In the figure shown, $\overline{US}$ and $\overline{UT}$ are line segments each of length 2, and $m\angle TUS = 60^\circ$. Arcs $\overarc{TR}$ and $\overarc{SR}$ are each one-sixth of a circle with radius 2. What is the area of the region shown? 

$\textbf{(A) }3\sqrt{3}-\pi\qquad\textbf{(B) }4\sqrt{3}-\frac{4\pi}{3}\qquad\textbf{(C) }2\sqrt{3}\qquad\textbf{(D) }4\sqrt{3}-\frac{2\pi}{3}\qquad\textbf{(E) }4+\frac{4\pi}{3}$
Let the centers of the circles containing arcs $\overarc{SR}$ and $\overarc{TR}$ be $X$ and $Y$, respectively. Extend $\overline{US}$ and $\overline{UT}$ to $X$ and $Y$, and connect point $X$ with point $Y$. 

We can clearly see that $\triangle UXY$ is an equilateral triangle, because the problem states that $m\angle TUS = 60^\circ$. We can figure out that $m\angle SXR= 60^\circ$ and $m\angle TYR = 60^\circ$ because they are $\frac{1}{6}$ of a circle. The area of the figure is equal to $[\triangle UXY]$ minus the combined area of the $2$ sectors of the circles (in red). Using the area formula for an equilateral triangle, $\frac{a^2\sqrt{3}}{4},$ where $a$ is the side length of the equilateral triangle, $[\triangle UXY]$ is $\frac{\sqrt 3}{4} \cdot 4^2 = 4\sqrt 3.$ The combined area of the $2$ sectors is $2\cdot\frac16\cdot\pi r^2$, which is $\frac 13\pi \cdot 2^2 = \frac{4\pi}{3}.$ Thus, our final answer is $\boxed{\textbf{(B)}\ 4\sqrt{3}-\frac{4\pi}{3}}.$