Find the area of the region inside the first curve and outside the 2nd curve:

1st: $\displaystyle r = 6cos(\theta)$

2nd:$\displaystyle r = 3$

Ok, so I have a circle w/ radius 3 with center (0,0) and a circle with radius 3 with center (3,0), and I found that they intersect at $\displaystyle \frac{\pi}{3}$ & $\displaystyle \frac{5\pi}{3}$ and the area I need to find is the right side of $\displaystyle r=6cos\theta$

*Hope I am okay up to here*

So, for my area:

$\displaystyle A = \frac{1}{2} \int^{\frac{\pi}{3}}_{\frac{5\pi}{3}} 36 cos^2\theta d\theta - \frac{1}{2} \int^{\frac{\pi}{3}}_{\frac{5\pi}{3}} 3^2 d\theta$

Did I set this up correctly?

Thanks!!