### 1.26. Find the center of gravity of a hemisphere

$$
0 \leqslant z \leqslant \sqrt{R^{2}-x^{2}-y^{2}},
$$

where the volume density \( p \) of the material at each point is proportional to the distance from the origin.
S o l u t i o n. The coordinates of the center of gravity of the body $\Omega$ are calculated by the formulas

$$
X_{c}=\frac{m_{y z}}{m} ; \quad Y_{c}=\frac{m_{z x}}{m} ; \quad Z_{c}=\frac{m_{x y}}{m}
$$

where $m_{x y}=\iiint_{\Omega} z \cdot p(M) d x d y d z$ (static moment of the body $\Omega$ relative to the plane $X O Y$); $m_{y z}=\iiint_{\Omega} x \cdot p(M) d x d y d z$ (static moment of the body $\Omega$ relative to the plane YOZ); $m_{z x}=\iiint_{\Omega} y \cdot p(M) d x d y d z$ (static moment of the body $\Omega$ relative to the plane $\mathrm{XOZ}$); $m=\iiint_{\Omega} p(M) d x d y d z$ (mass of the body $\Omega$).

Since the shape of the body $\Omega$ and the distribution of mass are symmetric relative to the $z$-axis, the center of gravity will also lie on this axis, i.e., $X_{c}=Y_{c}=0$.

For the calculations in this problem, it is convenient to switch to spherical coordinates $x=r \sin \theta \cos \varphi ; y=r \sin \theta \sin \varphi$; $z=r \cos \theta$. In this case, the region $\Omega$ corresponds to the region

$$
D=\left\{(r, \varphi, \theta): \quad r \in[0, R] ; \quad \varphi \in[0,2 \pi) ; \quad \theta \in\left[0, \frac{\pi}{2}\right)\right\} .
$$

The Jacobian of this transformation $J$ is

$$
J=\frac{D(x, y, z)}{D(r, \theta, \varphi)}=r^{2} \sin \theta
$$

Let's proceed to the corresponding calculations (given $p=k r, k=$ const $)$ :

$$
\begin{gathered}
m_{x y}=\iiint_{\Omega} z \cdot p(M) d x d y d z=k \iiint_{D} r^{4} \cdot \sin \theta \cos \theta d r d \theta d \varphi= \\
=k \int_{0}^{2 \pi} d \varphi \int_{0}^{\frac{\pi}{2}} \sin \theta \cos \theta d \theta \int_{0}^{R} r^{4} d r=\frac{k \pi R^{5}}{5} \\
m=\iiint_{\Omega} p(M) d x d y d z=k \iiint_{D} r^{3} \sin \theta d r d \theta d \varphi= \\
=k \int_{0}^{2 \pi} d \varphi \int_{0}^{\frac{\pi}{2}} \sin \theta d \theta \int_{0}^{R} r^{3} d r=\frac{k \pi R^{4}}{2} \\
Z_{c}=\frac{m_{x y}}{m}=\frac{2}{5} R
\end{gathered}
$$