Area between inner and outer regions of $r(\theta) = 3(\theta)\sin(\theta)$

Question

Every example in existence on the internet or in my textbook has something like $r(\theta)=1+\sin(\theta)$ or some variation of that where their are two distinct points on the interval $0\leq\theta\leq2\pi$ where the answer equals $0$. However in this case there are 3 answers, $0$, $\pi$, $2\pi$. So how do I know where the inner circle begins so I can begin solving for the area? I need 2 points, not 3.

Answer 1

Calculate "area of outer region" - "area of inner region".

• Inner region: when $\theta=0$ $r=0$, the next time this happens is when $r(\theta)=3\theta\sin(\theta)=0$, which is when $\theta=\pi$.

• Outer region: we can not look at $\pi\leq \theta\leq2\pi$, because then $r(\theta)\leq 0$. We therefore have to look at the interval $3\pi\leq\theta\leq 4\pi$

Answer 2

Most calculus texts allow $r$ to be negative. This problem is lacking a domain for $\theta$. If you graph the equation over $-4\pi<\theta <4\pi$, then there are 4 circle-oids above and 4 below the $x$-axis. The graph looks like Ruben Clamzo's final nemesis on the attack.

If you graph from $0 \leq \theta \leq 2\pi$ there are only 2 circle-oids. The inner one is formed on $0\leq \theta \leq \pi$ and the outer one from $\pi \leq \theta \leq 2\pi.$

If your text or instructor doesn't allow negative $r$, then the outer circle changes to $2\pi \leq \theta \leq 3\pi$ and is much larger.