Let's rotate the unit-edged cube around one of its body diagonals. What is the volume of the resulting solid of revolution?
Lettering according to the figure.

![](https://cdn.mathpix.com/cropped/2024_05_02_d151537a78ba24541d8cg-1.jpg?height=523&width=616&top_left_y=178&top_left_x=749)

Since the edge length of the cube is 1, the face diagonal is $\sqrt{1^{2}+1^{2}}=\sqrt{2}$ and the space diagonal is $\sqrt{1^{2}+(\sqrt{2})^{2}}=\sqrt{3}$.

If we rotate the cube around the space diagonal $A H$, then the 3 edges starting from point $A$ (or each of the 3 edges separately) describe a cone, whose vertex is $A$, the base circle is the circumcircle of the equilateral triangle $B C D$ with center $K$, and the height is $m=A K$. Since the side length of the equilateral triangle $B C D$ is $\sqrt{2}$, the radius of the circumcircle (which is $\frac{2}{3}$ of the height of the equilateral triangle) is $r=\frac{2}{3} \cdot \frac{\sqrt{2}}{2} \sqrt{3}=\frac{\sqrt{6}}{3} \cdot (A z=B K$ distance can also be calculated from the right triangle $A B H$ as the height to the hypotenuse: $B K: A B=B H: A H$ or $r: 1=\sqrt{2}: \sqrt{3}$, from which $\left.r=\frac{\sqrt{2}}{\sqrt{3}}=\frac{\sqrt{6}}{\sqrt{3}}\right)$. The height of the cone $m=A K$ can be calculated from the right triangle $A B H$ as the projection of the leg $A B$ on the hypotenuse $A H$. Thus, $A K: A B=A B: A H$, or $m: 1=1: \sqrt{3}$, from which $m=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$. The volume of the cone is

$$
V_{1}=\frac{1}{3} \cdot r^{2} \pi \cdot m=\frac{1}{3} \cdot\left(\frac{\sqrt{6}}{3}\right)^{2} \pi \frac{\sqrt{3}}{3}=\frac{6 \sqrt{3}}{81} \pi=\frac{2 \sqrt{3}}{27} \pi
$$

This is also true for the 3 edges starting from the vertex $H$, and thus the volume of these two cones is

$$
2 V_{1}=\frac{4 \sqrt{3}}{27} \pi
$$

The other 6 edges of the cube (or each of these 6 edges separately) describe a one-sheet hyperboloid. The volume of the solid of revolution described by the edge, as a segment, is bounded by the two circular bases described by the endpoints of the edge (these two circular bases are the bases of the two cones mentioned earlier): $V_{2}$, which we calculate by integration. For this, we place the meridian of the entire solid of revolution $A B E^{*}$ in the plane of a Cartesian coordinate system such that the midpoint of the cube, i.e., the midpoint of $A H$, is at the origin $O$ and the axis of rotation $A H$ coincides with the $x$-axis. Then $y$ is the symmetry axis of the meridian, on which lies the real axis of the hyperbola. $E^{*}$ is the 180-degree rotation of point $E$. As we have seen, $A K=H K_{1}=\frac{\sqrt{3}}{3}$ and thus $K K_{1}=\sqrt{3}-\frac{2 \sqrt{3}}{3}=\frac{\sqrt{3}}{3}=A K=H K_{1}$, which means that points $K$ and $K_{1}$ divide the axis of rotation $A H$ into 3 equal parts. $O K=O K_{1}=\frac{\sqrt{3}}{6}$. Let the semi-transverse axis of the hyperbola be denoted by $O M_{0}$. $O M_{0}$ is nothing else but the distance from the center of the cube $O$ to any of the 6 edges, for example, $B F$. This distance is the same as the distance from $O$ to the midpoint of $B F$, $M$, because $O M \perp B F$. Furthermore, $O M$ is the height of the isosceles triangle $A M H$, which is perpendicular to $A H$. (The realists will recognize that $O M$ is the normal-transversal of the axis of rotation $A H$ and the edge $B F$.) From the right triangle $M O H$, $O M^{2}=A M^{2}-A O^{2}$, but $A M^{2}=1+\left(\frac{1}{2}\right)^{2}=\frac{5}{4}$, and $A O^{2}=\left(\frac{\sqrt{3}}{2}\right)^{2}=\frac{3}{4}$, so $O M^{2}=\frac{5}{4}-\frac{3}{4}=\frac{2}{4}=\frac{1}{2}$, or $O M=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_d151537a78ba24541d8cg-2.jpg?height=444&width=816&top_left_y=125&top_left_x=641)

(Those who have studied descriptive geometry can also arrive at this result by recalling the representation of the cube, if they consider that the projection of the 6 edges describing the hyperboloid from a direction parallel to the axis $A H$ onto a plane perpendicular to the axis is a regular hexagon, the radius of the circumscribed circle of which is the previously mentioned $r=\frac{\sqrt{3}}{6}$. The radius of the circle with radius $O M$ is congruent to the projection of this regular hexagon, which is the inscribed circle of the regular hexagon, and thus its radius is

$$
\left.O M=O M_{0}=\frac{1}{2} r \sqrt{3}=\frac{1}{2} \cdot \frac{\sqrt{6}}{3} \cdot \sqrt{3}=\frac{\sqrt{18}}{6}=\frac{3 \sqrt{2}}{6}=\frac{\sqrt{2}}{2}\right) .
$$

The equation of the hyperbola in the meridian plane is:

$\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$, where $a$ represents half of the real axis, i.e., $a=O M_{0}=\frac{\sqrt{2}}{2}$. Substituting this value, we get

$$
2 y^{2}-\frac{x^{2}}{b^{2}}=1
$$

$b^{2}$ is determined by substituting the coordinates of a point on the hyperbola, for example, the point $B$ with coordinates $\left(-\frac{\sqrt{3}}{6}, \frac{\sqrt{6}}{3}\right)$ into the equation of the hyperbola:

$$
2 \cdot \frac{2}{3}-\frac{1}{12 b^{2}}=1
$$

from which

$$
\frac{1}{b^{2}}=\left(\frac{4}{3}-1\right) \cdot 12=4
$$

Thus, the equation of the hyperbola is

$$
2 y^{2}-4 x^{2}=1
$$

or

$$
y^{2}=\frac{4 x^{2}+1}{2}=2 x^{2}+\frac{1}{2}
$$

The volume of the hyperboloid of revolution is:

$$
\begin{gathered}
V_{2}=\pi \int_{-\frac{\sqrt{3}}{6}}^{\frac{\sqrt{3}}{6}} y^{2} d x=2 \pi \int_{0}^{\frac{\sqrt{3}}{8}} y^{2} d x=2 \pi \int_{0}^{\frac{\sqrt{3}}{6}}\left(2 x^{2}+\frac{1}{2}\right) d x= \\
=2 \pi\left[\frac{2}{3} x^{3}+\frac{1}{2} x\right]_{0}^{\frac{\sqrt{3}}{6}}=2 \pi\left(\frac{2}{3} \cdot \frac{3 \sqrt{3}}{216}+\frac{1}{2} \cdot \frac{\sqrt{3}}{6}\right)=2 \pi\left(\frac{\sqrt{3}}{108}+\frac{\sqrt{3}}{12}\right)= \\
=2 \pi \frac{\sqrt{3}+9 \sqrt{3}}{108}=2 \pi \frac{10 \sqrt{3}}{108}=\frac{5 \sqrt{3}}{27} \pi
\end{gathered}
$$

Thus, the volume of the entire solid of revolution is

$$
V=2 V_{1}+V_{2}=\frac{4 \sqrt{3}}{27} \pi+\frac{5 \sqrt{3}}{27} \pi=\frac{9 \sqrt{3}}{27} \pi=\frac{\sqrt{3}}{3} \pi
$$

Remark: The volume of our solid of revolution is exactly half the volume of the cylinder of revolution whose radius is $B K=r=\frac{\sqrt{6}}{3}$ and height is $A H=\sqrt{3}$. (The cylinder circums