Chapter 10 - Section 10.5 - Areas and Lengths in Polar Coordinates - Exercises - Page 585: 7

$2$

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