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?
Best Answer
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.