Answer
$$\frac{9\pi }{2}$$
Work Step by Step
The area is bounded by $r=2 -\cos \theta$ for $\theta=0$ to $\theta=2\pi$.
$$
\begin{aligned}
A & =\int_0^{2\pi} \frac{1}{2} r^2 d \theta\\
&=\frac{1}{2} \int_0^{2\pi}(2- \cos \theta)^2 d \theta\\
&= \frac{1}{2} \int_0^{2\pi}(4-4 \cos \theta+\cos^2\theta) d \theta\\
&= \frac{1}{2} \int_0^{2\pi}(\frac{9}{2}-4 \cos \theta+\cos2\theta) d \theta\\
&= \frac{1}{2}\left(\frac{9}{2}\theta-4 \sin \theta+\frac{1}{2}\sin2\theta\right)\bigg|_0^{2\pi}\\
&= \frac{9\pi }{2}
\end{aligned}
$$
