Volumes by Disks and Washers
Volumes by disks and washers are strategies for finding the volume of a solid of revolution. A solid of revolution is formed when a cross
sectional strip (Figure 1a) is rotated around the xy-plane. In the disk and washer method, a strip that is perpendicular to the axis of rotation is
taken so that when it is rotated around the xy-plane it forms a disk or washer, respectively.
In most cases either method works, however, generally, the disk method is reserved for problems where the area bounded by only one
equation/curve is rotated around the xy-plane, whereas the washer method is generally used for problems that involve revolving a region
bounded by two different equations/curves about the xy-plane.
Figure 1a shows the graph of one line with a cross section. When the cross section is rotated around the x-axis it forms a disk (Figure 1b).
Figure 2a shows the graph of two different lines with a cross section between them. When the cross section is rotated around the x-axis, it
forms a ring or washer (Figure 2b). The key to understanding how to set up problems is to understand when each method should be used.
Figure 1a
Figure 1b
Figure 2a
Figure 2b
The formulas for the disk and washer method are shown below. Depending on which axis the curve is being rotated around, the respective
equation will be used.
Volume by Disk Method
Axis of Rotation is Horizontal
Axis of Rotation is Vertical
b
b
V= π ⋅ ∫ [ f ( x)] dx
V= π ⋅ ∫ [ f ( y )]2 dy
2
a
a
V is the volume, a and b are the limits of integration such that 0 ≤ a ≤ b , and f(x)/f(y) is the function/equation given in the problem.
Volume by Washer Method
Axis of Rotation is Horizontal
Axis of Rotation is Vertical
b
b
V= π ⋅ ∫ [( f ( x)) 2 − ( g ( x)) 2 ]dx
V= π ⋅ ∫ [( f ( y )) 2 − ( g ( y )) 2 ]dy
a
a
V is the volume, a and b are the limits of integration such that 0 ≤ a ≤ b , f(x)/f(y) and g(x)/g(y) are the two equations given such that
f ( x) ≥ g ( x) and f ( y ) ≥ g ( y ) .
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Figure 3a and 3b illustrates what it means for f ( x) ≥ g ( x) and f ( y ) ≥ g ( y ) .
Figure 3a
f ( x) ≥ g ( x)
Figure 3b
f ( y) ≥ g ( y)
Generally, for the horizontal axis of rotation, f(x) is the graph of the equation on top and g(x) is the graph of the equation on the bottom. For a
vertical axis of rotation, f(y) is the equation on the right and g(y) is the equation on the left. When setting up problems for the washer method
a good strategy is to think “the top minus the bottom” or “the right minus the left”.
Example 1
The equation y = x 2 revolved about the x-axis. 0 ≤ x ≤ 2
(Note: Graphing the equation helps to set up the problem.)
This problem could be solved easily with the disk method. It asks to revolve the equation about the x-axis, so the cross section is perpendicular
to the x-axis. Notice how, if the cross section is rotated, it forms a disk. The limits of integration are given by 0 ≤ x ≤ 2 . So a=0 and b=2.
f(x) is the given equation x 2 . The setup and solution for the problem should be:
2
( )
2
V= π ⋅ ∫ x 2 dx
Square the Function.
0
2
V= π ⋅ ∫ x 4 ⋅ dx
Integrate.
0
2
⎛ x5 ⎞
V= π ⋅ ⎜⎜ ⎟⎟
⎝ 5 ⎠0
Plug in the limits of integration.
⎡⎛ (2 )5 ⎞ ⎛ (0)5 ⎞⎤
⎟−⎜
⎟
V= π ⋅ ⎢⎜⎜
⎟ ⎜ 5 ⎟⎥
5
⎠ ⎝
⎠⎦⎥
⎣⎢⎝
32
V= ⋅ π
5
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Example 2
Find the volume of the area bounded by the curves y = x 2 and y = x 3 revolved about the x-axis.
Graphing the equations shows that the washer method should be used for this equation. The top equation ( y = x 2 ) is labeled f ( x) and the
bottom equation ( y = x 3 ) is labeled g ( x) . This information is useful to setup the problem. To get the limits of integration, set the two
equations equal to each other and solve for x as shown:
x3 = x2
Subtract x 2 from both sides.
x3 − x2 = 0
Factor x 2 out of the equation.
x 2 ( x − 1) = 0
Set both pieces equal to zero.
x 2 = 0 and (x − 1) = 0
Solve both equations for x to get
x = 0 and x = 1
These are the limits of integration. The problem asks to revolve about the x-axis. The setup and solution of the equation is as follows:
1
[
]
V= π ⋅ ∫ (x 2 ) − (x 3 ) ⋅ dx
2
2
Square both functions.
0
1
[
]
V= π ⋅ ∫ x 4 − x 6 ⋅ dx
Integrate
0
1
⎛ x5 x7 ⎞
V= π ⋅ ⎜⎜ − ⎟⎟
7 ⎠0
⎝ 5
Plug in the limits
⎡⎛ (1)5 (1)7 ⎞ ⎛ (0)5 (0)7 ⎞⎤
⎟
⎟−⎜
V= π ⋅ ⎢⎜⎜
−
⎟ ⎜ 5 − 7 ⎟⎥
5
7
⎠⎦⎥
⎠ ⎝
⎣⎢⎝
2
V= ⋅ π
35
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