Chapter 10 - Section 10.2 - Areas and Lengths in Polar Coordinates - 10.4 Exercises - Page 673: 22

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

$r=2cos\theta-sec\theta$ $r^{2}=2cos(2\theta)-2+sec^{2}\theta$ $A=2.\frac{1}{2}\int_{0}^{\pi/4}[2cos(2\theta)-2+sec^{2}\theta]d \theta$ $=[sin(2\theta)-2\theta+tan\theta]_{0}^{\pi/4}$ $=2-\frac{\pi}{2}$