I'm having some trouble with this problem:
Find the length of the entire perimeter of the region inside $r = 11 \sin(\theta)$ but outside $r = 3$.
I am using the formula
$$\int_a^b\! \sqrt{r^2 + (\frac{dr}{d\theta})^2} \mathrm{d}\theta$$.
How do I find the perimeter outside of $r=3$?
Thank you!
Find the intervals where the radius is greater than or equal to $3$, then find the arclength over these intervals.
This amounts to solving $11\sin\theta\geq 3$ for $\theta$. There are two points on the interval $[0,2\pi]$ where there is an equality, namely $\sin^{-1}(3/11)$ and $\pi-\sin^{-1}(3/11)$ (where this is the principal inverse sine, with values in the range $[-\pi/2,\pi/2]$). We can see that $[\sin^{-1}(3/11),\pi-\sin^{-1}(3/11)]$ is the interval along which $r=11\sin(\theta)$ is at least $3$ (at least when $\theta$ is restricted to $[0,2\pi]$).
For your arclength integral, $\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}=\sqrt{11^2\sin^2\theta+11^2\cos^2\theta}=11$, so the definite integral evaluates to $11(\pi-\sin^{-1}(3/11))-11\sin^{-1}(3/11)=11(\pi-2\sin^{-1}(3/11))$.