Circles centered at $A$ and $B$ each have radius $3$, as shown. Point $O$ is the midpoint of $\overline{AB}$, and $OA=6$. Segments $OC$ and $OD$ are tangent to the circles centered at $A$ and $B$, respectively, and $EF$ is a common tangent. What is the area of the shaded region $ECODF$?

1. Connect $AE, BF, AC, BD$. $EF$ is tangent to both circles at $E$ and $F$, so $AE\perp EF$ and $BF\perp EF$, with $AE=BF=3$. Hence $ABFE$ is a rectangle with $AB=2\cdot OA=12$, and $[ABFE]=12\cdot3=36$.
2. $OC$ is tangent at $C$, so $AC\perp OC$. In the right triangle $ACO$, $AC=3$ and $AO=6$, so $\cos\angle CAO=\dfrac36=\dfrac12$, i.e. $\angle CAO=60^\circ$, and $CO=\sqrt{6^2-3^2}=3\sqrt3$. Then $[ACO]=\dfrac12\cdot3\cdot3\sqrt3=\dfrac{9\sqrt3}2$.
3. Since $\angle EAB=90^\circ$, $\angle EAC=90^\circ-60^\circ=30^\circ$, so the sector $AEC$ has area $\dfrac{30}{360}\cdot\pi\cdot3^2=\dfrac{3\pi}4$. The same holds on the side of $B$.
4. The shaded region is the rectangle minus two triangles and two sectors: $$[ECODF]=36-2\cdot\frac{9\sqrt3}2-2\cdot\frac{3\pi}4=\boxed{36-9\sqrt3-\frac{3\pi}2}\approx15.70.$$