Find the area between $r=1$ and $r=3\cos\theta$ 
Find the area between $r=1$ and $r=3\cos\theta$.


I squared both sides to get $r^2 = 1$, then did $r^2(\cos^2 \theta + \sin^2 \theta) = (r \cos \theta)^2 + (r \sin \theta)^2$$ = x^2+y^2 = 1$ to get $x^2+y^2=1$.
For $r = 3 \cos \theta$, I multiplied by $r$ on both sides to get $r^2 = 3r \cos \theta$, then substituted $x = r \cos \theta$ to get $x^2+y^2 =3x$. However, I don't know if it is easier to do it this way.
If not, how can I find this area?

The area under a polar function $r(\theta)$ is:
$$\frac{1}{2} \int_a^b \big(r(\theta) \big)^2  \ d \theta$$
Here $r(\theta) = 3 \cos \theta - 1$. Now find the limits of integration, and use the identity $\cos^2 \theta = \frac{1}{2} (\cos 2 \theta+1)$.
