Surface Area of Revolution, calculus lesson

How much glaze covers a vase, or how much metal goes into a satellite dish? Both surfaces come from spinning a curve around an axis. Volume needed the area of slices. Surface area needs the length of the curve as well, because a slanted band of surface is wider than its shadow on the axis. Approximating the surface by thin bands of cones gives the right formula.

Key terms

Surface of revolution

Definition Let $C$ be a curve in a plane and let $\ell$ be a line in the same plane that $C$ does not cross. The surface of revolution is the set of points traced by $C$ as the plane rotates one full turn about $\ell$ . Each point of $C$ at distance $\rho$ from $\ell$ traces a circle of radius $\rho$ centered on $\ell$ .

In plain words Spin a wire shaped like the curve around the axis and look at the thin shell it sweeps. Only the skin counts, not the inside.

Example Rotating the segment $y = 2$ , $0 \le x \le 3$ about the $x$ -axis gives the side of a circular cylinder of radius $2$ and length $3$ . Unrolled it is a rectangle, with area $2\pi \cdot 2 \cdot 3 = 12\pi$ .

Frustum approximation

Definition A frustum of a cone is the part of a circular cone between two planes perpendicular to its axis. If its circular edges have radii $r_1$ and $r_2$ and its slant height (the length of a straight segment on the surface from one edge to the other) is $\ell$ , its lateral surface area is

A = \pi (r_1 + r_2)\, \ell = 2\pi \bar r \, \ell, \qquad \bar r = \frac{r_1 + r_2}{2}.

Rotating an inscribed polygon of a curve about the axis gives a stack of such frustums, and the area of the surface is defined as the limit of their total area as the polygon’s segments get shorter.

In plain words A band of a cone is a lampshade. Its area is the average circumference times the slanted width. Many short lampshades stacked together approximate the curved surface.

Example Rotating the segment of $y = x$ from $x = 1$ to $x = 2$ about the $x$ -axis gives a frustum with $r_1 = 1$ , $r_2 = 2$ and $\ell = \sqrt{2}$ . Its area is $\pi \cdot 3 \cdot \sqrt{2} = 3\sqrt{2}\,\pi \approx 13.33$ .

Surface area about a horizontal axis

Definition Let $f$ be nonnegative on $[a, b]$ with $f'$ continuous on $[a, b]$ . The surface obtained by rotating the graph of $f$ about the $x$ -axis has area

S = \int_a^b 2\pi f(x) \sqrt{1 + f'(x)^2} \, dx.

In short,

S = \int 2\pi \rho \, ds,

where $\rho = f(x)$ is the distance to the axis and $ds = \sqrt{1 + f'(x)^2}\,dx$ is the arc length element. If $f$ is continuous on $[a, b]$ but $f'$ is continuous only on $(a, b]$ , the area is defined by the improper integral

S = \lim_{t \to a^+} \int_t^b 2\pi f(x) \sqrt{1 + f'(x)^2} \, dx,

and likewise at $b$ .

In plain words Each tiny piece of curve, of length $ds$ , sweeps a thin ring of circumference $2\pi f(x)$ . Add the rings.

Example Rotate $y = \sqrt{x}$ on $[0, 2]$ about the $x$ -axis. Here $f'(x) = \frac{1}{2\sqrt{x}}$ is continuous on $(0, 2]$ but unbounded near $0$ , so this is the improper case. For $x > 0$ the integrand simplifies to

f(x)\sqrt{1 + f'(x)^2} = \sqrt{x}\sqrt{1 + \frac{1}{4x}} = \sqrt{x + \frac{1}{4}},

and

S = \lim_{t \to 0^+} 2\pi\int_t^2 \sqrt{x + \frac{1}{4}} \, dx = \lim_{t \to 0^+} \frac{4\pi}{3}\left(\frac{27}{8} - \left(t + \frac{1}{4}\right)^{3/2}\right) = \frac{4\pi}{3}\left(\frac{27}{8} - \frac{1}{8}\right) = \frac{13\pi}{3}.

The simplified integrand $\sqrt{x + \frac{1}{4}}$ is continuous on all of $[0, 2]$ , which is why the limit simply equals its value at $t = 0$ .

Surface area about a vertical axis

Definition Let $f'$ be continuous on $[a, b]$ with $0 \le a < b$ . Rotating the graph of $y = f(x)$ about the $y$ -axis gives a surface of area

S = \int_a^b 2\pi x \sqrt{1 + f'(x)^2} \, dx.

Here the distance to the axis is $x$ . If the curve is given as $x = g(y)$ for $c \le y \le d$ with $g \ge 0$ and $g'$ continuous, the same surface area is

S = \int_c^d 2\pi g(y)\sqrt{1 + g'(y)^2} \, dy.

In plain words The arc length element $ds$ stays the same. Only the radius changes, because now it is the horizontal distance to the vertical axis.

Example Rotate $y = x^2$ on $[0, 1]$ about the $y$ -axis, which gives a bowl. Its area is

S = 2\pi\int_0^1 x\sqrt{1 + 4x^2} \, dx = 2\pi \cdot \frac{1}{12}\left[(1 + 4x^2)^{3/2}\right]_0^1 = \frac{\pi}{6}\left(5\sqrt{5} - 1\right) \approx 5.330.

Nonnegative distance to the axis

Definition For rotation about any line in the plane, the radius $\rho$ in the formula for $S$ is the distance $\rho \ge 0$ from the point of the curve to the axis. For the axis $y = k$ and a graph $y = f(x)$ on one side of it,

S = \int_a^b 2\pi \lvert f(x) - k \rvert \sqrt{1 + f'(x)^2} \, dx.

Using $f(x) - k$ without the absolute value gives a negative answer when the curve lies below the axis.

In plain words A radius is a distance, so it is never negative. Measure from the axis to the curve, whichever side the curve is on.

Example Rotate the segment $y = x$ , $0 \le x \le 1$ , about the line $y = -1$ . The distance to the axis is $x + 1$ and $ds = \sqrt{2}\,dx$ , so

S = 2\pi\sqrt{2}\int_0^1 (x + 1) \, dx = 2\pi\sqrt{2} \cdot \frac{3}{2} = 3\sqrt{2}\,\pi.

It is a frustum with radii $1$ and $2$ and slant height $\sqrt{2}$ , the same one as in the frustum example.

Regularity and overlap

Definition The surface area formulas above require a curve with a continuous derivative on the closed interval (or on finitely many closed pieces, with the areas added), so that the arc length element is a continuous function. Where the derivative blows up at an endpoint, the integral must be read as an improper integral. The formulas also assume the curve does not cross the axis and that different parts of the curve do not sweep the same part of the surface. If two parts of the curve trace the same surface, the integral counts that surface more than once.

In plain words The curve should be smooth and stay on one side of the axis, and you should rotate each piece of the surface only once.

Example Rotating $y = x^2$ on $[-1, 1]$ about the $y$ -axis gives the same bowl as rotating it on $[0, 1]$ , because the left half is the mirror image of the right half. The correct area is $\frac{\pi}{6}(5\sqrt{5} - 1)$ . The integral

\int_{-1}^1 2\pi \lvert x \rvert \sqrt{1 + 4x^2} \, dx

gives twice that, because it paints the bowl twice.

Check yourself

Find the area of the sphere of radius $r$ by rotating $y = \sqrt{r^2 - x^2}$ , $-r \le x \le r$ , about the $x$ -axis.

Show the answer

For $-r < x < r$ the derivative is

f'(x) = -\frac{x}{\sqrt{r^2 - x^2}},

so $f(x)\sqrt{1 + f'(x)^2} = r$ . Since $f'$ is unbounded near $\pm r$ , the integral is improper at both ends, but the simplified integrand is the constant $2\pi r$ , so the limits give

S = \int_{-r}^r 2\pi r \, dx = 4\pi r^2.

The part of a sphere of radius $2$ between the planes $x = -1$ and $x = 1$ is a band. Find its area.

Show the answer

As in the first exercise the integrand is $2\pi \cdot 2$ , so

S = \int_{-1}^1 4\pi \, dx = 8\pi,

which is half of the sphere’s area $16\pi$ .

Use the integral to find the area of the cone made by rotating $y = x/2$ , $0 \le x \le 4$ , about the $x$ -axis.

Show the answer

The slope is $1/2$ , so $ds = \sqrt{1 + 1/4}\,dx = \frac{\sqrt{5}}{2}\,dx$ and

S = 2\pi \cdot \frac{\sqrt{5}}{2}\int_0^4 \frac{x}{2} \, dx = \pi\sqrt{5} \cdot 4 = 4\sqrt{5}\,\pi.

This matches $\pi r \ell$ with $r = 2$ and slant height $\ell = 2\sqrt{5}$ .

Surface Area of Revolution

Key terms

Surface of revolution

Frustum approximation

Surface area about a horizontal axis

Surface area about a vertical axis

Nonnegative distance to the axis

Regularity and overlap

Check yourself

How this lesson connects

Builds on