## University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{\pi}{2}-1$
The area of a shaded region can be found as: $A=\dfrac{1}{2}\int_p^q r^2 d \theta$ Here, we have $A_1=2 \int_{0}^{\pi/4} \sin^2 \theta d \theta=2 \int_{0}^{\pi/4}[\dfrac{1- \cos 2\theta}{2}] =\dfrac{\pi}{4}-\dfrac{1}{2}$ 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, $A=\dfrac{\pi}{4}-\dfrac{1}{2}+\dfrac{\pi}{4}-\dfrac{1}{2}=\dfrac{\pi}{2}-1$