College Math Teaching

December 22, 2015

Multi leaf polar graphs and total area…

Filed under: calculus, elementary mathematics, integrals — Tags: , — collegemathteaching @ 4:07 am

I saw polar coordinate calculus for the first time in 1977. I’ve taught calculus as a TA and as a professor since 1987. And yet, I’ve never thought of this simple little fact.

Consider $r(\theta) = sin(n \theta), 0 \theta \ 2 \pi$. Now it is well know that the area formula (area enclosed by a polar graph, assuming no “doubling”, self intersections, etc.) is $A = \frac{1}{2} \int^b_a (r(\theta))^2 d \theta$

Now the leaved roses have the following types of graphs: $n$ leaves if $n$ is odd, and $2n$ leaves if $n$ is even (in the odd case, the graph doubles itself).

So here is the question: how much total area is covered by the graph (all the leaves put together, do NOT count “overlapping”)?

Well, for $n$ an integer, the answer is: $\frac{\pi}{4}$ if $n$ is odd, and $\frac{\pi}{2}$ if $n$ is even! That’s it! Want to know why?

Do the integral: if $n$ is odd, our total area is $\frac{n}{2}\int^{\frac{\pi}{n}}_0 (sin(n \theta)^2 d\theta = \frac{n}{2}\int^{\frac{\pi}{n}}_0 \frac{1}{2} + cos(2n\theta) d\theta =\frac{\pi}{4}$. If $n$ is even, we have the same integral but the outside coefficient is $\frac{2n}{2} = n$ which is the only difference. Aside from parity, the number of leaves does not matter as to the total area!

Now the fun starts when one considers a fractional multiple of $\theta$ and I might ponder that some.