Finding area enclosed by two curves (circles) 
Given information : F : r=18cosθ, rg=6,G : r=6,rf=9
The question is: Find the area of region A in Figure 1. I know that I need to subtract the smaller area enclosed by the two circles. I tried finding the intersection points first to do this, but the angle was complicated, so I am not sure if I need to use the angle to solve this problem...
18cosθ = 6  θ=cos-1(1 ⁄ 3)  Do I use this as the angle? Also, the picture is small and I can't see how the intersection lines would make the smaller area into two areas (dividing the top part from the bottom) or not. I am really confused with how to interpret this problem. 
 A: Draw a common chord of circles $G$ and $F$. The plan  is to calculate the areas of segments $CEDB$ and $CEDO$ separately.

Let the distance of this chord from centre of $G$ be $x$. Upon using pythagorean theorem on $\triangle CEO$ and $\triangle CEA$, we will get $x = 2$.
To calculate the area of segment $CEDB$, we subtract the area of triangle $COD$ from sector $CODB$. Note that $CE = \sqrt{36-x^2} = 4\sqrt{2}$. So angle $COD$ can be written as $2 \arctan (\tfrac{CE}{OE}) = 2 \arctan{2\sqrt2}$.
Area of sector $CODB$ is simply $\frac{2\arctan(2\sqrt 2)}{2\pi} \pi \cdot 6^2 = 36 \arctan(2\sqrt 2)$. The area of triangle $COD$ is $8\sqrt 2$. 
Therefore area of segment $CEDB$ is $\color{red}{36 \arctan(2\sqrt 2)-8\sqrt 2}$
Repeat the calculations to get area of segment $CEDO$. That is subtract area of triangle $CDA$ from area of sector $CODA$. I get upon calculation, area of segment $CEDO $ as $ \color{red}{81\arctan(\tfrac{4\sqrt 2}{ 7})-28\sqrt 2}$.
Now area of the shaded region in question is simply the area of circle $F$ minus the area of segments $CEDB$ and $CEDO$. So the required answer is:
$$81 \pi-36\arctan(2\sqrt2) - 81 \arctan(\tfrac{4\sqrt2}{7}) + 36\sqrt 2$$
