# Find the area inside $r=3\cos\Theta$ and outside $r = 1+\cos\Theta$

In here, there are two polar curves. I first graphed both of them. I also shaded the area that I am going to find.

I assume that $$r = 3$$ is at $$\Theta = 0$$. I am unsure about this, but I proceeded anyway. I did not know the $$\Theta$$ at the intersection, which will be my upper bound, so I solved it by equating both $$r$$'s:

\begin{align*} 3 \cos\Theta &= 1 + \cos\Theta \\ 2\cos\Theta &= 1 \\ \cos\Theta &= \frac{1}{2} \\ \Theta &= \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3} . \end{align*}

After that, I inputted the following in my calculator to find the area. I multiplied it to two because I recognized that I am just getting the area of one half:

$$A = 2\frac{1}{2}\int_{0}^{\frac{\pi }{3}}\left[(3\cos\Theta )^{2}-(1+\cos\Theta )^{2}\right]\,\mathrm{d}\Theta = \pi \,\text{units}^{2}$$

In my equation above, I also assumed that $$r_{2}$$ is $$3\cos\Theta$$ because it has the farthest circumference from the origin.

Is my solution correct?