Surface Area

Surface area here is simply the area of the outer skin of a shape you get by spinning a curve around a line, like a vase on a potter's wheel. All we're really asking is how much material it would take to cover that outside, which is what you need to price sheet metal, paint or heat shielding. The formula $\int 2\pi r\,ds$ looks like a lot, but it's just three simple parts multiplied and added. $r$ (how far that bit of curve is from the spinning axis), $2\pi r$ (the distance around the circle it traces, the same circumference formula you've known for years), and $ds$ (the width of the thin ring, the tiny piece of curve from the last lesson). Multiply around × width to get one ring, and let the integral add up the rings. It's not as bad as it looks.

This lesson builds right on the last one. Arc length gave us $ds$, the length of a tiny piece of curve. Now spin that piece around an axis. It sweeps out a thin band, like a ring cut from a lampshade (the official name is a frustum, a slice of a cone). Unroll that band and it's nearly a long thin strip: its length is the distance around, and its width is the slant width $ds$. So its area is (average circumference) × (slant width), and in the limit that's $2\pi r\,ds$, where

• $r$ is the distance from the curve to the axis of rotation: $f(x)$ for the x-axis, $x$ for the y-axis, $f(x) - c$ for the line $y = c$;
• $ds$ is the arc length element: $\sqrt{1 + [f'(x)]^2}\,dx$, or $\sqrt{1 + [g'(y)]^2}\,dy$ if the curve is $x = g(y)$.

So every surface area problem is really the same one: $S = \int 2\pi r\,ds$. Pick whichever variable makes $ds$ easier, and make sure $r$ is written in that same variable.

A common mistake. It's tempting to use $dx$ for the band's width, because that's what we did for volumes with disks. It isn't the same thing. A disk is a flat slice, so its thickness really is $dx$ and its volume is $\pi r^2\,dx$. A band on the surface is slanted, and where the curve is steep it's much wider than $dx$. Use $dx$ there and you'd get the wrong area for every surface except a cylinder. The width has to be $ds$.

Material cost is surface area: sheet metal for a tank, glaze on a vase, heat-shield tiles on a capsule. Gabriel's horn (rotate $y = 1/x$, $x \ge 1$, about the x-axis) famously holds a finite volume of paint but needs infinite paint to coat.