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?

Geometric solution:

Rewrite $x^2+y^2=3x$ as $x^2-3x+(\frac{3}{2})^2 + y^2 = (\frac{3}{2})^2 \Rightarrow (x-\frac{3}{2})^2 + y^2 = \frac{9}{4}$. Then solving for the intersections:
$$(x-\frac{3}{2})^2 + y^2 = \frac{9}{4} \tag{1}$$
$$x^2+y^2 = 1 \tag{2}$$
$(1) - (2)$ gives $-3x + \frac{9}{4} = \frac{5}{4} \Rightarrow x = \frac{1}{3}$. Substituting back into equation $(2)$, we get that $y = ±\frac{2 \sqrt2}{3}$.
The area of $\Delta AFC$ can be found in multiple ways, including the shoelace formula. Using this gives the area as $\frac{1}{\sqrt2}$.
Now the central angle of sector $ACF$ is just $\tan^{-1} \frac{2 \sqrt{2}/3}{3/2 - 1/3} = \tan^{-1} \frac{4 \sqrt2}{7}$ radians. Therefore its area is $\frac{1}{2} \cdot (\frac{3}{2})^2 \tan^{-1} \frac{4 \sqrt2}{7} = \frac{9}{8} \tan^{-1} \frac{4 \sqrt2}{7}$, so area $a$ is just $\frac{9}{8} \tan^{-1} \frac{4 \sqrt2}{7} - \frac{1}{\sqrt2}$.
Finally, the central angle of sector $AFB$ is $\tan^{-1} \frac{2 \sqrt{2}/3}{1/3} 
= \tan^{-1} 2 \sqrt{2}$, so the area of that is $\frac{1}{2} \tan^{-1} 2 \sqrt{2}$.
Therefore circular area $AFB$ is $\frac{9}{8} \tan^{-1} \frac{4 \sqrt2}{7}- \frac{1}{\sqrt2} + \frac{1}{2} \tan^{-1} 2 \sqrt{2}$, and by symmetry, the total area is just twice that, or:
$$\frac{9}{4} \tan^{-1} \frac{4 \sqrt2}{7} -\sqrt{2} + \tan^{-1} 2 \sqrt{2} \approx 1.346.$$
