How to find limits of integration of polar curves? Right now I am working on a problem that involves finding the area enclosed by a single loop given the equation $r=4\cos(3\theta)$. I know that the cosine is bounded from zero to $\pi$, but when using a lower limit of $0$, and a upper limit of $\pi/3$, I get the wrong answer (the answer is $4\pi/3$).  
Should I instead use zero to $2\pi/3$, since that would be a full period? I tried drawing the graph as well, but had no luck with that. What is the best way of finding the upper and lowers limits for these kind of problems?
 A: It helps if you have an idea of the graph, but even if you don't: it should be clear that at $\theta = 0$, you have $r(0) = 4$ so you're not at the beginning of a loop. A loop starts when $r=0$ and a single loop closes at the next root of $r(\theta)$.
The function $\cos x$ has a root at $-\tfrac{\pi}{2}$ and the next one is at $\tfrac{\pi}{2}$. For this polar curve $r = 4 \cos(3\theta)$, you get (with $x=3\theta$):
$$3\theta = \pm \frac{\pi}{2} \Rightarrow \theta = \pm \frac{\pi}{6}$$
so you go through exactly one loop if you let $\theta$ run from $-\tfrac{\pi}{6}$ to $\tfrac{\pi}{6}$. Using the formula for area:
$$\int_{\theta_1}^{\theta_2} \tfrac{1}{2}r^2 \,\mbox{d}\theta \to
\int_{-\tfrac{\pi}{6}}^{\tfrac{\pi}{6}} \tfrac{1}{2}\left( 4 \cos(3\theta) \right)^2 \,\mbox{d}\theta = \cdots = \frac{4\pi}{3}$$
Note that because the function is even, it is slightly easier to (manually) calculate :
$$\color{blue}{2}\int_{\color{red}{0}}^{\tfrac{\pi}{6}} \tfrac{1}{2}\left( 4 \cos(3\theta) \right)^2 \,\mbox{d}\theta = \int_{0}^{\tfrac{\pi}{6}} \left( 4 \cos(3\theta) \right)^2 \,\mbox{d}\theta = \cdots = \frac{4\pi}{3}$$
