jmc

Rectangle $ABFE$ has area $AE\cdot AB=2\cdot 4\sqrt{2}=8\sqrt{2}$. Right triangles $ACO$ and $BDO$ each have hypotenuse $2\sqrt{2}$ and one leg of length 2.



Hence they are each isosceles, and each has area $(1/2)\left(2^2\right)=2$. Angles $CAE$ and $DBF$ are each $45^{\circ }$, so sectors $CAE$ and $DBF$ each have area $$\frac{1}{8}\cdot \pi \cdot 2^2=\frac{\pi }{2}.$$ Thus the area of the shaded region is $$8\sqrt{2}-2\cdot 2-2\cdot \frac{\pi }{2}=\boxed{8\sqrt{2}-4-\pi }.$$