Chapter 11: Parametric Equations and Polar Coordinates - Section 11.5 - Areas and Lengths in Polar Coordinates - Exercises 11.5 - Page 670: 9

$(\dfrac{\pi}{2})-1$

Consider $A_1=2 \int_{0}^{(\pi/4)} (\sin^2 \theta) d \theta=2 \int_{0}^{(\pi/4)}[\dfrac{1- \cos 2\theta}{2}] =(\dfrac{\pi}{4})-\dfrac{1}{2}$ and $A_2=2 \int_{\pi/4}^{(\pi/2)} (\cos^2 \theta) d \theta= \int_{(\pi/4)}^{(\pi/2)}1+ \cos 2\theta =(\dfrac{\pi}{4})-\dfrac{1}{2}$ Now, $A=A_1+A_2=(\dfrac{\pi}{4})-\dfrac{1}{2}+(\dfrac{\pi}{4})-\dfrac{1}{2}$ or, $A=(\dfrac{\pi}{2})-1$