3. In a square $A B C D$ with side length 2, $E$ is the midpoint of $A B$. Now, fold the figure along segments $E C, E D$ so that segments $E A, E B$ coincide. This results in a tetrahedron $C D E A$ (point $B$ has coincided with point $A$). The volume of this tetrahedron is
Since $A C=A D=C D \Rightarrow \triangle A C D$ is an equilateral triangle. Let the midpoint of $C D$ be $F$, then $A F=\sqrt{3}, A E=1, E F=2$. Also, $A E \perp$ plane $A C D \Rightarrow C D \perp$ plane $A E F$, thus the height $h=\frac{\sqrt{3}}{2}$ of tetrahedron $C D E A$, and $S_{\triangle E C D}=2 \Rightarrow V_{C D E A}=\frac{1}{3} \cdot 2 \cdot \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{3}$.