$\fbox{B}$ The area of the entire outer shape is the area of sector $ABE$, plus the area of sector $ABF$, minus the area of triangle $ABD$ (since it is part of both sectors), plus the area of sector $DEF$. We know $AC = CD = 1$, so the sector angles for $ABE$ and $ABF$ are $45$ degrees, and the radius of both of them is $2$. The radius of $DEF$ is $DE = BE - BD = 2 - BD$, and $BD$ can be found using Pythagoras in triangle $BCD$, giving $BD = \sqrt{2}$ and $DE = 2 - \sqrt{2}$, so after doing all the calculations, the area of the entire outer shape is $\pi(\frac{5}{2} - \sqrt{2}) - 1$. To get the area of the smile, we need to subtract the area of semicircle $ABD$, which is $\frac{1}{2} \pi 1^2 = \frac{\pi}{2}$, so the answer is $\pi(\frac{5}{2} - \frac{1}{2} - \sqrt{2}) - 1$ = $2\pi - \pi \sqrt{2} - 1$.