Chapter 11: Parametric Equations and Polar Coordinates - Practice Exercises - Page 688: 50

$5\pi$

Here, $A=2 \int_{-(\pi/2)}^{(\pi/2)} (\dfrac{1}{2}) [2(1+\sin \theta)]^2 d\theta-\pi$ This gives: $A=[6 (\theta)-8 \cos (\theta)-\sin (2\theta)]_{-(\pi/2)}^{(\pi/2)}-\pi$ Hence, $A=3\pi -(-3 \pi)-\pi=6 \pi-\pi=5\pi$