I'm stuck with a problem that despite a good bit of searching and even toying around with wolfram|alpha, I can't find an answer to:
Find the area of the region inside r=5sinθ but outside r=4
I have seen how to find the area outside of one region but inside another, but I can't find a way to apply those to this problem. Any help would be appreciated!
The first step that you need to do is identify the measure of the angles that define the intersection point(s) of your curves. In this case, the intersections of $r=5\sin(\theta)$ and $r=4$. After equating equations and comparing the possible solutions that would make sense graphically, we see that $\theta=\arcsin(\frac{4}{5})$ and $\theta=\pi-\arcsin(\frac{4}{5})$ fit what is required.
Next, we see that the curve $r=5\sin(\theta)$ is above the curve $r=4$, so we will subtract $r=4$ from $r=5\sin(\theta)$ in our integral. This integral will be of the form $\frac{1}{2}\int_\alpha^\beta (5\sin(\theta))^2-4^2 d\theta$. By applying the angles identified above and evaluating, the integral is $$\frac{1}{2}\int_{\arcsin(\frac{4}{5})}^{\pi-\arcsin(\frac{4}{5})} 25\sin^2(\theta)-16\;d\theta=6-\frac{7\pi}{4}+\frac{7}{2}\arcsin\left(\frac{4}{5}\right)\approx3.7477$$