Chapter 10 - Section 10.2 - Areas and Lengths in Polar Coordinates - 10.4 Exercises - Page 673: 30

$=\frac{3\pi}{2}-4$

$A=\int_{0}^{\pi/2}\frac{(1-cos\theta)^{2}}{2}d \theta$ $=\int_{0}^{\pi/2}\frac{(1-2cos\theta+cos^{2}\theta)}{2}d \theta$ $=\frac{1}{4}[3\theta-4sin\theta+\frac{1}{2}sin2\theta]_{0}^{\pi/2}$ $=\frac{3\pi}{2}-4$