Answer
The area that the curve encloses
$$ r=3-2 \cos 4 \theta \quad \text {for } \theta =0 \quad \text {to} \quad \theta=2 \pi $$
is equal to
$$
A =\int_{0}^{2\pi} \frac{1}{2} r^{2} d \theta =11 \pi
$$
Work Step by Step
The area that the curve encloses
$$ r=3-2 \cos 4 \theta \quad \text {for } \theta =0 \quad \text {to} \quad \theta=2 \pi $$
is equal to
$$
\begin{aligned} A &=\int_{0}^{2 \pi} \frac{1}{2} r^{2} d \theta \\
&=\int_{0}^{2 \pi} \frac{1}{2}(3-2 \cos 4 \theta)^{2} d \theta \\
&=\frac{1}{2} \int_{0}^{2 \pi}\left(9-12 \cos 4 \theta+4 \cos ^{2} 4 \theta\right) d \theta \\
&=\frac{1}{2} \int_{0}^{2 \pi}\left[9-12 \cos 4 \theta+4 \cdot \frac{1}{2}(1+\cos 8 \theta)\right] d \theta \\
&=\frac{1}{2} \int_{0}^{2 \pi}(11-12 \cos 4 \theta+2 \cos 8 \theta) d \theta \\
& =\frac{1}{2}\left[11 \theta-3 \sin 4 \theta+\frac{1}{4} \sin 8 \theta\right]_{0}^{2 \pi} \\
&=\frac{1}{2}(22 \pi) \\
&=11 \pi \end{aligned}
$$
