2e2 * A cylindrical container with a base radius of $1 \mathrm{~cm}$ contains four solid iron balls, each with a radius of $\frac{1}{2} \mathrm{~cm}$. The four balls are pairwise tangent, with the two balls at the bottom touching the bottom of the container. Water is now poured into the container until the water level just covers all the iron balls. The amount of water needed is $\mathrm{cm}^{3}$.
Let the centers of four solid iron balls be $O_{1}, O_{2}, O_{3}, O_{4}$, where $O_{1}, O_{2}$ are the centers of the two balls on the lower layer, and $A, B, C, D$ are the projections of the four centers on the bottom plane. It is easy to see that $ABCD$ is a square with side length $\frac{\sqrt{2}}{2}$, so the height of the water added is $1+\frac{\sqrt{2}}{2}$. Therefore, the volume of water to be added is $\pi\left(1+\frac{\sqrt{2}}{2}\right)-4 \cdot \frac{4}{3} \pi \cdot\left(\frac{1}{2}\right)^{3}=$ $\left(\frac{1}{3}+\frac{\sqrt{2}}{2}\right) \pi$.