Chapter 10 - Parametric Equations and Polar Coordinates - 10.4 Exercises - Page 693: 31

$\frac{\pi}{2}-1$

$A=\frac{1}{2}\int_{0}^{\pi/8}(sin2\theta)^{2}d \theta+\frac{1}{2}\int_{\pi/8}^{\pi/4}(cos2\theta)^{2}d \theta$ $=2.\frac{1}{2}\int_{0}^{\pi/2}(sin2\theta)^{2}d \theta$ $=\frac{1}{2}[\theta-\frac{1}{4}sin4\theta]_{0}^{\pi/8}$ $=\frac{\pi}{16}-\frac{1}{8}$ Therefore the area of the region is, $A=8(\frac{\pi}{16}-\frac{1}{8})=\frac{\pi}{2}-1$