Example 3 (2003 National High School Competition Question) Place 8 balls, each with a radius of 1, in two layers inside a cylinder, such that each ball is tangent to its four adjacent balls, and is also tangent to one base and the side of the cylinder. Then the height of the cylinder is. $\qquad$
Solve $\sqrt[4]{8}+2$. Reason: If calculating directly from the diameter of the sphere seems difficult, can we find a more familiar geometric shape to calculate with? According to the problem, 8 spheres with a radius of 1 are placed in two layers. Since the distance from the center of the upper layer of spheres to the top base of the cylinder and the distance from the center of the lower layer of spheres to the bottom base of the cylinder are both 1, the key is to find the distance between the planes containing the centers of the two layers of spheres. As shown in Figure 1-2, the centers of the upper and lower four spheres $A^{\prime} 、 B^{\prime} 、 C^{\prime} 、 D^{\prime}$ and $A 、 B 、 C 、 D$ are the vertices of two squares with side lengths of 2, and the circumcircles $\odot O^{\prime}$ and $\odot O$ of these squares form the top and bottom bases of the cylinder. At the same time, the projection of $A^{\prime}$ on the lower base must be the midpoint $M$ of $\overparen{A B}$.

In $\triangle A^{\prime} A B$, $A^{\prime} A=A^{\prime} B=A B=2$. Let the midpoint of $A B$ be $N$, then $A^{\prime} N=\sqrt{3}$, and $O M=O A=\sqrt{2}, O N=1$, so,
$$
M N=\sqrt{2}-1, A^{\prime} M=\sqrt{\left(A^{\prime} N\right)^{2}-(M N)^{2}}=\sqrt[4]{8} \text {. }
$$

Therefore, the height of the original cylinder is $\sqrt[4]{8}+2$.