Question

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

Answer 1

The intersection pints of both the curves are given by :

$2 sin\theta = 1 \\ \theta = π/6 , 5π/6$

Let $r_1=1 , r_2=2sin\theta$

Area = $\frac{1}{2}\int_{π/6}^{5π/6} (r_2^2-r_1^2)d\theta\\$

$=\frac{1}{2}\int_{π/6}^{5π/6} (4sin\theta^2-1)d\theta\\ = =\frac{1}{2}\int_{π/6}^{5π/6} (2cos2\theta+1)d\theta\\ =\frac{1}{2} (sin2\theta+\theta)|_{π/6}^{5π/6}\\ = -\frac{-\sqrt3}{2}+\frac{π}{3}$