Area of the intersection of the region outside of a unit circle centered at origin and the region inside the unit circle centered at (1,0)? This problem was on my calculus final, and I had no idea how to set up the double integral to solve it. 
 A: $r=2\cos \theta$ is the circle centered at $(1,0)$ with radius $1$. (Why?)
$r=1$ is the unit circle.
Let's find the angles of intersections:
$\displaystyle 2 \cos \theta = 1 \implies \cos \theta = \frac 12 \implies \theta = \pm \frac \pi 3$
We want the area outside the unit circle $\implies r >1$
And the area inside the other circle $\implies r<2 \cos \theta$
$$A= \int_{- \frac \pi 3}^{\frac \pi 3} \int_1^{2\cos \theta}\underbrace{r}_{\text{Jacobian}} \ dr \ d \theta$$
A: Solution without integral: the intersection area of the two circles is formed by two symmetric equal parts, each corresponding to the area of a circular segment whose angle is $\frac {2 \pi}{3}$. The area of each of these two circular segments is $\frac {\pi}{3}-\frac {\sqrt {3}}{4} $. So the area of the intersection of the region outside the first circle and the region  inside the second circle is
$$\pi- \left(\frac {2\pi}{3}-\frac {\sqrt {3}}{2} \right)= \frac {\pi}{3}+\frac {\sqrt {3}}{2}$$
A: Using good old cut-and-paste geometry:

The straight vertical sides are $2\frac{\sqrt3}2$ tall and have a mutual distance of $1$.
Total area $\displaystyle \color{#380}{\frac{\pi}3} + \color{#003ce4}{\frac{\sqrt3}{2}}$.
