Chapter 10 - Parametric Equations and Polar Coordinates - 10.4 Areas and Lengths in Polar Coordinates - 10.4 Exercises - Page 712: 5

$$ \frac{1}{2}$$

Given $$r^2= \sin{2\theta},\ \ \ \ \ $$ From the given graph, the region bounded by $ 0\leq \theta \leq \pi/2 $. Then area given by \begin{aligned} A&= \frac{1}{2}\int_{a}^{b} r^2d\theta\\ &= \frac{1}{2}\int_{0}^{\pi/2} \sin{2\theta}d\theta \\ &= \frac{-1}{4}\cos 2\theta\bigg|_{0}^{\pi/2} \\ &=\frac{1}{4}+ \frac{1}{4}\\ &= \frac{1}{2} \end{aligned}