III A cylindrical container with a base radius of $1 \mathrm{~cm}$ contains four solid iron balls 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. Now water is poured into the container so that the water level just covers all the iron balls. Then the amount of water needed is $\qquad$ $\mathrm{cm}^{3}$.
Solution: Let the centers of the four solid iron balls be $O_{1}, O_{2}, O_{3}, O_{4}$, where $\mathrm{O}_{1}, \mathrm{O}_{2}$ are the centers of the two balls on the lower layer, and $A, B, C, D$ are the projections of the four ball centers on the bottom plane. Then $ABCD$ is a square with side length $\frac{\sqrt{2}}{2}$. Therefore, the height of the water added is $1+\frac{\sqrt{2}}{2}$. Hence, the volume of water to be added is $\pi\left(1+\frac{\sqrt{2}}{2}\right)-4 \times \frac{4}{3} \pi\left(\frac{1}{2}\right)^{3}=\left(\frac{1}{3}+\frac{\sqrt{2}}{2}\right) \pi$.