Chapter 11: Parametric Equations and Polar Coordinates - Section 11.5 - Areas and Lengths in Polar Coordinates - Exercises 11.5 - Page 670: 8

$\dfrac{2}{3}$

Consider $A=\int_{0}^{\pi/3} (\dfrac{1}{2}) (2 \sin 3 \theta)^2 d \theta$ $\implies A=\int_{0}^{\pi/3} (\dfrac{1}{2}) (2 \sin (3 \theta))^2 d \theta=(\dfrac{1}{2})\int_{0}^{\pi/3} 4 \sin (3 \theta) d \theta=$ Thus, $A=[- \dfrac{ \cos (3\theta)}{3})]_{0}^{(\pi/3)} =\dfrac{2}{3}$