Math Core

The disc method needs the region to touch the axis of rotation. When there is a gap between the region and the axis, the solid has a hole through the middle, like a bead or a bundt cake. Every cross section is then a washer: a disc with a smaller disc removed. This is the last and most complete volume tool in AP Calculus AB.

Outer radius and inner radius

Revolve the region between $y = f(x)$ and $y = g(x)$ , with $f(x) \ge g(x) \ge 0$ , around the $x$ -axis. A vertical slice at $x$ no longer starts on the axis. As it spins, its far end traces a circle of radius $R(x) = f(x)$ and its near end traces a circle of radius $r(x) = g(x)$ . The cross section is the ring between them, with area

A(x) = \pi R^2 - \pi r^2 = \pi\big(R(x)^2 - r(x)^2\big).

The washer method

When a region is revolved around an axis and each slice perpendicular to the axis leaves a gap next to the axis,

V = \pi\int_a^b \Big(\big[R(x)\big]^2 - \big[r(x)\big]^2\Big)\,dx,

where $R$ is the outer radius (axis to the farther boundary) and $r$ is the inner radius (axis to the nearer boundary). For a vertical axis, use $dy$ and write both radii in terms of $y$ .

If the region touches the axis, the inner radius is 0 and the washer formula becomes the disc formula. So the washer method includes the disc method as a special case.

Region between y = x and y = x², revolved around y = −1. At each x, the outer radius reaches the line and the inner radius reaches the parabola.Open in grapher →

Finding the radii for any axis

Each radius is a distance from the axis, so it is always "far minus near." For a horizontal axis $y = k$ :

If the region is above the axis: $R = (\text{top curve}) - k$ and $r = (\text{bottom curve}) - k$ .
If the region is below the axis: $R = k - (\text{bottom curve})$ and $r = k - (\text{top curve})$ .

Notice that when the region is below the axis, the bottom curve gives the outer radius. The same logic works for a vertical axis $x = h$ , with "right" and "left" in place of "top" and "bottom."

Common mistake

The integrand is $R^2 - r^2$ , not $(R - r)^2$ . Subtracting the radii first and then squaring computes the volume of a completely different solid. Square each radius separately, then subtract.

Worked examples

Worked example: Around the x-axis

The region between $y = x$ and $y = x^2$ is revolved around the $x$ -axis. Find the volume.

Solution. The curves meet at $x = 0$ and $x = 1$ , and $y = x$ is on top. Around the $x$ -axis, $R = x$ and $r = x^2$ .

V = \pi\int_0^1 \big(x^2 - x^4\big)\,dx = \pi\left(\frac{1}{3} - \frac{1}{5}\right) = \frac{2\pi}{15}.

Worked example: Around the y-axis

The same region is revolved around the $y$ -axis. Find the volume.

Solution. Use horizontal slices, $0 \le y \le 1$ . Rewrite the curves: $y = x$ becomes $x = y$ , and $y = x^2$ becomes $x = \sqrt{y}$ . For $0 \lt y \lt 1$ , $\sqrt{y} > y$ , so the parabola is farther from the $y$ -axis: $R = \sqrt{y}$ and $r = y$ .

V = \pi\int_0^1 \big(y - y^2\big)\,dy = \pi\left(\frac{1}{2} - \frac{1}{3}\right) = \frac{\pi}{6}.

Worked example: Around a line below the region

The same region is revolved around the line $y = -1$ . Find the volume.

Solution. The region is above the axis, so add 1 to each curve: $R = x + 1$ and $r = x^2 + 1$ .

\begin{aligned} V &= \pi\int_0^1 \Big((x + 1)^2 - (x^2 + 1)^2\Big)\,dx = \pi\int_0^1 \big(-x^4 - x^2 + 2x\big)\,dx \\ &= \pi\left(-\frac{1}{5} - \frac{1}{3} + 1\right) = \frac{7\pi}{15}. \end{aligned}

Worked example: Around a line above the region

The same region is revolved around the line $y = 2$ . Find the volume.

Solution. Now the region is below the axis. The farther curve is the lower one, $y = x^2$ , so $R = 2 - x^2$ and $r = 2 - x$ .

\begin{aligned} V &= \pi\int_0^1 \Big((2 - x^2)^2 - (2 - x)^2\Big)\,dx = \pi\int_0^1 \big(x^4 - 5x^2 + 4x\big)\,dx \\ &= \pi\left(\frac{1}{5} - \frac{5}{3} + 2\right) = \frac{8\pi}{15}. \end{aligned}

One region, four axes, four different volumes. The work is always the same: identify $R$ and $r$ as distances, then integrate $\pi(R^2 - r^2)$ .

Tip

A quick check: $R$ must be at least $r$ everywhere on the interval. If your integrand $R^2 - r^2$ is negative at a test point, you swapped the radii.

Practice

Practice 1

The region between $y = \sqrt{x}$ and $y = x$ is revolved around the $x$ -axis. Find the exact volume.