Answer
$\dfrac{\pi}{12}$
Work Step by Step
The area is given as follows:
$A=(1/2) \int_{0}^{\pi/3} \sin^2 3 \theta d\theta$
Then, $(1/2) \int_{0}^{\pi/3} (1-\dfrac{\cos 6 \theta}{2}) d\theta=(\dfrac{1}{4}) [\theta -(\dfrac{1}{6}) \sin 6 \theta]_{0}^{\pi/3} $
After solving, we get $A=\dfrac{\pi}{12}$
