Chapter 10 - Parametric Equations and Polar Coordinates - 10.4 Exercises - Page 693: 21

$\pi-\frac{3\sqrt 3}{2}$

Area enclosed by one loop is $A=\int_{7\pi/6}^{11\pi/6}\frac{1}{2}(1+2sin\theta)^{2}d \theta$ $=\int_{7\pi/6}^{11\pi/6}\frac{1}{2}(1+4sin^{2}\theta+4sin\theta)d \theta$ $=\pi-\frac{3\sqrt 3}{2}$