14. The record has the shape of a regular hexagon

![](https://cdn.mathpix.com/cropped/2024_05_21_27f86caa754616304e40g-100.jpg?height=284&width=427&top_left_y=1154&top_left_x=625)

Fig. 98.

$A B C D E F$ (Fig. 97) with side $a$, $O$ - the center of the hexagon. A square $O M A N$ is cut out from the record, having its diagonal as the segment $O A$. Determine the distance from $O$ to the center of gravity of the remaining plate. (O. 117, P. 209.)
14. We will rotate the plate around the axis $x x^{\prime}$ passing through $A$ and perpendicular to $A O$. Then the volume $V$ of the body formed by the rotation of the plate can be found by subtracting the volume of the body formed by the rotation of the square from the volume of the body formed by the rotation of the hexagon. Therefore,

$$
V=2 \pi a \cdot 6 \frac{a^{2} \sqrt{3}}{4}-2 \pi \frac{a}{2}\left(\frac{a}{\sqrt{2}}\right)^{2}=\pi a^{3}\left(3 \sqrt{3}-\frac{1}{2}\right)
$$

Let $Z$ be the center of gravity of the plate. Due to the symmetry of the plate relative to the line $A D$, the center of gravity of the plate lies on $A D$. Since the area of the plate is

$$
6 \cdot a^{2} \frac{\sqrt{3}}{4}-\left(a \frac{\sqrt{2}}{2}\right)^{2}=(3 \sqrt{3}-1) \frac{a^{2}}{2}
$$

then by the second theorem of Guldin,

$$
V=2 \pi \cdot A Z \cdot(3 \sqrt{3}-1) \frac{a^{2}}{2}
$$

Comparing the two expressions for $V$, we find

$$
A Z=\frac{6 \sqrt{3}-1}{6 \sqrt{3}-2} a
$$

Therefore,

$$
O Z=A Z-a=\frac{1}{6 \sqrt{3}-2} a=\frac{3 \sqrt{3}+1}{52} a
$$