I have tried to solve this problem by subtracting the area of the whole by the smaller area. I got up to
$$\int_{1/2}^{11\pi/6} 12 (\cos (\theta)-6)^2 \mathrm{d}\theta- \frac12 \int_{-\pi/6}^{\pi/6} 12 (\cos (\theta)-6)^2 \mathrm{d}\theta $$
Are these the right integrals? Any help here would be very appreciated. Maybe I am doing it the wrong way? Thanks!
So I figured it out for anyone who wants to see what the answer is. I was using $\pi/6$ instead of $\pi/3$. So instead the integral should have been $$\int_\pi^{\frac{5\pi}{3}} 12(\cos\theta-6)^2 d\theta- \int_{-\frac{\pi}{3}}^{\frac\pi3} 12( \cos \theta - 6)^2 d\theta $$. The answer is $\sim 300.1584$.