Chapter 10 - Section 10.5 - Areas and Lengths in Polar Coordinates - Exercises - Page 585: 6

$\dfrac{\pi}{12}$

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, $A=\dfrac{\pi}{12}$