9. For a regular tetrahedron $ABCD$, the center of the base $BCD$ is $O$, and $AB=1$. If the tetrahedron is rotated $90^{\circ}$ about the axis $OA$, then the volume of the overlapping part of the resulting tetrahedron and the original tetrahedron is $\qquad$.
Solve:
As shown in the figure, $F C=M C-M F=\frac{1}{2}-\frac{\sqrt{3}}{6} \Rightarrow F G=\frac{\sqrt{3}}{2}-\frac{1}{2}$, then $S_{\triangle F G C}=\frac{1}{2} \cdot\left(\frac{1}{2}-\frac{\sqrt{3}}{6}\right) \cdot\left(\frac{\sqrt{3}}{2}-\frac{1}{2}\right)=\frac{\sqrt{3}}{6}-\frac{1}{4}$. Therefore, $S_{E F G H J K}=S_{\triangle A B C}-3 S_{\triangle F G C}=\frac{3-\sqrt{3}}{4}$, so the volume of the overlapping part of the two regular tetrahedrons is $V=\frac{1}{3} \cdot \frac{3-\sqrt{3}}{4} \cdot \frac{\sqrt{6}}{3}=\frac{\sqrt{6}-\sqrt{2}}{12}$.