Answer
$A=11\pi$
Work Step by Step
Given:
$r=3+2cos\theta$
First, sketch a graph by plugging in values for $\theta$ (see below), then plug into the polar area formula:
$A=0.5\int_a^br^2d\theta=\int_o^\pi(3+2cos\theta)^2d\theta=\int_o^\pi(9+12cos\theta+4cos^2\theta)d\theta$
Since the graph is symmetric, you can just integrate from $0$ to $\pi$ and double the value.
Using half angle identities:
$A=\int_o^\pi(9+12cos\theta+2+2cos2\theta)\theta)d\theta=(11\theta+12sin\theta+sin2\theta{\vert}_o^\pi)=11\pi$
