Surface Area of Revolution

Revolve a curve and sum the bands: S = ∫ 2π r ds.

The idea

A surface of revolution is the surface a curve sweeps out when revolved about an axis: a line through the origin sweeps a cone, a semicircle a sphere. The problem is to measure the area of that surface — the outside skin of the solid, not the volume inside it.

Theorem (Area of a surface of revolution).

When a smooth curve is revolved about an axis, the surface it sweeps out has area $S = \int 2\pi r\,ds,$ where $r$ is the distance from the point of the curve to the axis and $ds$ is the arc-length element along the curve; for a curve $y = y(x)$, $ds = \sqrt{1 + \left(\frac{dy}{dx}\right)^{2}}\,dx$.

Cut the surface into thin bands, each traced by a short piece of the curve as it revolves. A band is very nearly a frustum, a slice of a cone: a circular strip of circumference $2\pi r$, where $r$ is the distance from that piece of curve to the axis, and of width $ds$, the length of the piece itself. Its area is circumference times width, and summing the bands gives the integral.

Two quantities must be supplied before the integral can be evaluated. The radius $r$ is the distance from a point on the curve to the axis being revolved about, so it depends on which axis that is. And $ds$ is longer than $dx$ because the curve slants. Replacing $ds$ by $dx$ would measure each band's horizontal shadow rather than the band, and would undercount the area by exactly the slant factor.

Ways to work on it

• Walkthrough. The band formula S = ∫ 2π r ds, built on a cone.
• Practice. Surface area of a cone from a revolved line.
• Hardest. Compute the surface area of a revolved curve whose slant factor genuinely varies.

Not sure where to start? Take the ten-question placement test.