Polar region between two circles

Polar. $\sin\theta\le1/2$ for the two-disk overlap; the region splits into $\theta\in[0,\pi/6]$, $[5\pi/6,\pi]$, $[\pi,2\pi]$.

Compute each piece.
- $I_1$ over $[\pi,2\pi]$, $0\le r\le1$: $I_1=\int_\pi^{2\pi}\int_0^1r^2\,dr\,d\theta=\frac{\pi}{3}$.
- $I_2$ over $[0,\pi/6]$, $2\sin\theta\le r\le1$: $I_2=\frac{\pi}{18}+\sqrt3-\frac{16}{9}$.
- $I_3$ over $[5\pi/6,\pi]$, by symmetry $I_3=I_2$.

Sum. $I=\frac{\pi}{3}+\frac{2\pi}{18}+2\sqrt3-\frac{32}{9}=\frac{4\pi-32}{9}+2\sqrt3$.

Conclusion. $I=\frac{4\pi-32}{9}+2\sqrt3$.