Answer
$\dfrac{2\pi}{3}-\dfrac{\sqrt 3}{2}$
Work Step by Step
Consider $A_1=(\dfrac{1}{2}) \int_{(\pi/6)}^{(3\pi/6)} (1) d\theta =\dfrac{\pi}{3}$
and $A_2=(\dfrac{1}{2}) \int_{0}^{\pi/6} (2\sin \theta)^2 d \theta= \int_{0}^{\pi/6}(1- \cos 2\theta) =\dfrac{\pi}{6}-\dfrac{\sqrt 3}{4}$
Now, $A_3=(\dfrac{1}{2}) \int_{(3 \pi/6)}^{\pi} (2)^2 (\sin \theta)^2 d \theta=\dfrac{\pi}{6}-\dfrac{\sqrt 3}{4}$
Total area: $A=A_1+A_2+A_3=\dfrac{\pi}{3}+2-(\dfrac{\pi}{6}-\dfrac{\sqrt 3}{4})$
or, $A=\dfrac{2\pi}{3}-\dfrac{\sqrt 3}{2}$