Chapter 10 - Parametric Equations and Polar Coordinates - 10.4 Areas and Lengths in Polar Coordinates - 10.4 Exercises - Page 713: 24

$$A=(2+\frac{\pi}{4})$$

Since, $r_{1}=1-sin\theta$ and $r_{2}=1$ Then $r_{1}^{2}-r_{2}^{2}=(1-sin\theta)^{2}-1=\frac{1}{2}-\frac{1}{2}cos(2\theta)-2sin\theta$ So, $$A=2.\frac{1}{2}\int_{\pi}^{2\pi}(\frac{1}{2}-\frac{1}{2}cos(2\theta)-2sin\theta)d \theta=\frac{1}{2}[\frac{1}{2}\theta-\frac{1}{2}.\frac{1}{2}sin(2\theta)+2cos\theta]_{\pi}^{2\pi}$$ Hence, $$A=(2+\frac{\pi}{4})$$