Volumes of Revolution: Disk and Washer Methods
Compute volumes of solids of revolution by integrating cross-sectional areas. Apply the disk method V = π∫[f(x)]²dx and the washer method V = π∫([f(x)]²−[g(x)]²)dx for regions with a hole.
A lesson of Calculus, in Applications of Integration, about 10 minutes.
By the end of this lesson, you will be able to find volumes of solids of revolution using the Disk and Washer methods by integrating circular cross-sections.
The Big Idea
Rotating a region in the plane about a line creates a 3D solid. Each cross-section perpendicular to the axis of rotation is a circle (disk) or an annulus (washer). Integrating the area of these cross-sections gives the volume.
Disk Method (Revolution About $x$-Axis)
When the region is between $y = f(x)$ and the $x$-axis, rotated about the $x$-axis:
$$V = \int_a^b \pi[f(x)]^2,dx$$
Each cross-section is a disk of radius $f(x)$ and area $\pi[f(x)]^2$.
Washer Method (Two Curves, Revolution About $x$-Axis)
When the region is between $y = f(x)$ (outer) and $y = g(x)$ (inner), $f \geq g \geq 0$:
$$V = \int_a^b \pi\left([f(x)]^2 - [g(x)]^2\right),dx$$
Each cross-section is a washer with outer radius $f(x)$ and inner radius $g(x)$.
Revolution About $y$-Axis (Integrating with Respect to $y$)
For $x = g(y)$ rotated about the $y$-axis:
$$V = \int_c^d \pi[g(y)]^2,dy$$
Example 1 — Disk Method
Rotate $y = \sqrt{x}$ on $[0, 4]$ about the $x$-axis. Find the volume.
$V = \int_0^4 \pi(\sqrt{x})^2,dx = \pi\int_0^4 x,dx = \pi\left[\frac{x^2}{2}\right]_0^4 = \pi(8) = 8\pi$
Example 2 — Washer Method
Region between $y = x^2$ and $y = x$, $0 \leq x \leq 1$, rotated about the $x$-axis.
Outer radius: $R = x$ (the line is above the parabola on $[0,1]$); Inner radius: $r = x^2$.
$V = \pi\int_0^1(x^2 - x^4),dx = \pi\left[\frac{x^3}{3} - \frac{x^5}{5}\right]_0^1 = \pi\left(\frac{1}{3} - \frac{1}{5}\right) = \frac{2\pi}{15}$
Example 3 — Revolution About $y = k$
Rotate $y = x^2$ on $[0, 2]$ about the line $y = -1$.
The axis of rotation is shifted. The radius from the axis to the curve is the distance from $y = -1$ to $y = x^2$, which is $x^2 - (-1) = x^2 + 1$.
$V = \pi\int_0^2(x^2+1)^2,dx = \pi\int_0^2(x^4+2x^2+1),dx$
$= \pi\left[\frac{x^5}{5} + \frac{2x^3}{3} + x\right]_0^2 = \pi\left[\frac{32}{5} + \frac{16}{3} + 2\right] = \pi\left[\frac{96+80+30}{15}\right] = \frac{206\pi}{15}$
Example 4 — Washer About $y$-Axis
Region between $x = \sqrt{y}$ and $x = y^2$, rotated about the $y$-axis ($0 \leq y \leq 1$).