#### Answer

$\dfrac{\pi}{12}$

#### 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 $A=\int_{-\pi/6}^{\pi/6} (\dfrac{1}{2})\cos^2 3 \theta d \theta$
$ \int_{0}^{\pi/6} (2) \dfrac{1}{2}\cos^2 3 \theta d \theta=(\dfrac{1}{2})[ \dfrac{\pi}{6}+\dfrac{ \sin 6 \theta}{6}]_{0}^{\pi/6} $
Thus, $A=\dfrac{\pi}{12}$