Finding area between two polar curves using double integrals

I have a homework question that is asking me to find the area that lies:

• Inside the curve $r=2+cos(2\theta)$
• But outside the curve $r=2+sin(\theta)$

I think I'm supposed to be using a double integral to solve this (the chapter was on multiple integrals, so I don't think I should use the $\frac{1}{2}\int thing$). For the region R, I think I should use $2+sin(\theta)\le r \le 2+cos(2\theta)$

So, for the final integral, I have: $$A=\int_{0}^{2\pi}\int_{2+sin(\theta)}^{2+cos(2\theta)}rdrd\theta$$ I'm not sure what, but it seems wrong, especially since I don't really have an integrand (aside from the $rdrd\theta$). Would this be correct or what would I use?

• Not having an integrand is fine - in fact, exactly what you'd expect from a "calculate the area" problem. In this sort of problem, you need to integrate the area element, which is precisely $rdrd\theta$. Your limits of integration seem to be correct. In general, if you're unsure of what the limits of integration should be, your first problem-solving step should be to draw a picture of the domain you need to consider. (Wolfram is great.) Actually, this is a pretty good first step regardless of whether or not you're unsure about the domain. – Gyu Eun Lee May 3 '13 at 4:47
• You'll find that what you get after carrying out the radial integration is just what you get in single-variable calculus with $\int_{\theta_1}^{\theta_2} \frac{1}{2} \cdot [r_{outer}^2 - \ r_{inner}^2] \ d\theta$ . We just never told you before that this came from a double integral... – colormegone May 3 '13 at 4:59

There isn't much difference between doing area integration in polar coordinates as a double integral and in the way you may have encountered it earlier in single-variable calculus. It is still important to have an idea of what the regions look like (here, you have a limacon and a "peanut").

Your radial portion is correct, but you really need to look at the angles where the curves intersect: you'll want to solve $\ 2 + \sin \theta \ = \ 2 + \cos (2 \theta)$ to get the range of angle integration. There are two zones to cover, but you can make use of symmetry here and just integrate over one of them. The red curve is the limacon $2 + \sin\theta$ , the blue curve, $2 + \cos(2 \theta)$ .

Incidentally, in the integral you wrote, it's not that there is no integrand, but rather that the integrand is "1". You are doing a surface integral where all infinitesimal patches are equally "weighted" at 1 , so the result is simply the area of the region. Surface integrals with some "weighting function" $f(r,\theta)$ or $g(x,y)$ turn up in various applications.

• Ooh, okay. So, I solved it and got $\theta=\frac{\pi}{6}, \frac{5\pi}{6}$. Can I set the bounds of the integral to this and multiply the entire thing by 2? – Kevin Evans May 3 '13 at 5:13
• Note that there is a third intersection point from solving $2 \sin^2\theta + \sin\theta - 1 \ = \ 0$ , as the graph also indicates. That point will be one endpoint of your angle integration. (And, yes, you will multiply the area of one region by 2.) – colormegone May 3 '13 at 5:20
• Oh, $\frac{3\pi}{2}$. So, $\int_{\frac{5\pi}6}^{\frac{3\pi}2}\int_{y_1}^{y_2} rdrd\theta$? – Kevin Evans May 3 '13 at 5:23
• In your angle integration for one of these "crescents", you may want to use a range of either $-\frac{\pi}{2}$ to $\frac{\pi}{6}$ or $\frac{3\pi}{2}$ to $\frac{13\pi}{6}$, since you are going to have to swing across the positive x-axis. (Of course, you could also use $\frac{5\pi}{6}$ to $\frac{3\pi}{2}$ , going across the negative x-axis to avoid that issue...) – colormegone May 3 '13 at 5:26
• Just saw your comment revision now: I believe that should be good. Happy integrating! – colormegone May 3 '13 at 5:28