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?
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 : $$\color{blue}{2}\int_{\color{red}{0}}^{\tfrac{\pi}{6}} \tfrac{1}{2}\left( 4 \cos(3\theta) \right)^2 \,\mbox{d}\theta = \int_{0}^{\tfrac{\pi}{6}} \left( 4 \cos(3\theta) \right)^2 \,\mbox{d}\theta = \cdots = \frac{4\pi}{3}$$