A circle of radius $1$ is internally tangent to two circles of radius $2$ at points $A$ and $B$, where $AB$ is a diameter of the smaller circle. What is the area of the region, shaded in the picture, that is outside the smaller circle and inside each of the two larger circles? 
$\mathrm{(A) \ } \frac{5}{3} \pi - 3\sqrt 2 \qquad  \mathrm{(B) \ } \frac{5}{3} \pi - 2\sqrt 3 \qquad  \mathrm{(C) \ } \frac{8}{3} \pi - 3\sqrt 3 \qquad  \mathrm{(D) \ } \frac{8}{3} \pi - 3\sqrt 2 \qquad  \mathrm{(E) \ } \frac{8}{3} \pi - 2\sqrt 3$
The area of the small circle is $\pi$. We can add it to the shaded region, compute the area of the new region, and then subtract the area of the small circle from the result.
Let $C$ and $D$ be the intersections of the two large circles. Connect them to $A$ and $B$ to get the picture below:

We can see that the triangles $\triangle ABC$ and $\triangle ABD$ are both equilateral with side $2$.
Take a look at the lower circle. The angle $ABC$ is $60^\circ$, thus sector $ABC$ is $1/6$ of the circle. The same is true for sector $ABD$ of the lower circle, and sectors $CAB$ and $BAD$ of the upper circle. 
If we now sum the areas of these four sectors, we will almost get the area of the new shaded region - except that each of the two equilateral triangles will be counted twice. The area of an equilateral triangle given its side, $s,$ is $\frac{\sqrt3}4s^2.$
Therefore, the area of the new shaded region is $4\cdot \left( \frac 16 \cdot \pi\cdot 2^2 \right) - 2 \cdot \left( 4 \cdot \frac{\sqrt3}4 \right) = \frac 83 \pi - 2\sqrt 3$. Lastly, we must subtract the area of the circle that we added earlier, $\pi$, and we get
$\left( \frac 83 \pi - 2\sqrt 3 \right) - \pi = \boxed{\mathrm{(B)\ } \frac 53 \pi - 2\sqrt 3 }$.