Double Integral Between Two Circles What is the area confined by these two circles?
$y^2=1-(x-1)^2$
$y^2=1-x^2$
I've set up the integral:
$\int_{\pi/3}^{2\pi/3}\int_{2\cos\theta}^1r\,\mathrm{d}r\mathrm{d}\theta$
Unfortunately, my answer is off so I believe that this is incorrect, I got $-\pi/6+\sqrt{3}/2$ when the correct answer is $\pi/3+\sqrt3/2$. I've checked my work so I don't believe the algebra is off. My logic behind the $r$ integral is that after the equations are converted to polar coordinates the $r$ value goes from $2\cos\theta$ to $1$. I believe that $\theta$ goes from $\pi/3$ to $2\pi/3$ also.
 A: The problem here is that $r = 2\cos\theta$ isn't really the lower bound. Instead, it is the upper bound if $\theta \in [\pi/3, 2\pi/3]$. So we should instead split up into two cases for when the upper bound changes:
$$
\int_{-\pi/3}^{\pi/3}\int_0^1 r \, \mathrm{d}r \, \mathrm{d}\theta
+ \int_{\pi/3}^{2\pi/3}\int_0^{2\cos\theta} r \, \mathrm{d}r \, \mathrm{d}\theta
$$
Note that due to the symmetry with respect to the $x$-axis, we can alternatively compute:
$$
2\left[\int_{0}^{\pi/3}\int_0^1 r \, \mathrm{d}r \, \mathrm{d}\theta
+ \int_{\pi/3}^{\pi/2}\int_0^{2\cos\theta} r \, \mathrm{d}r \, \mathrm{d}\theta \right]
= \frac{2\pi}{3} - \frac{\sqrt 3}{2}
$$

Note that I've interpreted "the area confined by these two circles" to mean "the area of the two circles' overlapping region". If we instead want "the area in the first circle but not the second", then the desired integral would be $\pi$ minus the above expression, or simply:
$$
\int_{-\pi/3}^{\pi/3}\int_1^{2\cos\theta} r \, \mathrm{d}r \, \mathrm{d}\theta
= \frac{\pi}{3} + \frac{\sqrt 3}{2}
$$
