Chapter 11: Parametric Equations and Polar Coordinates - Practice Exercises - Page 688: 48

$\dfrac{\pi}{12}$

Here, $A=\dfrac{1}{2} \int_{0}^{\pi/3} \sin^2 (3 \theta) d\theta$ Then, $A=\dfrac{1}{2} \int_{0}^{(\pi/3)} (1-\dfrac{1}{2}\cos (6 \theta)) d\theta$ This implies that $A=(1/4) [\theta -(\dfrac{\sin (6 \theta)}{6}) ]_{0}^{(\pi/3)} =\dfrac{\pi}{12}$