7.3 day 2
Disk and Washer Methods
Limerick Nuclear Generating Station, Pottstown, Pennsylvania
Photo by Vickie Kelly, 2003
Greg Kelly, Hanford High School, Richland, Washington
2
Suppose I start with this curve.
y x
My boss at the ACME Rocket
Company has assigned me to
build a nose cone in this shape.
1
0
1
2
3
4
So I put a piece of wood in a
lathe and turn it to a shape to
match the curve.

2
How could we find the volume
of the cone?
y 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 2  the thickness

 x
2
dx
In this case:
r= the y value of the function
thickness = a small change
in x = dx

2
The volume of each flat
cylinder (disk) is:
y x
 r 2  the thickness
1

0
1
2
3
 x
2
dx
4
If we add the volumes, we get:
   x
2
4
0
dx
4
   x dx
0


2
4
 8
x2
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:
b
V    y 2 dx
a
Since we will be using the disk method to rotate shapes
about other lines besides the x-axis, we will not have this
formula on the formula quizzes.
b
A shape rotated about the y-axis would be:
V    x 2 dy
a

1
The region between the curve x 
, 1  y  4 and the
y
y-axis is revolved about the y-axis. Find the volume.
y
x
1
1
2
3
4
1
 .707
2
1
 .577
3
1
2
We use a horizontal disk.
The thickness is dy.
4
3
2
The radius is the x value of the
1
function 
.
dy
y
1
2
 1 
V  
dy
 y 
1


4
 
4
0
1
1
1
dy
y
volume of disk
0
2
  ln y 1    ln 4  ln1   ln 2  2 ln 2
4

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:
500 ft
x
x  .000574 y 2  .439 y  185
The volume can be calculated using the disk method with
a horizontal disk.

500
0
.000574 y
2
 .439 y  185 dy  24, 700, 000 ft 3
2

4
3
y  2x
2
y  x2
The region bounded by
y  x 2 and y  2 x is
revolved about the y-axis.
Find the volume.
1
If we use a horizontal slice:
yx
y  2x
y
x
2
2
yx
0
1
2
The volume of the washer is:

V  
0


4
 y
2
 y
 
2
2

 dy

1 

V     y  y 2  dy
0
4 

4
V 
4
0
1 2
y  y dy
4
The “disk” now has a hole in
it, making it a “washer”.
 R   r   thickness
  R  r  dy
2
2
2
2
outer
radius
4
1 
1
   y 2  y3 
12  0
2
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:
b
V    R 2  r 2 dx
a
Like the disk method, this formula will not be on the
formula quizzes. I want you to understand the formula.

y  x2
4
3
y  2x
2
1
0
1
r
y  2x
y
x
2
y  x2
yx
2
r  2 y
y2
   4  2 y   4  4 y  y dy
0
4
1
4
1 2
   3 y  y  4 y 2 dy
0
4
4
V    R2  r 2 dy
0
2

y

  2   2 y
0
2

 dy
2


4
 3 2 1 3 8 
    y  y  y 
12
3 0
 2
16 64 
8

    24    
3 3
3

3
2

y2 
    4  2 y    4  4 y  y dy
0
4 

4
The outer radius is:
y
R  2
2
The inner radius is:
R
4
4
If the same region is
rotated about the line x=2:
