Chapter 10 - Parametric Equations and Polar Coordinates - 10.4 Areas and Lengths in Polar Coordinates - 10.4 Exercises - Page 712: 4

$$\frac{3}{4\pi }$$

Given $$r= \frac{1}{\theta},\ \ \ \ \ \pi/2\leq \theta \leq 2\pi $$ Then area given by \begin{aligned} A&= \frac{1}{2}\int_{a}^{b} r^2d\theta\\ &= \frac{1}{2}\int_{\pi/2}^{2\pi} \frac{1}{\theta^2}d\theta \\ &= \frac{1}{2}\int_{\pi/2}^{2\pi} \theta^{-2}d\theta\\ &= \frac{-1}{2\theta }\bigg| _{\pi/2}^{2\pi} \\ &= \frac{-1}{4\pi}+\frac{1}{\pi}\\ &=\frac{3}{4\pi } \end{aligned}