#### Answer

$5\pi-8 $

#### Work Step by Step

The area of between the polar curve can be found as: $A=\dfrac{1}{2}\int_p^q [(r_2(\theta))^2-(r_1(\theta))^2] d \theta$
Here, we have $A=(1/2)\int_{-\pi/2}^{\pi/2} [2^2-(2-2\cos \theta)^2] d\theta =(1/2)\int_{-\pi/2}^{\pi/2} [8 \cos \theta-4 \cos^2 \theta] d\theta$
Thus, $A=\dfrac{1}{2}[8 \sin \theta-4 (1/2)\sin 2 \theta/2+\theta]_{-\pi/2}^{\pi/2}=\pi(2)^2-(8-\pi)=5\pi-8 $