Calculus 2

Surface area of a solid of revolution

Spin a curve about a line and it sweeps out a thin shell, like a bowl spun from sheet metal. How much area does the shell have?

You need

Arc lengthCalculus 2 The disk methodCalculus 1

Where you are 10 in Applications of integrals

Calculus8

Applications of integrals10

▴ 5 above Arc length Average value of a function The disk method

Surface area of a solid of revolution

The washer method

When you reach for it

In a metal spinning workshop a flat disk of sheet metal is pressed against a form turning on a lathe until it takes the form’s shape: a reflector, a bowl, a funnel. The finished piece is a surface of revolution, what a curve sweeps out when it makes one full turn about a line, the axis. The disk method measures the solid such a shape encloses. This lesson measures the shell itself: its area, which is what the metal, the polish or the plating must cover.

Geometry gives the area of a cylinder, a cone and a sphere. For any other profile, reach for an integral built, like arc length, on the short piece of curve ds.

The procedure

The band. Replace a short piece of the profile by a straight segment. Turned about the axis, a segment sweeps a band: the side of a cone with its tip cut off, called a frustum. Its area comes from the cone’s. A cone with base radius r and slant height \ell has side area \pi r \ell (the lateral area of a cone).

Let a band have end radii r_1 < r_2 and slant width \ell. Extend it to a whole cone. The tip that was cut off is a small cone of slant \ell_1, and the whole cone has slant \ell_1 + \ell. The two cones are similar, so \dfrac{r_1}{\ell_1} = \dfrac{r_2}{\ell_1 + \ell}. Cross-multiplying gives r_2 \ell_1 = r_1 \ell_1 + r_1 \ell, that is, (r_2 - r_1)\ell_1 = r_1 \ell. The band is the big cone minus the tip:

\begin{aligned} \pi r_2(\ell_1 + \ell) - \pi r_1 \ell_1 &= \pi(r_2 - r_1)\ell_1 + \pi r_2 \ell \\ &= \pi r_1 \ell + \pi r_2 \ell \\ &= 2\pi \cdot \frac{r_1 + r_2}{2} \cdot \ell \end{aligned}

So a band has area 2\pi r \ell, where r is its average radius: the circumference at the middle times the slant width. When r_1 = r_2 the band is a piece of a cylinder, and the same formula holds directly.

Figure 1One piece of the profile in Example 1, replaced by a straight segment of length \ell. Turned about the x-axis it sweeps a band with end radii r_1 and r_2.

Adding the bands. Let f(x) \ge 0 with f' continuous on [a, b], and turn its graph about the x-axis. Split [a, b] into n pieces of width \Delta x and join the points P_i = (x_i, f(x_i)) by segments, exactly as for arc length. The segment from P_{i-1} to P_i sweeps a band (Figure 1) with:

slant width \displaystyle \sqrt{1 + [f'(x_i^*)]^2}\,\Delta x, for some x_i^* in the piece, by the mean value theorem as in the arc length lesson;
average radius \displaystyle \frac12\big(f(x_{i-1}) + f(x_i)\big), which lies between two values of the continuous f, so by the intermediate value theorem it equals f(x_i^{**}) for some x_i^{**} in the piece.

The banded surface therefore has area

\sum_{i=1}^{n} 2\pi f(x_i^{**})\sqrt{1 + [f'(x_i^*)]^2}\,\Delta x

This is almost a Riemann sum, with two sample points in each piece instead of one. Because f and f' are continuous on a closed interval, the sums still converge to the integral as n \to \infty (stated here without proof; it rests on uniform continuity). The surface area S is defined as this limit, just as length was defined by inscribed paths:

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

In short, S = \int 2\pi r\,ds: a band of circumference 2\pi r and width ds, added along the curve. Here r is the distance from the axis to the curve.

About the y-axis. The radius becomes the distance to the y-axis, and ds may be written in whichever variable describes the curve. For x = g(y) \ge 0 on [c, d],

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

and for y = f(x) on [a, b] with a \ge 0,

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

The axis decides the radius; the curve decides ds. About a horizontal line y = k the radius is \lvert f(x) - k \rvert.

Why ds and not dx. The integral \int_a^b 2\pi f(x)\,dx adds bands of width \Delta x, the rims of the disks in the disk method. A tilted band’s true width is its slant, \displaystyle \sqrt{1 + [f'(x)]^2} times its run, and that factor does not shrink when the pieces do. So \int 2\pi f(x)\,dx falls short of S wherever f' \ne 0, by an amount that refining never removes (Example 2). Volume behaves differently: the disk method’s error on one piece is about 2\pi f(x)\,\lvert\Delta y\rvert\,\Delta x, a product of two small quantities, and those errors vanish in the limit.

Worked examples

Example 1

A root that cancels the radius

Vahagn spins a lampshade whose profile is \displaystyle y = \sqrt{2x + 1} for 0 \le x \le 4, in decimeters, turned about the x-axis. Find its area.

Solution

By the chain rule \displaystyle f'(x) = \dfrac{1}{\sqrt{2x + 1}}, so

1 + [f'(x)]^2 = 1 + \frac{1}{2x + 1} = \frac{2x + 2}{2x + 1}

The radius is f(x), and it cancels the denominator of the root:

\begin{aligned} f(x)\sqrt{1 + [f'(x)]^2} &= \sqrt{2x + 1}\cdot\frac{\sqrt{2x + 2}}{\sqrt{2x + 1}} \\ &= \sqrt{2x + 2} \end{aligned}

Then

\begin{aligned} S &= 2\pi\int_0^4 \sqrt{2x + 2}\,dx = 2\pi\cdot\frac13 (2x + 2)^{3/2}\Big|_0^4 \\ &= \frac{2\pi}{3}\left(10\sqrt{10} - 2\sqrt2\right) \approx 60.31 \end{aligned}

square decimeters. Differentiating \displaystyle \frac13(2x + 2)^{3/2} gives \displaystyle \frac13 \cdot \frac32 \cdot 2\,(2x + 2)^{1/2} = \sqrt{2x + 2}, as it should. As a check, the stretch factor is at least 1, so S must exceed \displaystyle 2\pi\int_0^4 \sqrt{2x + 1}\,dx = \frac{52\pi}{3} \approx 54.45, and it does.

Example 2

A straight profile, by one band and by the integral

Bakhyt spins a funnel whose profile is the segment y = \tfrac34 x + 2, 0 \le x \le 4, in centimeters, about the x-axis. Find its area as a single band and by the integral, and compare with \int_0^4 2\pi y\,dx.

Solution

The end radii are f(0) = 2 and f(4) = 5. The segment runs 4 and rises 3, so its slant width is \displaystyle \ell = \sqrt{4^2 + 3^2} = 5. As one band, \displaystyle S = 2\pi \cdot \frac{2 + 5}{2} \cdot 5 = 35\pi.

By the integral, f'(x) = \tfrac34 and \displaystyle \sqrt{1 + \tfrac{9}{16}} = \tfrac54:

\begin{aligned} S &= \int_0^4 2\pi\left(\tfrac34 x + 2\right)\cdot\tfrac54\,dx \\ &= \frac{5\pi}{2}\left(\tfrac38 x^2 + 2x\right)\Big|_0^4 = \frac{5\pi}{2}(6 + 8) = 35\pi \end{aligned}

about 109.96 square centimeters. The two agree, as they must: a segment is its own inscribed path. With dx in place of ds, the integral 2\pi\int_0^4 \left(\tfrac34 x + 2\right)dx is 2\pi(6 + 8) = 28\pi \approx 87.96, exactly \displaystyle \frac45 of the true area. The ratio \displaystyle \frac45 is run over slant. Every piece of this line has the same tilt, so cutting it finer changes nothing.

Example 3

A paraboloid reflector, about the y-axis

Hasmik spins a reflector whose profile is y = x^2, 0 \le x \le 2, in decimeters, turned about the y-axis. Find its area with ds written in x, then again in y.

Solution

In x: the radius is the distance to the y-axis, x, and f'(x) = 2x:

S = \int_0^2 2\pi x\sqrt{1 + 4x^2}\,dx

Substitute u = 1 + 4x^2, so du = 8x\,dx, and u runs from 1 to 17:

\begin{aligned} S &= \frac{2\pi}{8}\int_1^{17} \sqrt{u}\,du = \frac{\pi}{4}\cdot\frac23 u^{3/2}\Big|_1^{17} \\ &= \frac{\pi}{6}\left(17\sqrt{17} - 1\right) \approx 36.18 \end{aligned}

square decimeters.

In y: the curve is \displaystyle x = g(y) = \sqrt{y} for 0 \le y \le 4, the radius is g(y), and \displaystyle g'(y) = \dfrac{1}{2\sqrt{y}}, so 1 + [g'(y)]^2 = \dfrac{4y + 1}{4y} and

\begin{aligned} g(y)\sqrt{1 + [g'(y)]^2} &= \sqrt{y}\cdot\frac{\sqrt{4y + 1}}{2\sqrt{y}} \\ &= \frac{\sqrt{4y + 1}}{2} \end{aligned}

\begin{aligned} S &= \pi\int_0^4 \sqrt{4y + 1}\,dy = \frac{\pi}{6}(4y + 1)^{3/2}\Big|_0^4 \\ &= \frac{\pi}{6}\left(17\sqrt{17} - 1\right) \end{aligned}

The same area. Here g'(y) is unbounded at y = 0, but the product with the radius stays bounded, so the integral gives no trouble. The flat disk across the mouth has area 4\pi \approx 12.57; the reflector needs almost three times as much metal.

Example 4

A zone of a sphere

A sphere of radius r is cut by two parallel planes a distance h apart, both meeting the sphere. Show that the area between them is 2\pi r h. Then compare, on Vahagn’s lamp globe of radius 10 centimeters, the band from the center plane to 2 centimeters above it with the cap that is 2 centimeters tall.

Solution

The sphere is the half circle \displaystyle y = \sqrt{r^2 - x^2} turned about the x-axis, and the planes are x = a and x = a + h with -r \le a < a + h \le r. Here \displaystyle f'(x) = \dfrac{-x}{\sqrt{r^2 - x^2}}, so

1 + [f'(x)]^2 = \frac{r^2 - x^2 + x^2}{r^2 - x^2} = \frac{r^2}{r^2 - x^2}

so \displaystyle f(x)\sqrt{1 + [f'(x)]^2} is \displaystyle \sqrt{r^2 - x^2}\cdot\dfrac{r}{\sqrt{r^2 - x^2}}, which is r. The integrand is the constant 2\pi r:

S = \int_a^{a + h} 2\pi r\,dx = 2\pi r h

Toward the poles the radius shrinks and the curve steepens in exact proportion, so each band of height dx has the same area. The zone’s area depends only on its height. The whole sphere, h = 2r, gives 4\pi r^2, the surface area of a sphere. Archimedes proved, in On the Sphere and Cylinder, that a cap of a sphere has the area of a circle whose radius is the segment from the cap’s top to its rim; the zone result follows from it by subtracting caps.

On the globe, both pieces have height 2, so both have area 2\pi(10)(2) = 40\pi \approx 125.66 square centimeters, though the cap looks far smaller. At x = \pm r the derivative is unbounded, so a zone reaching a pole is strictly an improper integral; its integrand is constant, so the limit is immediate.

Example 5

A bowl with no elementary antiderivative

Bakhyt spins a bowl whose profile is y = \dfrac{x^4}{16}, 0 \le x \le 2, in decimeters, about the y-axis: 1 decimeter deep with a mouth of radius 2. Estimate its area.

Solution

The radius is x and f'(x) = \dfrac{x^3}{4}, so

S = \int_0^2 2\pi x\sqrt{1 + \frac{x^6}{16}}\,dx

The substitution u = x^2 turns this into \displaystyle \pi\int_0^4 \sqrt{1 + \frac{u^3}{16}}\,du, a square root of a cubic, which has no elementary antiderivative (stated without proof). So estimate. Simpson’s rule with n = 4 uses \displaystyle \Delta x = \frac12 and the values of \displaystyle g(x) = 2\pi x\sqrt{1 + x^6/16}:

x	g(x)
0	0
0.5	3.1431
1	6.4766
1.5	12.3314
2	28.0993

\begin{aligned} S &\approx \frac{0.5}{3}\big(0 + 4(3.1431) + 2(6.4766) \\ &\qquad + 4(12.3314) + 28.0993\big) \approx 17.158 \end{aligned}

A computer’s quadrature gives S \approx 17.1427 square decimeters. As a check, the stretch factor is at least 1, so S \ge \int_0^2 2\pi x\,dx = 4\pi \approx 12.57.

Example 6

A horn with finite volume and infinite area

Turn the graph of y = \dfrac{1}{x}, x \ge 1, about the x-axis (Figure 2). Show that the solid has finite volume but the surface has infinite area.

Solution

By the disk method, with an improper integral:

V = \pi\int_1^{\infty} \frac{dx}{x^2} = \lim_{b \to \infty} \pi\left(1 - \frac1b\right) = \pi

Figure 2The horn of Example 6. Its radius \displaystyle \frac{1}{x} shrinks, but not fast enough for its surface area to stay finite.

For the surface, f'(x) = -\dfrac{1}{x^2}, so

S = \int_1^{\infty} \frac{2\pi}{x}\sqrt{1 + \frac{1}{x^4}}\,dx

The root is at least 1, so the integrand is at least \dfrac{2\pi}{x} > 0, and \displaystyle \int_1^{\infty} \frac{2\pi}{x}\,dx = \lim_{b \to \infty} 2\pi\ln b does not exist as a number. Both integrands are continuous on [1, \infty), so by the direct comparison test S diverges: S = \infty.

The solid, often called Gabriel’s horn, holds \pi cubic units, yet its surface has no finite area. There is no contradiction: volume adds the square of the radius, \displaystyle \frac{1}{x^2}, which falls fast enough, while area adds the radius itself, \displaystyle \frac{1}{x}, which does not.

Boundaries

The radius is a distance. About the x-axis it is \lvert f(x) \rvert; the formula with f(x) assumes f \ge 0. About any axis, measure from that axis.

f' must be continuous on the closed interval. At a vertical tangent the integral is improper; it may still converge (Example 4). A corner splits the interval into pieces.

ds, not dx. \int 2\pi r\,dx equals the area only where the curve runs parallel to the axis; elsewhere it is too small by the stretch factor.

Each piece of surface once. A whole circle turned about a diameter sweeps the sphere twice, once from each half. Turn only a part of the curve that sweeps each point of the surface once.

An infinite interval gives an improper integral, which can diverge even when the volume converges (Example 6).

Exact answers are the exception. A decimal from numerical integration is a complete answer.

Where this goes next

For a curve traced by a parameter t, \displaystyle ds = \sqrt{[x'(t)]^2 + [y'(t)]^2}\,dt and the area is again \int 2\pi r\,ds (arc length for parametric curves). The shell method uses the same circumference 2\pi r, times a height instead of a slant, to build volumes. In several-variable calculus, 2\pi r\,ds becomes the surface area element of a surface integral.

Exercises

About the x-axis

Find the area of the surface swept out by each curve.

\displaystyle y = 2\sqrt{x}, 0 \le x \le 3.
y = x^3, 0 \le x \le 1.
y = 2x, 0 \le x \le 3. Check against the cone’s \pi r \ell.
y = x, 0 \le x \le 2, turned about the line y = -1. Check with one band.
y = \dfrac{x^3}{3} + \dfrac{1}{4x}, 1 \le x \le 2.

About the y-axis

y = \dfrac{x^2}{4}, \displaystyle 0 \le x \le 2\sqrt2.
y = 2x - 2, 1 \le x \le 3. Check with one band.
x = \dfrac{y^3}{3}, 0 \le y \le 1.

Spheres and zones

Find the area swept out by \displaystyle y = \sqrt{9 - x^2}, -2 \le x \le 1, about the x-axis.
A sphere of radius 5 is cut by a plane 3 from its center. Find the area of the smaller cap, and confirm Archimedes’ circle gives the same.
What fraction of a sphere’s surface lies within \displaystyle \frac{r}{2} of a plane through its center?

Numerical and improper

Use a computer to estimate the area swept out by y = \dfrac{1}{x}, 1 \le x \le 3, about the x-axis, and explain why it must exceed 2\pi\ln 3.
Find the area swept out by y = e^{-x}, x \ge 0, about the x-axis. You may use this antiderivative, from Trigonometric substitution:

\begin{aligned} &\int \sqrt{1 + u^2}\,du \\ &\quad = \tfrac12\Big(u\sqrt{1 + u^2} + \ln\big(u + \sqrt{1 + u^2}\big)\Big) + C \end{aligned}

Reasoning

Turn y = x^{-p}, x \ge 1, p > 0, about the x-axis. For which p is the volume finite, and for which p is the area finite?
For y = 3 - x, 0 \le x \le 3, about the x-axis, compute S and \int_0^3 2\pi y\,dx, and explain their ratio.
Show that S \ge 2\pi\int_a^b f(x)\,dx for f \ge 0, with equality when f is constant.
Show that scaling a profile by k > 0 in every direction multiplies its surface area by k^2.
Turn y = \dfrac{r}{h}x, 0 \le x \le h, about the x-axis, and show the integral gives the cone’s \pi r\ell.

Answers

\displaystyle \frac{56\pi}{3} \approx 58.64. \displaystyle f'(x) = \frac{1}{\sqrt{x}}, so \displaystyle f\sqrt{1 + f'^2} = 2\sqrt{x}\cdot\frac{\sqrt{x + 1}}{\sqrt{x}} = 2\sqrt{x + 1} and \displaystyle S = 4\pi\int_0^3 \sqrt{x + 1}\,dx = 4\pi\cdot\frac23(8 - 1).
\displaystyle \frac{\pi}{27}\left(10\sqrt{10} - 1\right) \approx 3.5631. \displaystyle S = \int_0^1 2\pi x^3\sqrt{1 + 9x^4}\,dx; with u = 1 + 9x^4, du = 36x^3\,dx, \displaystyle S = \frac{\pi}{18}\int_1^{10} \sqrt{u}\,du.
\displaystyle 18\sqrt5\,\pi \approx 126.45. \displaystyle \int_0^3 2\pi(2x)\sqrt5\,dx = 4\sqrt5\,\pi \cdot \frac92. The cone has r = 6, \displaystyle \ell = \sqrt{9 + 36} = 3\sqrt5, and \displaystyle \pi r\ell = 18\sqrt5\,\pi.
\displaystyle 8\sqrt2\,\pi \approx 35.54. The radius is x + 1 and \displaystyle ds = \sqrt2\,dx: \displaystyle 2\sqrt2\,\pi\int_0^2 (x + 1)\,dx = 2\sqrt2\,\pi\cdot 4. One band: radii 1 and 3, \displaystyle \ell = 2\sqrt2, \displaystyle 2\pi\cdot 2\cdot 2\sqrt2.
\displaystyle \frac{515\pi}{64} \approx 25.28. \displaystyle f'(x) = x^2 - \frac{1}{4x^2}, so \displaystyle 1 + f'^2 = \left(x^2 + \frac{1}{4x^2}\right)^2 and the integrand of S is 2\pi times \displaystyle \frac{x^5}{3} + \frac{x}{3} + \frac{1}{16x^3}, so \displaystyle S = 2\pi\left(\frac72 + \frac12 + \frac{3}{128}\right).
\displaystyle \frac{8\pi}{3}\left(3\sqrt3 - 1\right) \approx 35.15. \displaystyle S = \int_0^{2\sqrt2} 2\pi x\sqrt{1 + \frac{x^2}{4}}\,dx; with \displaystyle u = 1 + \frac{x^2}{4}, \displaystyle du = \frac{x}{2}\,dx, \displaystyle S = 4\pi\int_1^3 \sqrt{u}\,du.
\displaystyle 8\sqrt5\,\pi \approx 56.20. \displaystyle \int_1^3 2\pi x\sqrt5\,dx = \sqrt5\,\pi(9 - 1). One band: radii 1 and 3, \displaystyle \ell = \sqrt{4 + 16} = 2\sqrt5, \displaystyle 2\pi\cdot 2\cdot 2\sqrt5.
\displaystyle \frac{\pi}{9}\left(2\sqrt2 - 1\right) \approx 0.6382. \displaystyle S = \int_0^1 2\pi\frac{y^3}{3}\sqrt{1 + y^4}\,dy; with u = 1 + y^4, \displaystyle S = \frac{\pi}{6}\int_1^2 \sqrt{u}\,du.
18\pi \approx 56.55. A zone of a sphere with r = 3 and height h = 3: 2\pi(3)(3).
20\pi \approx 62.83. The cap has height 5 - 3 = 2, so 2\pi(5)(2). Its top is (5, 0) and its rim passes through (3, 4), \displaystyle \sqrt{4 + 16} = \sqrt{20} apart; a circle of radius \displaystyle \sqrt{20} has area 20\pi.
\displaystyle \frac12. The zone has height r, so area 2\pi r\cdot r, half of 4\pi r^2.
S \approx 7.6031, from \displaystyle \int_1^3 \frac{2\pi}{x}\sqrt{1 + x^{-4}}\,dx. The root exceeds 1 for every x, so \displaystyle S > \int_1^3 \frac{2\pi}{x}\,dx = 2\pi\ln 3 \approx 6.9028.
\displaystyle \pi\left(\sqrt2 + \ln\left(1 + \sqrt2\right)\right) \approx 7.2118. \displaystyle S = \int_0^{\infty} 2\pi e^{-x}\sqrt{1 + e^{-2x}}\,dx; with u = e^{-x}, du = -e^{-x}\,dx, this is \displaystyle 2\pi\int_0^1 \sqrt{1 + u^2}\,du, which converges.
Volume finite exactly when \displaystyle p > \frac12; area finite exactly when p > 1. V = \pi\int_1^{\infty} x^{-2p}\,dx converges when 2p > 1. For the area, \displaystyle x^{-p} \le x^{-p}\sqrt{1 + p^2x^{-2p-2}} \le \sqrt{1 + p^2}\,x^{-p} on x \ge 1, so by comparison S converges exactly when \int_1^{\infty} x^{-p}\,dx does, p > 1. For \displaystyle \frac12 < p \le 1 the volume is finite and the area is not.
\displaystyle S = 9\sqrt2\,\pi \approx 39.99 and \int_0^3 2\pi(3 - x)\,dx = 9\pi \approx 28.27. Here \displaystyle ds = \sqrt2\,dx everywhere, so the dx integral measures every band by its run \Delta x instead of its slant \displaystyle \sqrt2\,\Delta x, and comes out \displaystyle \frac{1}{\sqrt2} of the area.
\displaystyle \sqrt{1 + [f'(x)]^2} \ge 1 and 2\pi f(x) \ge 0, so the integrand of S is at least 2\pi f(x) at every x, and integrals keep the order. If f is constant, f' = 0 and the two integrands are equal.
The scaled profile is \displaystyle y = k f\!\left(\frac{x}{k}\right) on [ka, kb]. Its derivative is \displaystyle f'\!\left(\frac{x}{k}\right), so its area is the integral of \displaystyle 2\pi k f(x/k)\sqrt{1 + [f'(x/k)]^2} over [ka, kb]. With \displaystyle t = \frac{x}{k} and dx = k\,dt, the factor k in the radius and the factor k in dx multiply, and the integral becomes k^2 times \displaystyle \int_a^b 2\pi f(t)\sqrt{1 + [f'(t)]^2}\,dt, which is k^2 S.
\displaystyle f' = \frac{r}{h}, so \displaystyle ds = \frac{\sqrt{h^2 + r^2}}{h}\,dx and \displaystyle S = \frac{2\pi r\sqrt{h^2 + r^2}}{h^2}\int_0^h x\,dx, which is \displaystyle \pi r\sqrt{h^2 + r^2} = \pi r\ell since the slant height is \displaystyle \ell = \sqrt{h^2 + r^2}.