Surface Area of Revolution

Suppose y=f(x) is some function you’d like to rotate around the x-axis (over some interval I), and that you’d like to calculate the surface area of the resulting surface. Dig up your favorite calculus reference, and you’ll find something like the formula:

\int_I 2\pi f(x)\sqrt{1+(f'(x))^2}\ dx.

If you read through the proof, you’ll see that this formula comes from breaking up the surface into lots of little end bits of cones. The way I explained those shapes to my class this semester was: take an ice cream cone and dip the “top” (where the ice cream goes) into chocolate. The bit that’s covered in chocolate is then the sort of shape we’re considering, on each little interval. These little bits give us a reasonable approximations on each interval, and we take a limit and a sum and it becomes the above integral.

Recently, another of the calc instructors asked: Why don’t we approximate each little bit with just cylinders? Well, the first several answers he got (there were a few of us talking about this) were “you get the wrong answer”, essentially. “The cylinder and cone have different surface areas”. Of course they do, but in the limit, why does it matter? Don’t those cylinders get really close to approximating the length of the curve?

Let’s start getting some formulas going. Consider the line y=mx on the interval [a,a+h]. For convenience, take m,a>0. Also consider the horizontal line y=ma on the same interval. Rotating the line y=mx gives us the cone, and rotating the line y=ma gives us the cylinder. Notice that we chose the horizontal line to be the value of the diagonal line at the point x=a.

While we’re thinking about surface area, let’s also think about volumes, to see what is different between the two situations. The formulas for the cylinder are easy, the surface area of the cylinder we made in the previous paragraph is just 2\pi mah, and the volume is \pi (ma)^2h. The formulas for the bit of the cone are slightly more complicated, but doable. If you’ve still got your calc reference out and find these formulas, you should be able to get the surface area to be \pi m\sqrt{m^2+1}(2ah+h^2) and volume as \frac{1}{3}\pi m^2(3a^2h+3ah^2+h^3).

Now what we want to do is compare how the formulas for surface area (and volume) compare as h\to 0. The way to see how they compare is to take their ratio. In the case of volume, if we put the volume of the cone in the numerator, we can simplify the formula to


Taking the limit as h\to 0 we get 1. That means that the formulas are essentially equivalent in terms of using them to find volume of a solid of revolution. However, in the case of surface area, the quotient boils down to


which only goes to 1 in the limit if m=0 (i.e., the cone is really a cylinder).

Now, if a student asks about this in calc class, how do we justify this process of taking quotients and looking at the limit? Has anybody ever had a student ask them this question?


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One Response to “Surface Area of Revolution”

  1. Deepak Suwalka Says:

    Thanks. It’s a nice post about surface area calculus. I really like it :). It’s really helpful. Good job.

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