Answer
The area that encloses
$$ r=3+2 \cos \theta \quad \text {for } \theta =0 \quad \text {to} \quad \theta=2 \pi $$
is equal to
$$
A =\int_{0}^{2\pi} \frac{1}{2} r^{2} d \theta =11 \pi
$$
Work Step by Step
The area that encloses
$$ r=3+2 \cos \theta \quad \text {for } \theta =0 \quad \text {to} \quad \theta=2 \pi $$
is equal to
$$
\begin{aligned} A &=\int_{0}^{2\pi} \frac{1}{2} r^{2} d \theta \\
&=\int_{0}^{2 \pi} \frac{1}{2}(3+2 \cos \theta)^{2} d \theta\\
&=\frac{1}{2} \int_{0}^{2 \pi}\left(9+12 \cos \theta+4 \cos ^{2} \theta\right) d \theta \\
&=\frac{1}{2} \int_{0}^{2 \pi}\left[9+12 \cos \theta+4 \cdot \frac{1}{2}(1+\cos 2 \theta)\right] d \theta
\\
&=\frac{1}{2} \int_{0}^{2 \pi}(11+12 \cos \theta+2 \cos 2 \theta) d \theta
\\
&= \frac{1}{2}[11 \theta+12 \sin \theta+\sin 2 \theta]_{0}^{2 \pi} \\
&=\frac{1}{2}(22 \pi)\\
&=11 \pi
\end{aligned}
$$
