When a flat region spins around a line, it sweeps out a solid of revolution. Every slice perpendicular to that line is round, so volume by slicing gets an easy cross-section: a disk, or a disk with a hole (a washer).
Why it works
Revolve a thin vertical strip of the region around the $x$-axis. If the strip reaches from the axis up to the curve $y = R(x)$, it sweeps out a disk of radius $R(x)$, with area $\pi R(x)^{2}$. If the strip starts at an inner curve $r(x)$ instead, it sweeps out a washer: the big disk minus the hole,
$$\begin{aligned} A(x) &= \pi R(x)^{2} - \pi r(x)^{2} \\ &= \pi\left[R(x)^{2} - r(x)^{2}\right]. \end{aligned}$$
Then $V = \int A(x)\,dx$, exactly as in slicing.
Key idea
Revolving about a horizontal axis:
$$V = \pi\int_a^b \left[R(x)^{2} - r(x)^{2}\right]dx,$$
where $R$ is the outer radius and $r$ the inner radius ($r = 0$ for disks). Both radii are distances from the axis of rotation.
Worked example: A disk
Revolve the region under $y = \sqrt{x}$, $0 \le x \le 4$, about the $x$-axis. Find the volume.
- Each slice is a disk of radius $R(x) = \sqrt{x}$, so $R^{2} = x$.
- $V = \pi\int_0^4 x\,dx = \pi\cdot\frac{16}{2} = 8\pi$.
Answer: $8\pi$
Worked example: A washer
Revolve the region between $y = x$ and $y = x^{2}$ ($0 \le x \le 1$) about the $x$-axis.
- On $[0, 1]$, $x \ge x^{2}$, so the outer radius is $R = x$ and the inner radius is $r = x^{2}$.
- $$\begin{aligned} V &= \pi\int_0^1 \left(x^{2} - x^{4}\right)dx \\ &= \pi\left(\frac{1}{3} - \frac{1}{5}\right) \\ &= \frac{2\pi}{15}. \end{aligned}$$
Answer: $\frac{2\pi}{15}$
Common mistake
Squaring the difference: $\pi\int (R - r)^{2}\,dx$. For the washer above that gives $\pi\int_0^1 (x - x^{2})^{2}\,dx = \frac{\pi}{30}$, only a quarter of the true $\frac{2\pi}{15}$. A thin ring far from the axis has a large area (it goes all the way around), but $(R - r)^{2}$ treats it like a tiny disk whose radius is the ring's thickness. Check: area of a washer = big disk minus hole.
When the axis is not a coordinate axis
Radii are always distances from the axis of rotation. If the axis is the line $y = k$ below the region, a point at height $y$ is $y - k$ from the axis. If the axis is above the region, the distance is $k - y$, and the curve that is lower on the graph becomes the outer one.
Worked example: Spinning about y = −1
Revolve the region between $y = x$ and $y = x^{2}$ ($0 \le x \le 1$) about the line $y = -1$.
- Distances to the axis add 1: outer radius $R = x + 1$, inner radius $r = x^{2} + 1$.
- $$\begin{aligned} &R^{2} - r^{2} \\ &= (x^{2} + 2x + 1) - (x^{4} + 2x^{2} + 1) \\ &= -x^{4} - x^{2} + 2x. \end{aligned}$$
- $$\begin{aligned} V &= \pi\int_0^1 (-x^{4} - x^{2} + 2x)\,dx \\ &= \pi\left(-\frac{1}{5} - \frac{1}{3} + 1\right) \\ &= \frac{7\pi}{15}. \end{aligned}$$
Answer: $\frac{7\pi}{15}$, more than before, because the region now sweeps a wider circle.
To revolve about the $y$-axis with this method, slice horizontally: write the radii as functions of $y$ and integrate $\pi\int \left[R(y)^{2} - r(y)^{2}\right]dy$. The next topic, Cylindrical Shells, offers a way that keeps $x$.