Find the area of the region outside $r=4−3\sin \theta$ but inside $r=5\sin \theta$

Question

How will I do this? Will I get the points of intersection and integrate one by one?

Find the area of the region outside $r=4−3\sin \theta$ but inside $r=5\sin \theta$.

Answer 1

It can be verified that in $[0,2\pi]$ the inequality $$\rho_1(\theta):=4-3\sin(\theta)\leq \rho_2(\theta):=5\sin(\theta)$$ is verified in $[\theta_1,\theta_2]=[\pi/6,5\pi/6]$.

Thus by using the given here, $$\mbox{Area}=\frac{1}{2}\int_{\theta_1}^{\theta_2}\rho_2^2(\theta) d \theta-\frac{1}{2}\int_{\theta_1}^{\theta_2}\rho_1^2(\theta) d \theta\\ =\frac{1}{2}\int_{\pi/6}^{5\pi/6}\left((25-9)\sin^2(\theta)-16+24\sin(\theta)\right)d \theta. $$ The final result is $14\sqrt{3}-\frac{8\pi}{3}$.