859. Find the center of gravity of a spherical segment, if at each (y) point the volumetric density is proportional to its distance from the base of the segment.
Solution. Let the radius of the sphere be denoted by $R$, and the height of the segment by $h$; we place the origin of the rectangular coordinate system at the center of the sphere and direct the z-axis along the axis of the segment (Fig. 183). The equations of the sphere and the plane that limit the segment ($G$) will be, respectively, $x^{2}+y^{2}+z^{2}=R^{2}$ and $z=R-h$; the volume density at point $M(x, y, z)$ of the segment is expressed by the formula

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-296.jpg?height=344&width=424&top_left_y=473&top_left_x=761)

Fig. 18? Note $\delta=k(z-R+h)$. Any section of the given non-uniform segment by a plane parallel to its base is a homogeneous circle, the center of gravity of which lies at its center. Therefore, the center of gravity of the given segment lies on its axis, i.e., $x_{c}=y_{c}=0$.

To find the z-coordinate of the center of gravity, we apply formula (3). We will compute: 1) the static moment $I_{x y}$ and 2) the mass $m$ of the segment.

1) $I_{x y}=\iint_{i} \int_{i} z \delta d x d y d z=k \iiint_{i}(z-a) z d x d y d z, a=R-h$.

We transform the triple integral into a double integral of a simple one:

$$
I_{x y}==i \int_{i_{i}} \int_{x i j} d x d y \int_{a}^{z_{1}}\left(z^{2}-a z\right) d z
$$

where $G_{k y}$ is the circle $x^{2}+y^{2} \leqslant r^{2}, r^{2}=R^{2}-a^{2}: \quad z_{1}=\sqrt{R^{2}-\overline{x^{2}} \cdots y^{2}}$.

We compute the inner simple integral:

$$
r_{1}=\frac{z^{3}}{3}-\left.\frac{a z^{2}}{2}\right|_{1} ^{2_{1}}=\frac{1}{3}\left[\left(R^{2}-x^{2}-y^{2}\right)^{\frac{3}{2}}-a^{3} j-\frac{a}{2}\left(R^{2}-x^{2}-y^{2}-a^{2}\right) .\right.
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-296.jpg?height=43&width=1052&top_left_y=1711&top_left_x=134)
We transform the obtained double integral to polar coordinates:

$$
\begin{gathered}
I_{x y}=k \int_{x^{2}+y^{2}} l_{1} d x d y=k \int_{p} \int_{\beta=r}\left[\frac{1}{3}\left(R^{2}-\rho^{2}\right)^{\frac{3}{2}}+\frac{a^{3}}{6}-\right. \\
\left.-\frac{a}{2}\left(R^{2}-\rho^{2}\right)\right] \rho d \varphi d \rho=k \int_{0}^{2 \pi}\left[\frac{a^{3} \rho^{2}}{12}-\frac{1}{15}\left(R^{2}-\rho^{2}\right)^{\frac{5}{2}}+\right. \\
\left.+\frac{a}{8}\left(R^{2}-\rho^{2}\right)^{2}\right]\left.\right|_{0} ^{r} d \varphi=2 k \pi\left\{\frac{a^{3} r^{2}}{12}+\frac{1}{15}\left[R^{5}-\left(R^{2}-r^{2}\right)^{\frac{5}{2}}\right]+\right. \\
\left.+\frac{a}{8}\left[\left(R^{2}-r^{2}\right)^{2}-R^{4}\right]\right\}
\end{gathered}
$$

Excluding the auxiliary constants $r$ and $a$, after simplifications, we get

$$
I_{x y}=\frac{k \pi h^{3}}{60}\left(20 R^{2}-15 R h+3 h^{2}\right)
$$

2) $m=\iiint_{G} \delta d x d y d z=k \iint_{G}(z-a) d x d y d z=$

$$
=k \int_{C_{x y}} \int_{a} d x d y \int_{a}^{z_{1}}(z-a) d z=\left.\frac{k}{2} \int_{G_{x y y}} \int(z-a)^{2}\right|_{a} ^{z_{1}} d x d y=
$$

$$
=\frac{k}{2} \int_{i^{2}+y^{2} \leqslant r^{2}}\left(\sqrt{R^{2}-x^{2}-y^{2}}-a\right)^{2} d x d y=\frac{k}{2} \iint_{k \leqslant r}\left(\sqrt{R^{2}-\rho^{2}}-a\right)^{2} \rho d \varphi d \rho=
$$

$$
\begin{aligned}
& =\frac{k}{2} \int_{0}^{2 \pi} d \varphi \int_{0}^{r}\left[R^{2}-\rho^{2}-2 a\left(R^{2}-\rho^{2}\right)^{\frac{1}{2}}+a^{2}\right] \rho d \varphi d \rho= \\
= & \left.\frac{k}{2} \int_{0}^{2 \pi}\left[\frac{\rho^{2}}{2}\left(R^{2}+a^{2}\right)-\frac{\rho^{4}}{4}+\frac{2 a}{3}\left(R^{2}-\rho^{2}\right)^{\frac{3}{2}}\right]\right|_{0} ^{r}=\frac{k \pi / h^{3}(4 R-h)}{12}
\end{aligned}
$$

According to formula (3), the sought z-coordinate of the center of gravity of the given segment is

$$
z_{C}=\frac{1_{x y}}{m}=\frac{20 R^{2}-15 R h+3 h^{2}}{5(4 R-h)}
$$

For a hemisphere when $h=R$, we have

$$
I_{x y}=\frac{2 k \pi R^{5}}{15} ; \quad m=\frac{k \pi R^{4}}{4} ; \quad z_{C}=\frac{8 R}{15}
$$