Example 2. In a cylindrical barrel with a radius of $2 R$, there are 6 small balls with diameters all equal to $2 R$, and they are in a fixed state. Water is poured into the barrel. What is the minimum height of the water level to completely submerge these 6 balls? What is the volume of water in the barrel at this time?
(1) The diameter of the cylinder's base is $4 R$, and the diameter of the sphere is $2 R$. To ensure the small balls are in a stable state, the 6 small balls are arranged in three layers. The line connecting the centers of the balls in the first layer, $A B$, is parallel to the base, and the line connecting the centers of the balls in the second layer, $C D$, is also parallel to the base, with $C D \perp A B$ (as shown in the figure). The key is to find the distance between $A B$ and $C D$, which can be converted to the distance between opposite edges of a regular tetrahedron $A B C D$ with edge length $2 R$:
$$
\begin{aligned}
M N & =\sqrt{(\sqrt{3} R)^{2}-R^{2}} \\
& =\sqrt{2} R .
\end{aligned}
$$
(2) The line connecting the centers of the balls in the third layer, $E F$, is parallel to the base, and $E F \perp C D$. The distance between $E F$ and $C D$ is also $\sqrt{2} R$. Considering that the top and bottom layers each occupy a radius $R$ of the small balls, the minimum height of the water level $P Q=2(1+\sqrt{2}) R$.
At this time, the volume of water in the cylinder is
$$
\begin{aligned}
V & =\pi(2 R)^{2} \cdot 2(1+\sqrt{2}) R-6 \cdot \frac{4}{3} \pi R^{3} \\
& =8 \sqrt{2} \pi R^{3} .
\end{aligned}
$$