## Archive for July, 2009

### Polar Areas

July 20, 2009

This summer I had the opportunity to teach Calc 2 for the third time in a row. The first few topics I covered were, essentially, “Things you can do with integrals.” Like finding arc length, parametric area, surface area of revolution, polar area, and iterated integrals for volumes.

I tried to present these as all following the same basic principle. In each case, you have something you would like to calculate, but it’s a curvy thing, so it’s hard to do. So you split it into lots of little bits, and then approximate whatever it is you are looking for for each of those little bits. Then you take a limit, and it becomes an integral.

This seemed to work out ok, but I did run into a question when we got to polar areas.

When you break up your polar area into lots of little bits, the standard method is to approximate each of those little bits with sectors of circles, which have an easily calculated area. If your small bit is over an interval where $\theta$ changes by some small amount, $d\theta$, then the area is $\frac{1}{2}r^2d\theta$, where $r$ is the radius of any sample point in the interval.

What struck me (and at least one of my students) was that you should also be able to approximate this little area using a triangle. Two formulas I came up with were $\frac{1}{2}r^2\sin d\theta$ or $r^2\tan(d\theta/2)$. That’s a little bit of a problem, for an integral, because your differentials shouldn’t be inside other functions like that. Right?

Does anybody know of a different way to express the area of approximating triangles so that you can do the integral for polar area some other way? Or is there some reason I shouldn’t think that such a process might work and give the right integral?

Update 20090814: Several commenters pointed out a typo in the formula above with tangent. A power of two was missing from r, which has since been corrected.

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