We draw spheres with a radius of $\frac{1}{2}$ around the vertices of a unit cube. There are two such spheres that touch all eight of these spheres. Calculate the difference in volume between the two spheres.

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Cut the cube with a plane $S$ passing through the parallel diagonals of two opposite faces. $S$ is a symmetry plane of the cube, containing the center of the cube, which is also the center of the two tangent spheres we are looking for. $S$ cuts a rectangle out of the cube, the shorter side of which is 1 unit, the longer side is the face diagonal of the cube, with a length of $\sqrt{2}$, and the diagonal of the rectangle is the space diagonal of the cube, with a length of $\sqrt{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_68e94b36a400d5832d62g-1.jpg?height=618&width=631&top_left_y=336&top_left_x=736)

Let $r$ be the radius of the sphere tangent from the inside, and $R$ be the radius of the sphere tangent from the outside. Then $R=\frac{\sqrt{3}+1}{2}, r=\frac{\sqrt{3}-1}{2}$. The difference in volumes is:

$\frac{4 \pi}{3}\left(R^{3}-r^{3}\right)=\frac{4 \pi}{3}\left(\left(\frac{\sqrt{3}+1}{2}\right)^{3}-\left(\frac{\sqrt{3}-1}{2}\right)^{3}\right)=\frac{10}{3} \pi \approx 10.471$ volume units.