10. 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, and the two balls at the bottom are tangent to 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 $\qquad$ $\mathrm{cm}^{3}$.
As shown in the figure, $O_{1} O_{2} O_{3} O_{4}$ is a regular tetrahedron with edge length 1, and $O_{1} O_{2}$ and $O_{3} O_{4}$ are the diagonals of the top and bottom faces of a cube, respectively. The edge length of the cube is $\frac{\sqrt{2}}{2}$, so the height of the water level is $1+\frac{\sqrt{2}}{2}$. Therefore, the volume of water needed $V=\pi\left(1+\frac{\sqrt{2}}{2}\right)-4 \cdot \frac{4}{3} \pi\left(\frac{1}{2}\right)^{3}=\pi\left(\frac{1}{3}+\frac{\sqrt{2}}{2}\right)\left(\mathrm{cm}^{3}\right)$.