First of all, a surface of revolution is not obtained by rotating a region about some axis, but by rotating a segment of a curve (which might happen to be on the edge of the region). What Pantsless Bob describes up to and including step 3 of his procedure is the process for finding the volume of a solid of revolution. Secondly, he comes up with formulas in step 5 without giving any justification of why they are what they are. This is bad, folks. Everything in calculus should be derivable (in the proof sense) and pulling formulae out of your backside is, in fact, bad, no matter how much your teachers and/or friends may like memorizing the book text. So, here I shall attempt to describe, in a correct and explained manner, how to find the area of a surface of revolution.

We have to realize that with most surfaces of revolution, the surface curves in two ways: the curvature of the initial curve segment, and the curvature produced by the circular rotation about the axis. (The only solids of revolution with singular curvature are the ones that come from straight lines in the first place, that is, cylinders and cones.) This makes them impossible to flatten; imagine making an orange peel perfectly flat, without tearing, stretching, or otherwise mangling. Impossible, eh? This is because the orange peel is a rough estimate of a sphere, made possible by the rotation of a semicircle about the axis running through its center and endpoints.

Anyway, in order to find the area of the surface, like most other things in calculus, we slice it! Now, the smaller and smaller a selected section of the curve gets, the more and more it resembles a straight line. If we rotate this "straight line segment" about the axis of rotation, we get a frustum, or truncated cone. In "frustum", it is shown that the curved surface area of this frustum would be S = 2πrl, where r is the radius and l is the slant height of the frustum. Obviously, for the slice, r represents the distance from the slice to the axis of rotation; the slant length can be represented by the length of the slice, commonly represented as dL. This makes the formula for the slice of a surface of revolution

dS = 2πr dL *

When you think about it, this makes sense: it's like taking a certain length (dL) and rotating it with a certain radius (r) to make a circular band (2πr dL). To find the slice length, use the Pythagorean theorem,

dL = √(dy2 + dx2) *

dL = dx√(dy2/dx2 + 1)

dL = √[(y′)2 + 1] dx for an equation of the form y(x)

(Note: if you're like me, all you'll need to memorize are the two equations with * by them, as they make sense in a geometrical sense anyway.)

Also, find an equation for r. If it is rotated about the x-axis, the radius is going to be the distance to the x-axis, i.e. |y|. Vice-versa applies for if it's rotated about the y-axis. If some other line is the axis of rotation, find the distance from any point on your rotating line to the axis; for example, if the line is y = -3, the radius will be r = |y + 3|; if the line is y = x, the radius will be r = |(y - x)/√2|. Some of these are a bit tricky, eh? Note that, if the entire section of the graph to be rotated is on the positive side of the axis (as most problems are), no absolute value is required.

Then, substitute the slice length into the frustum surface area equation. This depends on what the nature of your graph is. Integrate with your given limits, stir, and enjoy!

Well, I suppose I'm obligated to give you all an example. Here we go:

The curve y = ex is rotated about the x-axis from x = 1 to x = 5. Find the area of the surface of revolution.
First, we notice that it is rotated about the x-axis, so r = y. In other words, r = ex. There's one part of it; now let's get dL.

y = ex

dy = ex dx

dL = √(e2x dx2 + dx2)

dL = √(e2x + 1) dx

Now for dS:

dS = 2πr dL

dS = 2πex √(e2x + 1) dx

Now simply (heh) set the limits of integration, and we're good to go!

S = ∫15 2πex √(e2x + 1) dx

Hoo, boy... perhaps I shouldn't have chosen that one as an example. Let's let that be an exercise for the reader. But, that's how you do the area of a surface of revolution. Simple, isn't it?

Third in a series of MathRants by PMDBoi.