Answer
$$\frac{27\pi }{2}$$
Work Step by Step
The area is bounded by $r=1+5\sin 6\theta$ 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}\left(1+5\sin 6\theta\right)^2 d \theta \\
& =\frac{1}{2} \int_0^{2 \pi}\left( 1+10\sin 6\theta +25\sin^2 \theta\right) d \theta\\
&=\frac{1}{2} \int_0^{2 \pi}\left[1+10\sin 6\theta+\frac{25}{2}(1-\cos 12 \theta)\right] d \theta \\
& =\frac{1}{2}\left[\frac{27}{2} \theta-\frac{10}{6} \cos 6 \theta+\frac{25}{24}\sin 12\theta\right]_0^{2 \pi}\\
&= \frac{27\pi }{2}
\end{aligned}
