Answer
The area that the curve encloses:
$$ r=2- \cos \theta \quad \text {for } \theta =0 \quad \text {to} \quad \theta=2 \pi $$
is equal to
$$
A =\int_{0}^{2\pi} \frac{1}{2} r^{2} d \theta =\frac{9}{2} \pi
$$
Work Step by Step
The area that the curve encloses:
$$ r=2- \cos \theta \quad \text {for } \theta =0 \quad \text {to} \quad \theta=2 \pi $$
is equal to
$$
\begin{aligned} A &=\int_{0}^{2\pi} \frac{1}{2} r^{2} d \theta \\
&=\int_{0}^{2 \pi} \frac{1}{2}(2- \cos \theta)^{2} d \theta\\
&=\frac{1}{2} \int_{0}^{2 \pi}\left(4-4\cos \theta+ \cos ^{2} \theta\right) d \theta \\
&=\frac{1}{2} \int_{0}^{2 \pi}\left[4-4 \cos \theta+ \frac{1}{2}(1+\cos 2 \theta)\right] d \theta
\\
&=\frac{1}{2} \int_{0}^{2 \pi}(\frac{9}{2}-4 \cos \theta+\frac{1}{2} \cos 2 \theta) d \theta
\\
&= \frac{1}{2}[\frac{9}{2} \theta-4 \sin \theta+\frac{1}{4}\sin 2 \theta]_{0}^{2 \pi} \\
&=\frac{1}{2}(9\pi)\\
&=\frac{9}{2} \pi
\end{aligned}
$$
