## University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{9}{2} \pi$
The area is given as follows: $A=(2) \int_{0}^{\pi} (\dfrac{1}{2}) r^2 d\theta$ Then, $(2) \int_{0}^{\pi} (2-\cos \theta)^2= \int_{0}^{\pi} [4-4 \cos \theta +\cos^2 \theta ]d\theta$ After solving, we get $A=\dfrac{9}{2} \pi$