Question

Find the area of the region inside the circle r=2sinθ and outside the circle r=1.

Answer 1

Since the two curves will intersect at

$2\sin \theta =1 \ \ \ \to \theta = \pi /6,\ \ \theta =5\pi/6$

Since

$1\geq 2\sin \theta ,\ \ \ \ \ \text{for } \ \ \ \pi/6\leq x\leq 5\pi/6$

Then area of the region inside the circle $r=2\sinθ$ and outside the circle r=1, given by

$\begin{aligned} A&= \frac{1}{2} \int_{\alpha}^{\beta}\left(f_{2}(\theta)^{2}-f_{1}(\theta)^{2}\right) d \theta\\ &= \frac{1}{2} \int_{\pi/6}^{5\pi/6}\left(4\sin^2\theta-1\right) d \theta\\ &= \frac{1}{2} \int_{\pi/6}^{5\pi/6}\left(2-2\cos2\theta-1\right) d \theta\\ &= \frac{1}{2} \int_{\pi/6}^{5\pi/6}\left(1-2\cos2\theta\right) d \theta\\ &= \frac{1}{2} \left(\theta-\sin2\theta\right)\bigg|_{\pi/6}^{5\pi/6}\\ &= \frac{1}{2} \left(\frac{2\pi}{3}+\sqrt{3} \right)\\ &=\frac{3\sqrt{3}+2\pi }{6} \end{aligned}$