[Circles](https://artofproblemsolving.com/wiki/index.php/Circle) centered at $A$ and $B$ each have radius $2$, as shown. Point $O$ is the [midpoint](https://artofproblemsolving.com/wiki/index.php/Midpoint) of $\overline{AB}$, and $OA = 2\sqrt {2}$. Segments $OC$ and $OD$ are tangent to the circles centered at $A$ and $B$, respectively, and $EF$ is a common [tangent](https://artofproblemsolving.com/wiki/index.php/Tangent_(geometry)). What is the area of the shaded region $ECODF$?

$\text{(A)}\ \frac {8\sqrt {2}}{3} \qquad \text{(B)}\ 8\sqrt {2} - 4 - \pi \qquad \text{(C)}\ 4\sqrt {2} \qquad \text{(D)}\ 4\sqrt {2} + \frac {\pi}{8} \qquad \text{(E)}\ 8\sqrt {2} - 2 - \frac {\pi}{2}$
The area we are trying to find is simply $ABFE-(\overarc{AEC}+\triangle{ACO}+\triangle{BDO}+\overarc{BFD})$.
Obviously, $\overline{EF}\parallel\overline{AB}$.  Thus, $ABFE$ is a [rectangle](https://artofproblemsolving.com/wiki/index.php/Rectangle), and so its area is $b\times{h}=2\times{(AO+OB)}=2\times{2(2\sqrt{2})}=8\sqrt{2}$.
Since $\overline{OC}$ is tangent to circle $A$, $\triangle{ACO}$ is a right triangle.  We know $AO=2\sqrt{2}$ and $AC=2$, so $\triangle{ACO}$ is an isosceles right triangle, and has $\overline{CO}$ with length $2$.  The area of $\triangle{ACO}=\frac{1}{2}bh=2$.  By symmetry, $\triangle{ACO}\cong\triangle{BDO}$, and so the area of $\triangle{BDO}$ is also $2$.
$\overarc{AEC}$ (or $\overarc{BFD}$, for that matter) is $\frac{1}{8}$ the area of its circle since $\angle{OAC}$ is 45 degrees and $\angle{OAE}$ forms a right [triangle](https://artofproblemsolving.com/wiki/index.php/Triangle).  Thus $\overarc{AEC}$ and $\overarc{BFD}$ both have an area of $\frac{\pi}{2}$.
Plugging all of these areas back into the original equation yields $8\sqrt{2}-(\frac{\pi}{2}+2+2+\frac{\pi}{2})=8\sqrt{2}-(4+\pi)=\boxed{8\sqrt{2}-4-\pi}\ \mathrm{(B)}$.