photo by Vickie Kelly, 2003
Greg Kelly, Hanford High School, Richland, Washington
2
y
Suppose I start with this curve.
x
And we want to make a cone
using this as the shape.
1
0
1
2
3
4

2
y
How could we find the volume
of the cone?
x
One way would be to cut it into a
series of thin slices (flat cylinders)
and add their volumes.
1
0
1
2
3
4
The volume of each flat
cylinder (disk) is:
 r  the thickness
2

 x
2
dx
In this case:
r= the y value of the function
thickness = a small change
in x = dx

2
y
The volume of each flat
cylinder (disk) is:
x
 r  the thickness
1
0
2
1
2
3
The Rectangle is
Perpendicular
to the Axis
4



x
2
dx
If we add the volumes, we get:

4


 x

0

2

4
2
dx
 x dx
0
4
x
 8
2
0

This application of the method of slicing is called the
disk method. The shape of the slice is a disk, so we
use the formula for the area of a circle to find the
volume of the disk.
If the shape is rotated about the x-axis, then the formula is:
Volume 

b
 f ( x ) dx
2
a
The Rectangle is Perpendicular to the X-Axis
If the Rectangle is
Perpendicular to the Y-Axis
V 

b
 f ( y ) dy
2
a

The region between the curve x 
1
y
,
1 y  4
and the
y-axis is revolved about the y-axis. Find the volume.
y
x
1
1
2
3
4
1
3
 .707
2
2
1
3
We use a horizontal disk.
The thickness is dy.
4
 .577
The radius is the x value of the
1
function 
.
dy
y
1
1
2
0
1
V 

4
1
 1 


 y 


2

dy

4
1

1
dy
y
volume of disk
0
  ln y 1    ln 4  ln 1 
4
  ln 4

Find the volume of rotating the area
enclosed by the three restrictions about:
the x-axis and x = 2
y x
y0
x2
2
Answers:
x-axis
32 
5
x=2
8
3


Food items that have a disk-based volume.
Hamburgers
French Bread
Cookies
Volumes using Shells
(Washers)
Limerick Nuclear Generating Station, Pottstown, Pennsylvania
Photo by Vickie Kelly, 2003
Greg Kelly, Hanford High School, Richland, Washington
y
The natural draft cooling tower
shown at left is about 500 feet
high and its shape can be
approximated by the graph of
this equation revolved about
the y-axis:
5 0 0 ft
x  .000574 y  .439 y  185
2
x
The volume can be calculated using the disk method with
a horizontal disk.

  .000574 y
500
0
2
 .439 y  185 
2
dy  24, 700, 000 ft 3

4
3
y  2x
2
y x
2
The region bounded by
2
y  x and y  2 x is
revolved about the y-axis.
Find the volume.
1
If we use a horizontal slice:
y x
y  2x
2
y
y  x
0
1
2
 x
2
The volume of the washer is:
V 

4
0
V 





4
0
V 

y

2
2
 y 
    dy
 2  
1 2

  y  y  dy
4



4
0
y
1
4
The “disk” now has a hole in
it, making it a “washer”.
 R
2
r
 R  r
2
outer
radius
4
2
y dy
1 3
1 2
  y 
y 
12
2
0
2
2
  thickness
 dy
inner
radius
16 

  8 
3 


8
3

This application of the method of slicing is called the
washer method. The shape of the slice is a circle
with a hole in it, so we subtract the area of the inner
circle from the area of the outer circle.
The washer method formula is:
V 

b
R r
2
2
dx
a
1) Draw the graphs
2) Draw two rectangles perpendicular to the axis
3) Integrate Outside - Inside

y x
4
3
2
y  2x
If the same region is
rotated about the line x=2:
The outer radius is:
2
R  2
1
0
1
2
2
The inner radius is:
r
y x
y  2x
2
y
y  x
R
r  2
 x
2
V 

4

R  r dy
2



4
0

0

4
4 2y 
y
y
2
44

0
y

2   2
2


2

y 
4 2y 
 44
4 


y

2
dy
y  y dy
4
0
2
2
4
y

4
0
3 y 
1
1
y  4 y 2 dy
2
4

y  y dy
4
 3 2
1 3 8 2
    y 
y  y 
12
3
 2
0
3
16 64 

     24 

3
3 


8
3

5
5
4
3
y  x 1
2
2
Find the volume
1  dthe
y  4region

1 4   y of
2
bounded by y  x  1 , x  2 ,
5
and y  0  revolved
the y5  y d y  4about

1
axis.
5
1 2

  5 y  y   4
2

1
1
0
2
1
We can use the washer method if we split
25  it into1 two
  parts:
y 1  x
5
 2 
2
1
outer
radius
2

x
y 1

inner
radius
2
   25 

y 1
dy    2  1
2
   5     4
2  
2 
 25 9 
 
   4
2
 2
cylinder
thickness
of slice

16
 4
2
8  4 
 1 2
