Find the area of the following region.
The region outside the circle $r=3$ and inside the circle $r=-6 \cos \theta$
The area of the region is square units. (Type an exact answer.)

Step 1 ：We are given two polar functions, \(r=3\) and \(r=-6\cos\theta\). The region we are interested in is outside the circle \(r=3\) and inside the circle \(r=-6\cos\theta\).

Step 2 ：To find the limits of the region, we need to find the values of \(\theta\) where the two functions intersect. This happens when \(3=-6\cos\theta\), or \(\cos\theta=-\frac{1}{2}\). This occurs at \(\theta=\frac{2\pi}{3}\) and \(\theta=\frac{4\pi}{3}\).

Step 3 ：The area of a polar region can be found using the formula \(\frac{1}{2}\int_{\alpha}^{\beta}(f(\theta))^2d\theta\) where \(\alpha\) and \(\beta\) are the limits of the region and \(f(\theta)\) is the polar function.

Step 4 ：The area of the region is then the difference between the area under the larger circle and the area under the smaller circle between these limits.

Step 5 ：Calculating the areas, we get \(area1 = 4.5\sqrt{3} + 6.0\pi\) and \(area2 = 3.0\pi\).

Step 6 ：Subtracting the smaller area from the larger one, we get \(area = 4.5\sqrt{3} + 3.0\pi\).

Step 7 ：Final Answer: The area of the region is \(\boxed{4.5\sqrt{3} + 3.0\pi}\) square units.