Answer
$18 \pi$
Work Step by Step
The area of a shaded region can be found as: $A=\dfrac{1}{2}\int_p^q r^2 d \theta$
Here, we have $A=\dfrac{1}{2}(2)\int_{0}^{\pi} (4+2 \cos \theta)^2 d \theta$
$\int_{0}^{\pi} 4\cos^2 \theta d \theta+\int_{0}^{\pi} 16 \cos \theta d \theta+\int_{0}^{\pi} 16 \cos \theta d \theta=\int_{0}^{\pi} (1/2) 4(1+\cos2 \theta) d \theta+\int_{0}^{\pi} 16 \cos \theta d \theta+\int_{0}^{\pi} 16 \cos \theta d \theta$
Thus, $A=18 \pi$
