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).

3leafedrose

4leafrose

6leafedrose

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.

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