Answer
$12\pi $
Work Step by Step
We need to find the area.
Our aim is to integrate the integral as follows:
$ Area=2\int^{\pi/6}_0 \int^{12 \cos (3\theta) }_0 r \space dr \space d\theta $
or, $=144 \times \int^{\pi/6}_0 cos^2 (3\theta ) \space d\theta $
or, $=144 \times \int^{\pi/6}_0 \dfrac{ \cos 6 \theta +1}{2} \space d\theta $
or, $=144 \times [\sin 6 \theta +6 \theta ]^{\pi/6}_0 $
or, $ Area=12\pi $
