The Washer Method The Washer Method
Concept
The washer method is used for solids with cross-sections which are discs with holes in the middle, like a washer. The outer ring of the washer is
f(x)
and the inner ring of the washer is
g(x)
.
Therefore,
\begin{align*}
A(x) &= \text{Area of }f(x) - \text{Area of }g(x) \\
&= \pi f(x)^2 - \pi g(x)^2
\end{align*}
Therefore,
\text{Volume} = \pi \int_{a}^{b} f(x)^2 - g(x)^2 \, dx
Examples
Example 1: Basic x-Axis Rotation
Find the volume formed by rotating the region between
y = x
and
y = x^2
around the x-axis (
0 \leq x \leq 1
).
\begin{align*}
V &= \pi \int_{0}^{1} \left[(x)^2 - (x^2)^2\right] dx \\
&= \pi \int_{0}^{1} (x^2 - x^4) dx \\
&= \pi \left[ \frac{x^3}{3} - \frac{x^5}{5} \right]_0^1 \\
&= \pi \left( \frac{1}{3} - \frac{1}{5} \right) = \frac{2\pi}{15} \approx 0.42 \, \text{units}^3
\end{align*}
The volume of the solid is
2\pi/15
cubic units.
Revolving Around Other Axes
Same as the disc method, we can find the volume of solids revolving around any line parallel to the axes by shifting our functions to have the x-axis or y-axis as their axis of rotation.
Horizontal Line: If we are revolving around
y = k
:
V = \pi \int_{a}^{b} \left[ (f(x) - k)^2 - (g(x) - k)^2 \right] dx
Vertical Line: If we are revolving around
x = h
:
V = \pi \int_{a}^{b} \left[ (f(y) - h)^2 - (g(y) - h)^2 \right] dx
Advanced Examples
Example 2: Rotation Around y = 2
Rotate the region between
y = x^2
and
y = 1
around
y = 2
(
0 \leq x \leq 1
).
\begin{align*}
V &= \pi \int_{0}^{1} \left[(x^2 - 2)^2 - (1 - 2)^2\right] dx \\
&= \pi \int_{0}^{1} (x^4 - 4x^2 + 4 - 1) dx \\
&= \pi \int_{0}^{1} (x^4 - 4x^2 + 3) dx \\
&= \pi \left[ \frac{x^5}{5} - \frac{4x^3}{3} + 3x \right]_0^1 \\
&= \pi \left( \frac{1}{5} - \frac{4}{3} + 3 \right) = \frac{28\pi}{15} \approx 5.89 \, \text{units}^3
\end{align*}
The volume of the solid is
28\pi/15
cubic units.
Example 4: Rotation Around x = 3
Rotate the region bounded by
x = \sqrt{y}
and
x = 1
around
x = 3
(
0 \leq y \leq 1
).
\begin{align*}
V &= \pi \int_{0}^{1} \left[(\sqrt{y} - 3)^2 - (1-3)^2\right] dy \\
&= \pi \int_{0}^{1} \left(y - 6\sqrt{y} + 9 - 4\right) dy \\
&= \pi \int_{0}^{1} (y - 6\sqrt{y} + 5) \, dy \\
&= \pi \left[\frac{y^2}{2} - 4y^{3/2} + 5y\right]_0^1 \\
&= \pi \left(\frac{1}{2} - 4 + 5\right) = \frac{3\pi}{2} \approx 4.71 \, \text{units}^3
\end{align*}
The volume is
3\pi/2
cubic units.