Surface Area of Revolution

Spin a curve (not a region) around the $x$-axis and it traces out a surface, like a lampshade or the skin of a vase. To find its area, combine two ideas you already have: the circles from disks and washers and the slanted pieces from arc length.

Why it works

Cut the curve into short pieces. A piece of length $\Delta s$ at height $y$ sweeps out a thin band, nearly a strip of a cone. Unrolled, the band is close to a rectangle whose length is the circumference $2\pi y$ and whose width is the slanted length $\Delta s$. So the band's area is about $2\pi y\,\Delta s$. From arc length, $\Delta s \approx \sqrt{1 + f'(x)^{2}}\,\Delta x$. Add the bands and let them shrink: $$S = \int_a^b 2\pi f(x)\sqrt{1 + f'(x)^{2}}\,dx.$$

Key idea

Revolving $y = f(x) \ge 0$, $a \le x \le b$, about the $x$-axis: $$\begin{aligned} S &= 2\pi\int_a^b f(x)\sqrt{1 + \left[f'(x)\right]^{2}}\,dx \\ &= 2\pi\int (\text{radius})\,ds. \end{aligned}$$

Worked example: A cone, checked by geometry

Revolve $y = \frac{3}{4}x$, $0 \le x \le 4$, about the $x$-axis. Find the area of the cone's side.

$f'(x) = \frac{3}{4}$, so $\sqrt{1 + f'^{2}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.
$$\begin{aligned} S &= 2\pi\int_0^4 \frac{3x}{4}\cdot\frac{5}{4}\,dx \\ &= 2\pi\cdot\frac{15}{16}\cdot 8 \\ &= 15\pi. \end{aligned}$$
Geometry agrees: the cone has radius $r = 3$ and slant height $\ell = 5$, and its side has area $\pi r\ell = 15\pi$.

Answer: $15\pi$

Common mistake

Using $dx$ instead of $ds$, as if the surface were made of flat rings: $2\pi\int f(x)\,dx$. For the cone above that gives $2\pi\int_0^4 \frac{3x}{4}\,dx = 12\pi$, too small, because each band is tilted and so is wider than $\Delta x$. Volumes can use $dx$ (a thin slab's volume barely depends on its tilt); surface areas need the slanted length $ds$.

Worked example: A band on a sphere

Revolve $y = \sqrt{r^{2} - x^{2}}$, $a \le x \le b$, about the $x$-axis (with $-r \le a < b \le r$).

$f'(x) = \frac{-x}{\sqrt{r^{2} - x^{2}}}$, so $1 + f'^{2} = \frac{r^{2}}{r^{2} - x^{2}}$.
Then $$\begin{aligned} f(x)\sqrt{1 + f'^{2}} &= \sqrt{r^{2} - x^{2}}\cdot\frac{r}{\sqrt{r^{2} - x^{2}}} \\ &= r. \end{aligned}$$
$S = 2\pi\int_a^b r\,dx = 2\pi r(b - a)$. With $a = -r$, $b = r$ this is the whole sphere: $4\pi r^{2}$.

Answer: $2\pi r(b - a)$: every band of the same width has the same area, wherever it is on the sphere.

Revolving about the $y$-axis instead, the radius of each band is $x$, so $S = 2\pi\int_a^b x\sqrt{1 + f'(x)^{2}}\,dx$. In every case: surface area is $2\pi\int(\text{radius})\,ds$, where the radius is the band's distance from the axis.

The curve $y = \sqrt{x}$, $0 \le x \le 2$, spun around the $x$-axis. The surface is the curved skin of this solid; its area is $\frac{13\pi}{3}$ (the practice problems work this kind out).

Worked example: A root under the root

Find the area of the surface made by revolving $y = \sqrt{x}$, $0 \le x \le 2$, about the $x$-axis.

$f'(x) = \frac{1}{2\sqrt{x}}$, so $$\begin{aligned} &f\sqrt{1 + f'^{2}} \\ &= \sqrt{x}\sqrt{1 + \frac{1}{4x}} \\ &= \sqrt{x + \frac{1}{4}}. \end{aligned}$$
$$\begin{aligned} S &= 2\pi\int_0^2 \sqrt{x + \tfrac{1}{4}}\,dx \\ &= \frac{4\pi}{3}\left[\left(x + \tfrac{1}{4}\right)^{3/2}\right]_0^2. \end{aligned}$$
$= \frac{4\pi}{3}\left(\frac{27}{8} - \frac{1}{8}\right) = \frac{13\pi}{3}$.

Answer: $\frac{13\pi}{3}$

Surface Area of Revolution

Practice

Builds on

Leads to

Sources and further reading