Chapter 10 - Section 10.2 - Areas and Lengths in Polar Coordinates - 10.4 Exercises - Page 673: 21

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

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