Chapter 10 - Parametric Equations and Polar Coordinates - 10.4 Areas and Lengths in Polar Coordinates - 10.4 Exercises - Page 713: 28

$$A=3\sqrt 3$$

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