Surface Area of Revolution

The surface area of revolution is the area of the surface made by spinning a curve around an axis. For y = f(x) from a to b spun around the x-axis, S = \int_a^b 2\pi f(x)\sqrt{1 + \left(f'(x)\right)^2}\,dx. Each tiny piece of the curve sweeps out a thin band whose area is about its circumference times its slanted width.

Before this lesson: Arc Length of a Curve and Disk Method

Thin bands around the axis

Spin a curve around the x-axis and it traces a surface, like the outside of a vase. Cut the surface into thin bands. Each band is almost a short, slanted strip wrapped around the axis.

A band’s length around is its circumference, 2π times its radius. Its width is a tiny piece of the curve, ds = \sqrt{1 + (f'(x))^2}\,dx, the same piece used for arc length. So each band has area about 2π · radius · ds.

Surface area of revolution formula

In both cases, the radius is the distance from the curve to the axis, and the square root is the arc length piece ds. Here f(x) ≥ 0 and f′ is continuous.

1. Find f′(x), then 1 + (f′(x))².
2. Multiply the square root by 2π times the radius.
3. Simplify. Often the radius cancels part of the root.
4. Integrate, often with a u-substitution.

Surface area of revolution examples

Example 1: Spinning y = √x

Find the area of the surface made by spinning y = \sqrt{x}, from x = 0 to x = 4, around the x-axis.

f'(x) = \dfrac{1}{2\sqrt{x}}

1 + (f'(x))^2 = 1 + \dfrac{1}{4x} = \dfrac{4x + 1}{4x}

2\pi\sqrt{x}\cdot\sqrt{\dfrac{4x + 1}{4x}} = 2\pi\cdot\dfrac{\sqrt{4x + 1}}{2} = \pi\sqrt{4x + 1}

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

= \pi\cdot\dfrac{1}{4}\cdot\dfrac{2}{3}(4x + 1)^{3/2}\Big|_0^4

= \dfrac{\pi}{6}\left(17^{3/2} - 1\right) \approx 36.18

Answer: S = \dfrac{\pi}{6}\left(17\sqrt{17} - 1\right) \approx 36.18.

Example 2: The surface of a sphere

Use the formula to find the surface area of a sphere of radius 3.

f(x) = \sqrt{9 - x^2}, \; f'(x) = \dfrac{-x}{\sqrt{9 - x^2}}

1 + \dfrac{x^2}{9 - x^2} = \dfrac{9}{9 - x^2}

2\pi\sqrt{9 - x^2}\cdot\dfrac{3}{\sqrt{9 - x^2}} = 6\pi

S = \int_{-3}^{3} 6\pi\,dx = 6\pi(6) = 36\pi

Answer: S = 36\pi, which matches the geometry formula 4πr² = 4π(9) (see surface area and volume of spheres).

Example 3: A cone

Spin the line y = 2x, from x = 0 to x = 3, around the x-axis. Find the surface area and check it.

f'(x) = 2, \; \sqrt{1 + 4} = \sqrt{5}

S = \int_0^3 2\pi(2x)\sqrt{5}\,dx = 4\sqrt{5}\,\pi\int_0^3 x\,dx

= 4\sqrt{5}\,\pi\cdot\dfrac{9}{2} = 18\sqrt{5}\,\pi

\pi r\ell = \pi(6)\left(\sqrt{3^2 + 6^2}\right) = 6\pi\cdot 3\sqrt{5} = 18\sqrt{5}\,\pi

Answer: S = 18\sqrt{5}\,\pi \approx 126.45.

Example 4: A curve that needs substitution

Find the area of the surface made by spinning y = x³, from x = 0 to x = 1, around the x-axis.

f'(x) = 3x^2, \; 1 + (f'(x))^2 = 1 + 9x^4

S = \int_0^1 2\pi x^3\sqrt{1 + 9x^4}\,dx

u = 1 + 9x^4, \; du = 36x^3\,dx; \; x = 0 \to u = 1, \; x = 1 \to u = 10

= \dfrac{2\pi}{36}\int_1^{10} u^{1/2}\,du = \dfrac{\pi}{18}\cdot\dfrac{2}{3}u^{3/2}\Big|_1^{10}

= \dfrac{\pi}{27}\left(10\sqrt{10} - 1\right) \approx 3.563

Answer: S = \dfrac{\pi}{27}\left(10\sqrt{10} - 1\right) \approx 3.563.

Example 5: Coating a satellite dish

A dish is made by spinning y = \dfrac{x^2}{4}, from x = 0 to x = 2, around the y-axis (in meters). It is 4 meters across and 1 meter deep. How much area must be coated?

f'(x) = \dfrac{x}{2}, \; 1 + \dfrac{x^2}{4}

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

u = 1 + \dfrac{x^2}{4}, \; du = \dfrac{x}{2}\,dx; \; x = 0 \to u = 1, \; x = 2 \to u = 2

= 2\pi\int_1^2 2\sqrt{u}\,du = 4\pi\cdot\dfrac{2}{3}u^{3/2}\Big|_1^2

= \dfrac{8\pi}{3}\left(2\sqrt{2} - 1\right) \approx 15.32

Answer: about 15.3 square meters must be coated.

Common mistakes

Related methods

Curves written as x = g(y)

If the curve is x = g(y) from y = c to y = d, use S = \int_c^d 2\pi g(y)\sqrt{1 + (g'(y))^2}\,dy around the y-axis, or replace g(y) with y as the radius for the x-axis.

Numerical estimates

Like arc length, most surface area integrals have no elementary antiderivative. Estimate them with Simpson’s rule.

Gabriel’s horn

Spin y = 1/x, for x ≥ 1, around the x-axis. The volume is π, but the surface area is infinite, because 2\pi\dfrac{1}{x}\sqrt{1 + \dfrac{1}{x^4}} \ge \dfrac{2\pi}{x}, whose integral diverges by the comparison test for improper integrals.

Practice problems

Find the surface area. Spin around the x-axis unless told otherwise.

1. y = 3x, from x = 0 to x = 2
2. y = √x, from x = 1 to x = 4
3. y = \sqrt{4 - x^2}, from x = −1 to x = 1
4. y = \dfrac{x^3}{3}, from x = 0 to x = 1
5. y = 1 − x, from x = 0 to x = 1, around the y-axis
6. y = x², from x = 0 to x = √2, around the y-axis
7. y = cosh x, from x = 0 to x = 1
8. y = sin x, from x = 0 to x = π. Use \int_{-1}^{1} \sqrt{1 + u^2}\,du = \sqrt{2} + \ln(1 + \sqrt{2}).
9. Show that a sphere of radius R has surface area 4πR².
10. A lampshade is made by spinning y = x + 1, from x = 0 to x = 2, around the x-axis (in feet). Find its area.

f'(x) = 3, \; \sqrt{1 + 9} = \sqrt{10}

S = \int_0^2 2\pi(3x)\sqrt{10}\,dx = 6\sqrt{10}\,\pi\cdot\dfrac{2^2}{2}

= 12\sqrt{10}\,\pi \approx 119.22

2\pi\sqrt{x}\sqrt{1 + \dfrac{1}{4x}} = \pi\sqrt{4x + 1}

S = \dfrac{\pi}{6}(4x + 1)^{3/2}\Big|_1^4 = \dfrac{\pi}{6}\left(17^{3/2} - 5^{3/2}\right)

\approx \dfrac{\pi}{6}(70.093 - 11.180) \approx 30.85

2\pi\sqrt{4 - x^2}\cdot\dfrac{2}{\sqrt{4 - x^2}} = 4\pi

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

S = \int_0^1 2\pi\cdot\dfrac{x^3}{3}\sqrt{1 + x^4}\,dx

u = 1 + x^4, \; du = 4x^3\,dx

= \dfrac{2\pi}{3}\cdot\dfrac{1}{4}\cdot\dfrac{2}{3}u^{3/2}\Big|_1^2 = \dfrac{\pi}{9}\left(2\sqrt{2} - 1\right)

f'(x) = -1, \; \sqrt{1 + 1} = \sqrt{2}

S = \int_0^1 2\pi x\sqrt{2}\,dx = 2\sqrt{2}\,\pi\cdot\dfrac{1}{2} = \sqrt{2}\,\pi

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

u = 1 + 4x^2, \; du = 8x\,dx

= \dfrac{2\pi}{8}\cdot\dfrac{2}{3}u^{3/2}\Big|_1^9 = \dfrac{\pi}{6}(27 - 1) = \dfrac{13\pi}{3}

2\pi\cosh x\sqrt{1 + \sinh^2 x} = 2\pi\cosh^2 x

= \pi(1 + \cosh 2x)

S = \pi\Big[x + \dfrac{\sinh 2x}{2}\Big]_0^1 = \pi\left(1 + \dfrac{\sinh 2}{2}\right)

\approx \pi(1 + 1.81343) \approx 8.839

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

u = \cos x, \; du = -\sin x\,dx

= 2\pi\int_{-1}^{1} \sqrt{1 + u^2}\,du

= 2\pi\left(\sqrt{2} + \ln(1 + \sqrt{2})\right) \approx 14.42

2\pi\sqrt{R^2 - x^2}\cdot\dfrac{R}{\sqrt{R^2 - x^2}} = 2\pi R

S = \int_{-R}^{R} 2\pi R\,dx = 2\pi R(2R) = 4\pi R^2

f'(x) = 1, \; \sqrt{2}

S = \int_0^2 2\pi(x + 1)\sqrt{2}\,dx = 2\sqrt{2}\,\pi\Big[\dfrac{x^2}{2} + x\Big]_0^2

= 2\sqrt{2}\,\pi(2 + 2) = 8\sqrt{2}\,\pi \approx 35.54

Surface area of revolution FAQ

What is the formula for surface area of revolution?

Around the x-axis, S = ∫ 2π f(x) √(1 + (f′(x))²) dx. Around the y-axis, replace f(x) with x. In both, 2π times the radius is multiplied by the arc length piece.

How is surface area different from volume of revolution?

Volume adds up disks, πr² dx, and needs no square root. Surface area adds up thin bands, 2πr ds, and uses the arc length piece ds.

Why is there a square root in the formula?

Each band is slanted, not flat. Its true width is a piece of the curve’s length, ds = √(1 + (f′)²) dx, not dx.

How do you find surface area around the y-axis?

Use the distance to the y-axis, x, as the radius: S = ∫ 2πx√(1 + (f′(x))²) dx, with x from a to b.

Summary

• Surface area of revolution: S = ∫ 2π · radius · ds, with ds = √(1 + (f′)²) dx.
• Around the x-axis, the radius is f(x); around the y-axis, it is x.
• The radius often cancels part of the root; u-substitution often finishes the job.
• Check answers against the sphere (4πr²) and cone (πrℓ) formulas.
• Hard cases need numerical methods.

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