5. Find the volume of the body obtained by rotating a cube with edge $a$ around the diagonal of one of its faces.
2.II.5. Answer: $\frac{7 \pi a^{3} \sqrt{2}}{6}$.

Solution. Let the given body be obtained by rotating around the line $A C$ of the cube $A B C D A_{1} B_{1} C_{1} D_{1}$, whose edge is equal to $a$. Drop perpendiculars $M N$ and $M K$ from point $M$ on edge $B_{1} C_{1}$ to the lines $A C$ and $B C$, respectively. Let $x = C N$ and find the length of segment $M N$ (figure).

![](https://cdn.mathpix.com/cropped/2024_05_21_46613852391e49da1fa9g-057.jpg?height=474&width=514&top_left_y=1710&top_left_x=774)

Since triangle $C K N$ is a right triangle with a right angle at vertex $N$ and $\angle K C N = 45^{\circ}$, then $N K = N C = x$. Therefore, $M N = \sqrt{x^{2} + a^{2}}$.

Introduce a coordinate system with the origin at point $C$, the x-axis of which coincides with the line $A C$ (as in the figure above). Since point $M$ is the farthest point from the line $A C$ in the section of the given cube by the plane $K M N$ (left figure), the volume of the body formed by rotating the prism $B C D B_{1} C_{1} D_{1}$ is equal to the volume of the body formed by rotating around the x-axis the curvilinear trapezoid bounded by the x-axis, the y-axis, the line $x = \frac{a}{\sqrt{2}}$, and the graph of the function $y = \sqrt{x^{2} + a^{2}}$ (right figure).
![](https://cdn.mathpix.com/cropped/2024_05_21_46613852391e49da1fa9g-058.jpg?height=394&width=1044&top_left_y=840&top_left_x=518)

Since the volume of this body of revolution is half of the desired volume, using the formula for the volume of a body of revolution, we get

$$
\begin{aligned}
& V = 2 \pi \int_{0}^{\frac{a}{\sqrt{2}}}\left(\sqrt{x^{2} + a^{2}}\right)^{2} d x = 2 \pi \int_{0}^{\frac{a}{\sqrt{2}}}\left(x^{2} + a^{2}\right) d x = \\
& \quad = \left.2 \pi \left(\frac{x^{3}}{3} + a^{2} x\right)\right|_{0}^{\frac{a}{\sqrt{2}}} = 2 \pi \left(\frac{a^{3} \sqrt{2}}{12} + \frac{a^{3} \sqrt{2}}{2}\right) = \frac{7 \pi a^{3} \sqrt{2}}{6}
\end{aligned}
$$

Remark. In fact, we can limit ourselves to the first step of the given solution. The point is that in the formula $V = \pi \int_{a}^{b} f^{2}(x) d x$ for the volume of a body of revolution, the expression $\pi f^{2}(x)$ is the area of a circle, which is a cross-section of the body of revolution by a plane perpendicular to the axis of rotation. In our case, the cross-section is a circle with radius $M N = \sqrt{a^{2} + x^{2}}$, the area of which is $\pi (a^{2} + x^{2})$, from which it follows that the volume of the desired body is $2 \pi \int_{0}^{\frac{a}{\sqrt{2}}}\left(x^{2} + a^{2}\right) d x$.