In the coordinate space, define a square $S$, defined by the inequality $|x|\leq 1,\ |y|\leq 1$ on the $xy$-plane, with four vertices $A(-1,\ 1,\ 0),\ B(1,\ 1,\ 0),\ C(1,-1,\ 0), D(-1,-1,\ 0)$. Let $V_1$ be the solid by a rotation of the square $S$ about the line $BD$ as the axis of rotation, and let $V_2$ be the solid by a rotation of the square $S$ about the line $AC$ as the axis of rotation.

(1) For a real number $t$ such that $0\leq t<1$, find the area of cross section of $V_1$ cut by the plane $x=t$.

(2) Find the volume of the common part of $V_1$ and $V_2$.
### Part (1)

To find the area of the cross-section of \( V_1 \) cut by the plane \( x = t \), we need to understand the geometry of the solid formed by rotating the square \( S \) about the line \( BD \).

1. **Equation of the Cones:**
   - The rotation of the square \( S \) about the line \( BD \) forms two cones with vertices at \( B \) and \( D \).
   - The equation of the cone with vertex \( B \) is given by:
     \[
     \frac{1}{2}z^2 = xy - x - y + 1
     \]
   - The equation of the cone with vertex \( D \) is given by:
     \[
     \frac{1}{2}z^2 = xy + x + y + 1
     \]

2. **Intersection of Cones:**
   - Both cones intersect in the plane \( x + y = 0 \).

3. **Cross-section at \( x = t \):**
   - Cutting the cones with the plane \( x = t \) results in:
     \[
     z = \pm \sqrt{2(1-t)(1-y)}
     \]
     and
     \[
     z = \pm \sqrt{2(1+t)(1+y)}
     \]
   - At \( y = -t \), the equations simplify to:
     \[
     z = \pm \sqrt{2(1-t)(1+t)}
     \]

4. **Area Calculation:**
   - The area \( S \) between these parabolas is given by:
     \[
     S = 2 \left[ \int_{-1}^{-t} \sqrt{2(1+t)(1+y)} \, dy + \int_{-t}^{1} \sqrt{2(1-t)(1-y)} \, dy \right]
     \]
   - Evaluating the integrals:
     \[
     S = 2 \left[ \frac{2\sqrt{2}}{3} \cdot \sqrt{1+t}(1-t)\sqrt{1-t} + \frac{2\sqrt{2}}{3} \cdot \sqrt{1-t}(1+t)\sqrt{1+t} \right]
     \]
   - Simplifying the expression:
     \[
     S = \frac{8\sqrt{2}}{3} \cdot \sqrt{1-t^2}
     \]

The final answer is \( \boxed{ \frac{8\sqrt{2}}{3} \cdot \sqrt{1-t^2} } \)