Chapter 10 - Section 10.2 - Areas and Lengths in Polar Coordinates - 10.4 Exercises - Page 673: 28

$=3\sqrt 3$

$r_{1}=3sin\theta$ and $r_{2}=2-sin\theta$ $r_{1}^{2}-r_{2}^{2}=(3sin\theta)^{2}-(2-sin\theta)^{2}=4(sin\theta-cos(2\theta))$ $A=\frac{1}{2}\int_{\pi/6}^{5\pi/6}(4(sin\theta-cos(2\theta)))d \theta$ $=2[-cos\theta-\frac{1}{2}sin(2\theta)]_{\pi/6}^{5\pi/6}$ $=3\sqrt 3$