Volumes of Revolution — the Washer Method

When a region doesn’t touch the axis it’s spinning around, the solid has a hole through the middle, like a bagel or a pipe. Each slice is then a ring, called a washer (like the flat metal washers in a hardware store). The disc method still works with one change: subtract the area of the hole.

Key ideas

Each slice is a washer

A washer is a disc with a smaller disc removed. If the outer radius is $R$ and the inner radius is $r$, its area is

$$\pi R^2 - \pi r^2 = \pi\left( R^2 - r^2 \right)$$

So the volume is

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

(or the same with $dy$ for a vertical axis).

• $R$ = outer radius: distance from the axis to the edge of the region farther from the axis.
• $r$ = inner radius: distance from the axis to the edge closer to the axis.

Revolving the region between $y = x$ and $y = x^2$ about the $x$-axis: outer radius $R = x$, inner radius $r = x^2$.

Finding R and r

Draw one slice perpendicular to the axis, from the axis out through the region. It crosses the near edge first ($r$) and then the far edge ($R$). Both are distances, so each is bigger minus smaller:

Axis	Variable	Distance from the axis to a curve
$x$-axis	$dx$	the curve’s $y$-value
$y = k$	$dx$	$\lvert y_{\text{curve}} - k \rvert$
$y$-axis	$dy$	the curve’s $x$-value (written in $y$)
$x = h$	$dy$	$\lvert x_{\text{curve}} - h \rvert$

When the axis is above the region (or to its right), the lower curve (or the left curve) is farther away, so it gives $R$. This flips many students; draw the picture.

R² − r², not (R − r)²

The washer’s area is $\pi(R^2 - r^2)$. Squaring the difference, $\pi(R - r)^2$, gives a different (wrong) number. Square each radius first, then subtract.

AP questions typically ask for a washer volume about the $x$-axis or about a line like $y = k$ in the same free-response question. Show the integral setup clearly; a correct setup earns credit even if a later step slips.

Worked examples

Example 1: About the x-axis

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

Solution. The curves meet at $x = 0$ and $x = 1$, with $y = x$ on top. Vertical slices: the far edge is the line, so $R = x$; the near edge is the parabola, so $r = x^2$.

$$V = \pi \int_0^1 \left( x^2 - x^4 \right) dx = \pi\left( \frac{1}{3} - \frac{1}{5} \right) = \frac{2\pi}{15} \approx 0.419$$

Example 2: The same region about the y-axis

Revolve the region between $y = x$ and $y = x^2$ about the $y$-axis instead.

Solution. The axis is vertical, so use horizontal slices and $dy$. In terms of $y$: the line is $x = y$ and the parabola is $x = \sqrt{y}$, for $0 \le y \le 1$. At $y = 0.25$, $y = 0.25$ and $\sqrt{y} = 0.5$, so the parabola is farther from the $y$-axis: $R = \sqrt{y}$ and $r = y$.

$$V = \pi \int_0^1 \left( (\sqrt{y})^2 - y^2 \right) dy = \pi \int_0^1 (y - y^2)\, dy = \pi\left( \frac{1}{2} - \frac{1}{3} \right) = \frac{\pi}{6} \approx 0.524$$

Example 3: About a line below the region

Revolve the same region about the line $y = -1$.

Solution. Vertical slices again. The axis is $1$ unit below the $x$-axis, so every distance gets $1$ added:

$$R = x - (-1) = x + 1, \qquad r = x^2 - (-1) = x^2 + 1$$ $$\begin{aligned} V &= \pi \int_0^1 \Big( (x + 1)^2 - (x^2 + 1)^2 \Big)\, dx = \pi \int_0^1 \left( x^2 + 2x + 1 - x^4 - 2x^2 - 1 \right) dx \\ &= \pi \int_0^1 \left( 2x - x^2 - x^4 \right) dx = \pi\left( 1 - \frac{1}{3} - \frac{1}{5} \right) = \frac{7\pi}{15} \approx 1.466 \end{aligned}$$

Example 4: About a vertical line to the right

Revolve the same region about the line $x = 2$.

Solution. Horizontal slices, $0 \le y \le 1$. The line $x = 2$ is to the right of the region, so the left curve, $x = y$, is farther away:

$$R = 2 - y, \qquad r = 2 - \sqrt{y}$$ $$\begin{aligned} V &= \pi \int_0^1 \Big( (2 - y)^2 - (2 - \sqrt{y})^2 \Big)\, dy = \pi \int_0^1 \left( 4 - 4y + y^2 - 4 + 4\sqrt{y} - y \right) dy \\ &= \pi \int_0^1 \left( y^2 - 5y + 4\sqrt{y} \right) dy = \pi\left( \frac{1}{3} - \frac{5}{2} + \frac{8}{3} \right) = \frac{\pi}{2} \approx 1.571 \end{aligned}$$

Notice how the same region gives four different volumes, depending on the axis.

Common mistakes

Writing (R − r)² instead of R² − r². Square each radius, then subtract.

Swapping R and r. If your integrand is negative, the radii are backwards. $R$ goes to the edge farther from the axis. When the axis is above the region or to its right, that is the lower or left curve.

Forgetting to shift the radii for a different axis. About $y = -1$, add $1$ to each height; about $y = 3$, the radius to a curve is $3 - y_{\text{curve}}$. Measure from the axis, not from the $x$-axis.

Using dx for a vertical axis. Slices must be perpendicular to the axis: vertical axis means horizontal slices and $dy$.

Using discs when there’s a hole. If the region doesn’t touch the axis along its whole length, there’s an inner radius. Even a region bounded by the axis only part of the way may need washers.

Practice

1. (Warm-up) A washer has outer radius $5$ and inner radius $3$. Find its area.

Solution

$$\pi(5^2 - 3^2) = \pi(25 - 9) = 16\pi$$

(Not $\pi(5 - 3)^2 = 4\pi$.)

2. (Warm-up) The region between $y = 2$ and $y = 1$, for $0 \le x \le 3$, is revolved about the $x$-axis. Find the volume.

Solution

$R = 2$ and $r = 1$:

$$V = \pi \int_0^3 (4 - 1)\, dx = 9\pi$$

The solid is a thick-walled pipe: a cylinder of radius $2$ with a cylinder of radius $1$ removed, both of length $3$. Check: $\pi(2)^2(3) - \pi(1)^2(3) = 12\pi - 3\pi = 9\pi$. ✓

3. (Warm-up) The region between $y = \sqrt{x}$ and $y = \dfrac{x}{2}$ (they meet at $x = 0$ and $x = 4$) is revolved about the $x$-axis. Find $R$ and $r$, then the volume.

Solution

$\sqrt{x}$ is on top, so it is farther from the $x$-axis: $R = \sqrt{x}$ and $r = \dfrac{x}{2}$.

$$V = \pi \int_0^4 \left( x - \frac{x^2}{4} \right) dx = \pi\left( 8 - \frac{16}{3} \right) = \frac{8\pi}{3}$$

4. (Core) The region bounded by $y = x^2$ and $y = 4$ is revolved about the $x$-axis. Find the volume.

Solution

$x$ runs from $-2$ to $2$. The line $y = 4$ is farther from the $x$-axis: $R = 4$ and $r = x^2$.

$$V = \pi \int_{-2}^{2} \left( 16 - x^4 \right) dx = \pi\left( 64 - \frac{64}{5} \right) = \frac{256\pi}{5}$$

5. (Core) The region bounded by $y = x^2$ and $y = 4$ is revolved about the line $y = -1$. Find the volume.

Solution

Distances from $y = -1$: to the line $y = 4$ is $5$, and to the parabola is $x^2 + 1$. So $R = 5$ and $r = x^2 + 1$.

$$\begin{aligned} V &= \pi \int_{-2}^{2} \left( 25 - (x^2 + 1)^2 \right) dx = \pi \int_{-2}^{2} \left( 24 - 2x^2 - x^4 \right) dx \\ &= \pi\left( 96 - \frac{32}{3} - \frac{64}{5} \right) = \frac{1088\pi}{15} \approx 227.870 \end{aligned}$$

6. (Core) The region between $y = x^2$ and $y = 2x$ is revolved about the line $x = -1$. Find the volume.

Solution

The curves meet at $x = 0$ and $x = 2$, that is, $y = 0$ and $y = 4$. The axis is vertical, so use $dy$. In terms of $y$: the parabola is $x = \sqrt{y}$ and the line is $x = \dfrac{y}{2}$. At $y = 1$, $\sqrt{y} = 1 \gt \tfrac{1}{2}$, so the parabola is on the right, farther from $x = -1$.

$$R = \sqrt{y} + 1, \qquad r = \frac{y}{2} + 1$$$$\begin{aligned} V &= \pi \int_0^4 \left( (\sqrt{y} + 1)^2 - \left( \frac{y}{2} + 1 \right)^2 \right) dy = \pi \int_0^4 \left( y + 2\sqrt{y} + 1 - \frac{y^2}{4} - y - 1 \right) dy \\ &= \pi \int_0^4 \left( 2\sqrt{y} - \frac{y^2}{4} \right) dy = \pi\left( \frac{32}{3} - \frac{16}{3} \right) = \frac{16\pi}{3} \end{aligned}$$

7. (Core) The region between $y = x$ and $y = x^2$ is revolved about the line $y = 1$. Find the volume.

Solution

The axis $y = 1$ is above the region, so the lower curve, $y = x^2$, is farther away:

$$R = 1 - x^2, \qquad r = 1 - x$$$$V = \pi \int_0^1 \left( (1 - x^2)^2 - (1 - x)^2 \right) dx = \pi \int_0^1 \left( x^4 - 3x^2 + 2x \right) dx = \pi\left( \frac{1}{5} - 1 + 1 \right) = \frac{\pi}{5}$$

8. (Challenge) (Calculator active.) Let $S$ be the region bounded by $y = 2\cos x$ and $y = x^2$ (radians).

• (a) Find the volume when $S$ is revolved about the $x$-axis.
• (b) Find the volume when $S$ is revolved about the line $y = 3$.

Solution

The curves meet at $x = \pm c$ with $c \approx 1.022$ (store it). On $[-c, c]$, $2\cos x$ is on top, and both curves are between $0$ and $3$.

(a) About the $x$-axis: $R = 2\cos x$, inner radius $x^2$.

$$V = \pi \int_{-c}^{c} \left( 4\cos^2 x - x^4 \right) dx \approx 17.034$$

(b) About $y = 3$, which is above the region: the far curve is the lower one, $y = x^2$. Outer radius $3 - x^2$, inner radius $3 - 2\cos x$.

$$V = \pi \int_{-c}^{c} \left( (3 - x^2)^2 - (3 - 2\cos x)^2 \right) dx \approx 33.878$$

9. (Challenge) Without integrating, explain why revolving the region between $y = x$ and $y = x^2$ about $y = -1$ gives a bigger volume than revolving it about the $x$-axis (Examples 1 and 3).

Solution

Moving the axis farther from the region makes every point of the region travel around a bigger circle. Each washer has the same thickness in the radial direction, $R - r = x - x^2$ in both cases, but its radii are larger, so its area $\pi(R^2 - r^2) = \pi(R - r)(R + r)$ is larger (because $R + r$ grows by $2$). Adding up bigger washers gives a bigger volume: $\dfrac{7\pi}{15} \gt \dfrac{2\pi}{15}$.