Chapter 10 - Section 10.4 - Areas and Lengths in Polar Coordinates - 10.4 Exercises - Page 699: 1

$\frac{\pi^2}{8}$

The area of the region is determined by the integral $A=\int_0^{\pi/2}\frac{1}{2}r^2d\theta$. Evaluate $A$: $A=\int_0^{\pi/2}\frac{1}{2}\cdot \sqrt{2\theta}^2d\theta$ $A=\int_0^{\pi/2}\frac{1}{2}\cdot 2\theta d\theta$ $A=\int_0^{\pi/2}\theta d\theta$ $A=[\theta^2/2]_0^{\pi/2}$ $A=(\pi/2)^2/2-0^2/2$ $A=\pi^2/8$ Thus, the area is $\frac{\pi^2}{8}$.