Chapter 11: Parametric Equations and Polar Coordinates - Section 11.5 - Areas and Lengths in Polar Coordinates - Exercises 11.5 - Page 670: 6

$ \dfrac{\pi}{12}$

Consider $A=\dfrac{1}{2}\int_m^n (r^2) d \theta$ $A=\int_{-\pi/6}^{\pi/6} (\dfrac{1}{2}) (\cos^2) (3 \theta) d \theta= (2) \dfrac{1}{2} \int_{0}^{\pi/6} \cos^2 (3 \theta) d \theta $ This implies that $A=(\dfrac{1}{2})[ \dfrac{\pi}{6}+\dfrac{ \sin (6 \theta)}{6}]_{0}^{(\pi/6)}= \dfrac{\pi}{12}$