Chapter 10 - Parametric Equations and Polar Coordinates - 10.4 Exercises - Page 693: 17

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

Area enclosed by one loop is $\int_{-\pi/6}^{-\pi/6}\frac{r^{2}}{2}d \theta=\int_{-\pi/6}^{-\pi/6}\frac{(4cos(3\theta))^{2}}{2}d \theta$ $=\int_{-\pi/6}^{-\pi/6}8cos^{2}(3\theta)d \theta$ $=8\int_{-\pi/6}^{-\pi/6}(\frac{(1+cos(6\theta))}{2})d \theta$ $=\frac{4 \pi}{3}$