Example 2 Rotate the closed figure formed by the parabola $y=x^{2}$ and the line $y=t$ around the $y$-axis once, and find the volume of the resulting solid of revolution.
Solution: Place the solid of revolution on a plane, and cut this solid with a plane parallel to the plane and at a distance $h$ from it (as shown in Figure 2). It is easy to see that the area of the circular cross-section is
$$
S_{\text {cut }}=\pi(\sqrt{h})^{2}=\pi h .
$$

This indicates that the cross-section of the constructed reference solid can be a rectangle, one side of which has a constant length $\pi$, and the other side has a variable length $h$. Consequently, it is easy to find the reference solid (as shown in Figure 3): a right prism with an isosceles right triangle as its base (with legs of length $t$) and a height of $\pi$. The volume of this prism is easily calculated as
$$
V=\frac{1}{2} t^{2} \cdot \pi=\frac{1}{2} \pi t^{2} .
$$