Solution:

Firstly, notice $FD = 1,$ so the arc ${XZ}$ must have length $$2 \arccos{\dfrac 12} = \dfrac {2\pi} 3 .$$ Since the area under semicircles is equal to the area of the arc minus the area of $FXZ,$ that area is $$\dfrac \theta {2} r^2 - \dfrac{FX\cdot FZ \sin XFZ}2$$$$= \dfrac{2\cdot 2^2 \pi}{3\cdot 2} - \dfrac{2\cdot 2 \frac {\sqrt{3}} 2}2$$$$= \frac 43 \pi - \sqrt 3 .$$ Then, the gray area above $EG$ is a semicircle $$\frac 12 r^2 \pi= \frac 12 \cdot 4 \pi = 2\pi.$$ Finally, the gray area consists of four of the following shapes. \t\t

The squares have side length $1$ so it has area $1.$ The quarter circle has area $$\frac \pi 4 r^2 = \frac \pi 4.$$ Therefore, the total amount of gray is $1- \frac \pi 4.$ We multiply this by $4$ since there are $4$ of these shapes, yielding an area of $4- \pi.$

The total area is $$\frac 43 \pi - \sqrt 3 + 2 \pi + 4 -\pi$$$$= \frac 73 \pi - \sqrt 3 + 4.$$ This makes our answer $$7+3+3+4 = 17.$$

Thus, the answer is E .