Example 3 In a cube with edge length 1, let the volume of the right prism with the midpoints of the edges of the top and bottom faces as vertices be $V_{1}$, and the volume of the right prism with the midpoints of the edges of the left and right faces as vertices be $V_{2}$. Try to find the volume $V$ of the common part of $V_{1}$ and $V_{2}$.
Solution: As shown in Figure 4, let $A B C D-A_{1} B_{1} C_{1} D_{1}$ be the given cube, and $P Q R S-P_{1} Q_{1} R_{1} S_{1}$ and $E P H P_{1}-F R G R_{1}$ be the right square prisms $V_{1}$ and $V_{2}$, respectively.

Notice that $E F$ intersects $Q Q_{1}$ at $K$, and $P, R$ are also common points of the plane $E F R P$ and $V_{1}$. Therefore, the section $K P R$ is the section of the plane $E F R P$ on $V_{1}$. The tetrahedron $Q-K P R$ is the part cut off by this plane from $V_{1}$, and its volume is
$$
\frac{1}{3}\left(\frac{1}{2} Q P \cdot P R\right) \cdot Q K=\frac{1}{24} \text {. }
$$

By symmetry, $V_{1}$ is cut by the 4 side faces of $V_{2}$ into 4 such tetrahedrons. Therefore, the volume of the common part of $V_{1}$ and $V_{2}$ is
$$
V=\frac{1}{2} \times 1-4 \times \frac{1}{24}=\frac{1}{3} .
$$