Answer
$5\pi$
Work Step by Step
The area is given as follows:
$A=(2) \int_{-\pi/2}^{\pi/2} (\dfrac{1}{2}) [2(1+\sin \theta)]^2 d\theta-\pi$
Then, $[6 \theta-8 \cos-\sin 2\theta]_{-\pi/2}^{\pi/2}-\pi=3\pi -(-3 \pi)-\pi$
After solving, we get $A=5\pi$
