## Polar Areas

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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### 5 Responses to “Polar Areas”

1. Michael Holroyd Says:

I believe Carl Gustav Jacob Jacobi has an opinion on that.

2. Jaime Says:

I think you are missing a 2 after the ‘r’ on your expression with the tangent…

If you now proceed to approximate either the sine of the tangent by its Maclaurin series, you end up with the original formula:

sin x = x – x3/3! + x5/5! – …

tan x = x + x3/3 + 2·x5/15 + …

Any other expression you can come up with will reduce to r2 dθ / 2 when you linearize it and drop all higher order terms.

This reminds me of my first physics exam in university, with some long problem of a boat steering towards a fixed point on land with a current parallel to the shore… I did realize that the proper setting where polar coordinates centered at the point the boat was steering towards, but decided that the length element in polar coordinates was dl2 = dr2 + dθ2 rather than dl2 = dr2 + r2dθ2… If I remember well, I turned it in with a side note along the lines of “..there must be some mistake somewhere, because it doesn’t make much sense that the starting velocity has to be a complex number…” which did not save me from not passing it…

3. Anonymous Says:

sin(dtheta) equals dtheta plus higher order terms (which don’t matter), so that gives the same answer. By contrast, the r tan(dtheta/2) formula isn’t dimensionally correct. (Should it be r^2? That would give the right answer.)

4. Qiaochu Yuan Says:

There’s a perfectly well-defined way to think about expressions like $\sin (d \theta)$, which is just to replace $\Delta x$ with $\sin \Delta x$ in the Riemann sum definition of an integral. And as Anonymous suggests this makes essentially no difference.

5. sumidiot Says:

Ok, thanks for pointing out the missing power of 2 in my tangent formula (and sorry about the delayed response), I’ll fix that momentarily. I had thought about the linear approximation bit you all mention, but for some reason didn’t like thinking about it that way. Probably for no better reason than: at this point in the semester, we hadn’t talked about Taylor approximations (though I guess Linearizations did show up in calc 1, hmm, perhaps I don’t mind thinking about it so much any more).

Anyway, thanks, all, for the comments. And for pushing me to think more about using the linear approximation idea.