Find the area of the axial section of a cylinder inscribed in a unit cube such that the axis of the cylinder lies on the diagonal of the cube, and each base touches three faces of the cube at their centers.

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Let the circle of radius $r$ of one of the bases of the cylinder with height $h$ touch the faces $ABCD$, $CC_1D_1D$, and $AA_1D_1D$ of the unit cube $ABCD A_1B_1C_1D_1$ at points $K$, $L$, and $M$ respectively. Since $K$, $L$, and $M$ are the midpoints of the diagonals $AC$, $CD_1$, and $AD_1$ of the corresponding faces of the cube, this circle is inscribed in the equilateral triangle $ACD_1$ with side length $\sqrt{2}$. Therefore,

$$
r = \sqrt{2} \cdot \frac{\sqrt{3}}{6} = \frac{\sqrt{6}}{6}
$$

The circle of the second base of the cylinder is inscribed in the equilateral triangle $BA_1C_1$. The axis of the cylinder passes through the centers of these circles and is perpendicular to the planes $ACD_1$ and $BA_1C_1$. It is known that these planes divide the diagonal of the cube into three equal parts. Therefore,

$$
h = \frac{1}{3} CB_1 = \frac{\sqrt{3}}{3}
$$

Thus, the area of the axial section of the cylinder is

$$
2 r h = 2 \cdot \frac{\sqrt{6}}{6} \cdot \frac{\sqrt{3}}{3} = \frac{\sqrt{2}}{3}
$$

## Answer

$\frac{\sqrt{2}}{3}$.