923. Find the center of gravity of a homogeneous surface of the paraboloid $y^{2}+z^{2}=10 x$, cut off by the plane $x=10$.
Solution. The given homogeneous surface ($\sigma$) is symmetric with respect to the x-axis (Fig. 196). Therefore, $y_{C}=z_{C}=0$.

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-324.jpg?height=377&width=501&top_left_y=1349&top_left_x=683)

Fig. 196

To find the abscissa of the center of gravity, we need to compute: 1) the static moment $m_{y z}$ and 2) the mass $m$ of the given surface:

1) $m_{y z}=\iint_{\sigma} x \delta d s=\delta \int_{\sigma_{y z}} x \sqrt{1+\left(x_{y}^{\prime}\right)^{2}+\left(x_{z}^{\prime}\right)^{2}} d y d z=$

$$
=\frac{\delta}{50} \int_{y^{2}+z^{2} \leqslant 100} \int_{i}\left(y^{2}+z^{2}\right) \sqrt{25+y^{2}+z^{2}} d y d z=
$$

$$
=\frac{\delta}{50} \iint_{\rho \leqslant 10} \rho^{2} \sqrt{25+\rho^{2}} \rho d \varphi d \rho=
$$

$$
=\frac{\delta}{50} \int_{0}^{2 \pi} d \varphi \int_{0}^{10} \rho^{2} \sqrt{25+\rho^{2}} \rho d \rho=\frac{50}{3} \pi \delta(1+25 \sqrt{5})
$$

Here, after transitioning to polar coordinates, the inner integral is computed using the substitution $\sqrt{25+\rho^{2}}=t$.
2) $m=\iint_{\sigma} \delta d s=\frac{\delta}{5} \iint_{y^{2}+z^{2}} \int_{100} \sqrt{25+y^{2}+z^{2}} d y d z=$

$=\frac{\delta}{5} \iint_{\rho \leqslant 10} \sqrt{25+\rho^{2}} \rho d \varphi d \rho=\frac{\delta}{10} \int_{0}^{2 \pi} d \varphi \int_{0}^{10}\left(25+\rho^{2}\right)^{\frac{1}{2}} d\left(25+\rho^{2}\right)=$

$$
=\left.\frac{\delta}{10} \int_{0}^{2 \pi} \frac{2}{3}\left(25+\rho^{2}\right)^{\frac{3}{2}}\right|_{0} ^{10} d \varphi=\frac{50}{3} \pi \delta(5 \sqrt{5}-1)
$$

By formula (3), $x_{C}=\frac{m_{y z}}{m}=\frac{25 \sqrt{5}+1}{5 \sqrt{5}-1}$.

Find the area of the part of the surface: