With disks and washers you slice across the axis of rotation. Sometimes that forces you to solve a curve for $x$ in terms of $y$, which can be hard or impossible. Cylindrical shells slice parallel to the axis instead.
Why it works
Take a thin vertical strip at $x$, of height $h(x)$ and width $\Delta x$, and spin it around the $y$-axis. It sweeps out a thin-walled tube: a shell. Cut the shell down one side and unroll it. You get a nearly flat slab with length $2\pi x$ (the circumference), height $h(x)$ and thickness $\Delta x$. So its volume is about $2\pi x\,h(x)\,\Delta x$, and adding shells gives
$$V = \int_a^b 2\pi x\,h(x)\,dx.$$
Key idea
$$V = 2\pi\int_a^b (\text{radius})(\text{height})\,dx,$$
where the radius is the distance from the strip to the axis of rotation and the height is top curve minus bottom curve.
Worked example: About the y-axis
Revolve the region between $y = 2x - x^{2}$ and the $x$-axis about the $y$-axis.
- The curve meets the axis at $x = 0$ and $x = 2$. A strip at $x$ has height $2x - x^{2}$ and radius $x$.
- $$\begin{aligned} V &= 2\pi\int_0^2 x(2x - x^{2})\,dx \\ &= 2\pi\left[\frac{2x^{3}}{3} - \frac{x^{4}}{4}\right]_0^2 \\ &= 2\pi\left(\frac{16}{3} - 4\right). \end{aligned}$$
- $V = \frac{8\pi}{3}$. With washers you would first have to solve $y = 2x - x^{2}$ for $x$, getting two branches $x = 1 \pm\sqrt{1 - y}$.
Answer: $\frac{8\pi}{3}$
Worked example: An axis to the left
Revolve the same region about the line $x = -1$.
- The axis is 1 unit to the left of the $y$-axis, so a strip at $x$ is $x + 1$ from it.
- $$\begin{aligned} V &= 2\pi\int_0^2 (x + 1)(2x - x^{2})\,dx \\ &= 2\pi\int_0^2 (2x + x^{2} - x^{3})\,dx. \end{aligned}$$
- $= 2\pi\left(4 + \frac{8}{3} - 4\right) = \frac{16\pi}{3}$.
Answer: $\frac{16\pi}{3}$
Common mistake
Using $x$ as the radius when the axis is not the $y$-axis. The radius is a distance from the axis, so it must be 0 on the axis and grow as you move away. For the axis $x = -1$ the radius is $x + 1$; for an axis $x = 3$ to the right of the region it is $3 - x$. Check by plugging in one point: the strip at $x = 0$ is 1 unit from $x = -1$, and $0 + 1 = 1$.
Worked example: An axis to the right
Revolve the same region about the line $x = 3$.
- Now the radius is $3 - x$.
- $$\begin{aligned} V &= 2\pi\int_0^2 (3 - x)(2x - x^{2})\,dx \\ &= 2\pi\int_0^2 (6x - 5x^{2} + x^{3})\,dx \\ &= 2\pi\left(12 - \frac{40}{3} + 4\right) \\ &= \frac{16\pi}{3}. \end{aligned}$$
- Same as for $x = -1$. The region is symmetric about $x = 1$, and both axes are 2 units from that center.
Answer: $\frac{16\pi}{3}$
| Disks and washers | Shells | |
|---|---|---|
| Slices | perpendicular to the axis | parallel to the axis |
| Piece | $\pi(R^{2} - r^{2})\,\Delta x$ | $2\pi(\text{radius})(\text{height})\,\Delta x$ |
| Axis vertical: integrate in | $y$ | $x$ |
| Axis horizontal: integrate in | $x$ | $y$ |