Answer
$$11\pi $$
Work Step by Step
The area is bounded by $r=3- 2\cos4 \theta$ for $\theta=0$ to $\theta=2\pi$.
\begin{aligned}
A & =\int_0^{2 \pi} \frac{1}{2} r^2 d \theta\\
&=\int_0^{2 \pi} \frac{1}{2}(3- 2\cos4 \theta)^2 d \theta\\
&=\frac{1}{2} \int_0^{2 \pi}\left(9- 12\cos4 \theta+ 4\cos^24\theta\right) d \theta \\
& =\frac{1}{2} \int_0^{2 \pi}\left[9- 12\cos4 \theta+ 2(1+\cos 8 \theta)\right] d \theta \\
& =\frac{1}{2} \int_0^{2 \pi}\left(11 - 12\cos4 \theta+2\cos 8 \theta\right) d \theta\\
&=\frac{1}{2}\left[11\theta - 3\sin4 \theta+\frac{1}{4}\sin 8 \theta\right]_0^{2 \pi} \\
& = 11\pi
\end{aligned}
