Chapter 10 - Parametric Equations and Polar Coordinates - 10.4 Areas and Lengths in Polar Coordinates - 10.4 Exercises - Page 713: 17

$$\frac{4 \pi}{3}$$

$$A =\int_{-\pi/6}^{\pi/6}\frac{r^{2}}{2}d \theta=\int_{-\pi/6}^{-\pi/6}\frac{(4cos(3\theta))^{2}}{2}d \theta=\int_{-\pi/6}^{\pi/6}8cos^{2}(3\theta)d \theta=8\int_{-\pi/6}^{\pi/6}(\frac{(1+cos(6\theta))}{2})d \theta=\frac{4 \pi}{3}$$