Answer
$\dfrac{2\pi}{3}-\dfrac{\sqrt 3}{2}$
Work Step by Step
The area of a shaded region can be found as: $A=\dfrac{1}{2}\int_p^q r^2 d \theta$
Here, we have $A_1=\int_{\pi/6}^{3\pi/6} \dfrac{1}{2}1 d\theta =\dfrac{\pi}{3}$
$A_2=(1/2) \int_{0}^{\pi/6} (2\sin \theta)^2 d \theta= \int_{0}^{\pi/6}(1- \cos 2\theta) =\dfrac{\pi}{6}-\dfrac{\sqrt 3}{4}$
Now, $A_3=(1/2) \int_{3 \pi/6}^{\pi} (2\sin \theta)^2 d \theta=\dfrac{\pi}{6}-\dfrac{\sqrt 3}{4}$
$A=\dfrac{\pi}{3}+2-(\dfrac{\pi}{6}-\dfrac{\sqrt 3}{4})=\dfrac{2\pi}{3}-\dfrac{\sqrt 3}{2}$
