Volumes of Revolution — the Disc Method

Spin a flat region around a line and it sweeps out a solid, the way a potter’s wheel turns a profile into a vase. These solids of revolution are a special case of volumes with known cross sections: every slice perpendicular to the axis is a circle, so its area is just $\pi R^2$. The only real job is finding the radius $R$.

Key ideas

Each slice is a disc

Revolve a region about a line that forms one of its edges. A thin slice perpendicular to the axis sweeps out a flat disc (a short, wide cylinder) with radius $R$ and thickness $dx$ or $dy$. Its volume is about $\pi R^2\, dx$. Add up the discs:

$$V = \pi \int_a^b \big( R(x) \big)^2\, dx \qquad \text{or} \qquad V = \pi \int_c^d \big( R(y) \big)^2\, dy$$

Revolving the region under $y = \sqrt{x}$ about the $x$-axis: the disc at $x$ has radius $R = \sqrt{x}$.

Which variable?

Slice perpendicular to the axis of rotation.

Axis of rotation	Slices	Integrate with	$R$ is written in terms of
Horizontal line (the $x$-axis, or $y = k$)	vertical	$dx$	$x$
Vertical line (the $y$-axis, or $x = h$)	horizontal	$dy$	$y$

The radius is a distance from the axis

$R$ is the distance from the axis of rotation to the far edge of the region, measured along the slice. Always “bigger minus smaller”:

• About the $x$-axis: $R = f(x)$ (the curve’s height).
• About $y = k$: $R = \lvert f(x) - k \rvert$. If the curve is above the line, $R = f(x) - k$; if it is below, $R = k - f(x)$.
• About the $y$-axis: $R = g(y)$, where $x = g(y)$ is the curve.
• About $x = h$: $R = \lvert g(y) - h \rvert$.

The disc method needs the region to touch the axis all the way along (no gap). If there’s a gap between the region and the axis, every slice is a ring with a hole, and you need the washer method.

On the AP exam, volume of revolution questions usually come as one part of a region question, often calculator active. Write $\pi\int R^2$ with the correct radius and limits before evaluating.

Worked examples

Example 1: About the x-axis

The region under $y = \sqrt{x}$ and above the $x$-axis, for $0 \le x \le 4$, is revolved about the $x$-axis. Find the volume.

Solution. Vertical slices; the radius at $x$ is $R = \sqrt{x}$.

$$V = \pi \int_0^4 (\sqrt{x})^2\, dx = \pi \int_0^4 x\, dx = \pi \Big[ \tfrac{x^2}{2} \Big]_0^4 = 8\pi \approx 25.133$$

Example 2: About the y-axis

The region bounded by $y = x^3$, $y = 8$, and the $y$-axis is revolved about the $y$-axis. Find the volume.

Solution. The axis is vertical, so use horizontal slices and $dy$. Solve for $x$: $x = y^{1/3}$. Each disc runs from the $y$-axis to the curve, so $R = y^{1/3}$, for $0 \le y \le 8$.

$$V = \pi \int_0^8 \left( y^{1/3} \right)^2 dy = \pi \int_0^8 y^{2/3}\, dy = \pi \left[ \frac{3}{5}y^{5/3} \right]_0^8 = \pi \cdot \frac{3}{5} \cdot 32 = \frac{96\pi}{5} \approx 60.319$$

Example 3: About a horizontal line other than the x-axis

The region bounded by $y = x^2$ and $y = 4$ is revolved about the line $y = 4$. Find the volume.

Solution. The region touches the axis $y = 4$ along its top edge, so discs work. The curves meet at $x = \pm 2$. The radius is the distance from the line $y = 4$ down to the parabola: $R = 4 - x^2$.

$$V = \pi \int_{-2}^{2} (4 - x^2)^2\, dx = \pi \int_{-2}^{2} (16 - 8x^2 + x^4)\, dx = \pi \cdot \frac{512}{15} = \frac{512\pi}{15} \approx 107.233$$

Example 4: About a vertical line other than the y-axis

The region bounded by $y = \sqrt{x}$, the $x$-axis, and $x = 4$ is revolved about the line $x = 4$. Find the volume.

Solution. The axis is vertical, so use horizontal slices. In terms of $y$, the curve is $x = y^2$, and $y$ runs from $0$ to $2$. Each slice runs from the curve to the line $x = 4$, so

$$R = 4 - y^2$$ $$V = \pi \int_0^2 (4 - y^2)^2\, dy = \pi \int_0^2 (16 - 8y^2 + y^4)\, dy = \pi\left( 32 - \frac{64}{3} + \frac{32}{5} \right) = \frac{256\pi}{15} \approx 53.617$$

Common mistakes

Forgetting π, or forgetting to square R. The disc’s area is $\pi R^2$. Leaving out either one is the most common error on the exam.

Slicing parallel to the axis. For a vertical axis you need horizontal slices and $dy$. Writing $\pi\int (x^3)^2\, dx$ in Example 2 gives the volume about the $x$-axis instead.

Using the curve’s height as the radius when the axis isn’t the x-axis. About $y = 4$, the radius is $4 - x^2$, not $x^2$. Draw the radius from the axis to the curve, and write it as bigger minus smaller.

Writing R² as a difference of squares. $(4 - x^2)^2$ is not $16 - x^4$. Expand the square properly.

Using discs when there’s a gap. If the region doesn’t touch the axis, the solid has a hole, and you need washers.

Practice

1. (Warm-up) The region under $y = 2x$, for $0 \le x \le 3$, is revolved about the $x$-axis. Find the volume. Then check your answer with the formula for the volume of a cone.

Solution

$$V = \pi \int_0^3 (2x)^2\, dx = \pi \Big[ \tfrac{4x^3}{3} \Big]_0^3 = 36\pi$$

The solid is a cone with radius $6$ and height $3$: $\dfrac{1}{3}\pi(6)^2(3) = 36\pi$. ✓

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

Solution

$$V = \pi \int_0^1 e^{2x}\, dx = \pi \left[ \frac{e^{2x}}{2} \right]_0^1 = \frac{\pi(e^2 - 1)}{2} \approx 10.036$$

3. (Warm-up) The region under $y = \dfrac{1}{x}$, for $1 \le x \le 3$, is revolved about the $x$-axis. Find the volume.

Solution

$$V = \pi \int_1^3 \frac{1}{x^2}\, dx = \pi \left[ -\frac{1}{x} \right]_1^3 = \pi\left( -\frac{1}{3} + 1 \right) = \frac{2\pi}{3}$$

4. (Core) The region under $y = \sec x$, for $0 \le x \le \dfrac{\pi}{4}$ (radians), is revolved about the $x$-axis. Find the volume.

Solution

$$V = \pi \int_0^{\pi/4} \sec^2 x\, dx = \pi \Big[ \tan x \Big]_0^{\pi/4} = \pi(1 - 0) = \pi$$

5. (Core) The region bounded by $y = \sqrt{x}$, the $y$-axis, and $y = 2$ is revolved about the $y$-axis. Find the volume.

Solution

Horizontal slices: $x = y^2$, so $R = y^2$, for $0 \le y \le 2$.

$$V = \pi \int_0^2 (y^2)^2\, dy = \pi \Big[ \tfrac{y^5}{5} \Big]_0^2 = \frac{32\pi}{5}$$

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

Solution

$2 - x^2 = 1$ at $x = \pm 1$. The region lies above the line $y = 1$, so $R = (2 - x^2) - 1 = 1 - x^2$.

$$V = \pi \int_{-1}^{1} (1 - x^2)^2\, dx = \pi \int_{-1}^{1} (1 - 2x^2 + x^4)\, dx = \pi\left( 2 - \frac{4}{3} + \frac{2}{5} \right) = \frac{16\pi}{15}$$

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

Solution

Horizontal slices: the curve is $x = \sqrt{y}$ and $y$ runs from $0$ to $4$. Each slice runs from the curve to $x = 2$, so $R = 2 - \sqrt{y}$.

$$V = \pi \int_0^4 (2 - \sqrt{y})^2\, dy = \pi \int_0^4 \left( 4 - 4\sqrt{y} + y \right) dy = \pi\left( 16 - \frac{64}{3} + 8 \right) = \frac{8\pi}{3}$$

(Here $\int_0^4 4\sqrt{y}\, dy = \tfrac{8}{3}y^{3/2}\Big|_0^4 = \tfrac{64}{3}$.)

8. (Challenge) Revolve the upper half of the circle $x^2 + y^2 = r^2$ about the $x$-axis to show that a sphere of radius $r$ has volume $\dfrac{4}{3}\pi r^3$.

Solution

The upper semicircle is $y = \sqrt{r^2 - x^2}$, for $-r \le x \le r$, so $R^2 = r^2 - x^2$.

$$V = \pi \int_{-r}^{r} (r^2 - x^2)\, dx = \pi \left[ r^2 x - \frac{x^3}{3} \right]_{-r}^{r} = \pi\left( \frac{2r^3}{3} + \frac{2r^3}{3} \right) = \frac{4}{3}\pi r^3$$

9. (Challenge) (Calculator active.) The region bounded by $y = \cos x$, $y = 1$, and $x = \dfrac{\pi}{2}$ (radians) is revolved about the line $y = 1$. Find the volume.

Solution

The region lies below $y = 1$, between $x = 0$ (where $\cos x = 1$) and $x = \dfrac{\pi}{2}$. The radius is $R = 1 - \cos x$.

$$V = \pi \int_0^{\pi/2} (1 - \cos x)^2\, dx \approx 1.119$$

(The exact value is $\pi\left( \dfrac{3\pi}{4} - 2 \right)$, but finding it needs an antiderivative of $\cos^2 x$, which is why this one is calculator active.)