smc

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 $\pi (\frac{5}{2}-\frac{1}{2}-\sqrt{2})-1$ = $2\pi -\pi \sqrt{2}-1$. $\boxed{2\pi -\pi \sqrt{2}-1}$