1.17. Find the centers of gravity of homogeneous bodies:

a) hemisphere $x^{2}+y^{2}+z^{2} \leqslant R^{2}, z \geqslant 0$

b) tetrahedron bounded by the planes $x+2 y+z=1$, $x=y=z=0$.
Solution.

a) For a homogeneous body $\Omega=\left\{(x, y, z):(x, y) \in \Omega_{\text {hom }}\right.$; $0 \leqslant z \leqslant f(x, y)\}$, the coordinates of the center of gravity are calculated using the formulas derived from (1.85) when $P(x, y, z)=1:$

$$
\begin{aligned}
& \iint x f(x, y) d x d y \\
& x_{c}=\frac{\Omega_{X O Y}}{V_{\Omega}} ; \\
& \iint y f(x, y) d x d y \\
& y_{c}=\frac{\Omega_{\text {hom }}}{V_{\Omega}} ; \\
& \iint f^{2}(x, y) d x d y \\
& z_{C}=\frac{\Omega_{\mathrm{XOY}}}{2 V_{\Omega}},
\end{aligned}
$$

where $V_{\Omega}=\iint_{\Omega_{\text {hom }}} f(x, y) d x d y$ is the volume of the homogeneous body.

By symmetry considerations for a hemisphere, $x_{C}=y_{C}=0$. Therefore, it remains to calculate the coordinate $z_{c}$. First, as follows from symmetry,

$$
V_{\Omega}=\frac{2}{3} \pi R^{3}
$$

Therefore,

$$
\begin{gathered}
z_{C}=\frac{3}{2 \pi R^{3}} \iint_{\Omega \times O}\left(R^{2}-x^{2}-y^{2}\right) d x d y=\frac{3}{2 \pi R^{3}} \int_{0}^{2 \pi} d \varphi \int_{0}^{R}\left(R^{2}-\rho^{2}\right) \rho d \rho= \\
=\left.\frac{3}{R^{3}} \int_{0}^{b} y^{2}\left(R^{2} \frac{\rho^{2}}{2}-\frac{\rho^{4}}{4}\right)\right|_{0} ^{R}=\frac{3}{4} R
\end{gathered}
$$

b) In the case of the tetrahedron specified in the problem, $\Omega_{\text {hom }}=\{(x, y)$ : $x \geqslant 0, y \geqslant 0,0 \leqslant x+2 y \leqslant 1\}$. Now we can perform the calculations for the coordinates of the center of gravity of the homogeneous tetrahedron:

$$
\begin{aligned}
& V_{\Omega}=\iint_{\Omega_{x O Y}}(1-x-2 y) d x d y=\int_{0}^{\frac{1}{2}} d y \int_{0}^{1-2 y}(1-x-2 y) d x= \\
& =\left.\int_{0}^{1}\left(x-\frac{x^{2}}{2}-2 y x\right)\right|_{0} ^{1-2 y} d y=\int_{0}^{\frac{1}{2}}\left(1-2 y-\frac{(1-2 y)^{2}}{2}-2 y(1-2 y)\right) d y= \\
& =\int_{0}^{1}\left(\frac{1}{2}-2 y+2 y^{2}\right) d y=\left.\left(\frac{1}{2} y-y^{2}+\frac{2}{3} y^{3}\right)\right|_{0} ^{\frac{1}{2}}=\frac{1}{4}-\frac{1}{4}+\frac{2}{3 \cdot 8}=\frac{1}{12} \\
& x_{c}=12 \iint_{\Omega_{x O Y}} x(1-x-2 y) d x d y=12 \int_{0}^{\frac{1}{2}} d y \int_{0}^{1-2 y}\left(x-x^{2}-2 y x\right) d x= \\
& =\left.12 \int_{0}^{1}\left(\frac{x^{2}}{2}-\frac{x^{3}}{3}-y x^{2}\right)\right|_{0} ^{1-2 y} d y=12 \int_{0}^{\frac{1}{2}}\left(\frac{(1-2 y)^{2}}{2}-\frac{(1-2 y)^{3}}{3}-y(1-2 y)^{2}\right) d y= \\
& =\left.12\left(-\frac{(1-2 y)^{3}}{12}+\frac{(1-2 y)^{4}}{24}-\frac{y^{2}}{2}+\frac{4}{3} y^{3}-y^{4}\right)\right|_{0} ^{\frac{1}{2}}=12\left(\frac{1}{12}-\frac{1}{24}-\frac{1}{8}+1-\frac{1}{16}\right)=\frac{1}{4} \\
& y_{c}=12 \iint_{\Omega_{X O Y}} y(1-x-2 y) d x d y=12 \int_{0}^{\frac{1}{2}} d y \int_{0}^{1-2 y} y(1-x-2 y) d x= \\
& =\left.12 \int_{0}^{\frac{1}{2}}\left(y x-\frac{y x^{2}}{2}-2 y^{2} x\right)\right|_{0} ^{1-2 y} d y=12 \int_{0}^{\frac{1}{2}}\left(y(1-2 y)-\frac{1}{2} y(1-2 y)^{2}-2 y^{2}(1-2 y)\right) d y= \\
& =12 \int_{0}^{\frac{1}{2}}\left(\frac{y^{2}}{2}-2 y^{2}+2 y^{3}\right) d y=\left.12\left(\frac{y^{2}}{4}-\frac{2}{3} y^{3}+\frac{y^{4}}{2}\right)\right|_{0} ^{\frac{1}{2}}=12\left(\frac{1}{16}-\frac{1}{12}+\frac{1}{32}\right)=\frac{1}{8} \\
& z_{c}=6 \iint_{\Omega_{X O Y}}(1-x-2 y)^{2} d x d y=6 \int_{0}^{\frac{1}{2}} d y \int_{0}^{1-2 y}(1-x-2 y)^{2} d x= \\
& =\left.6 \int_{0}^{\frac{1}{2}} \frac{-(1-x-2 y)^{3}}{3}\right|_{0} ^{1-2 y} d y=6 \int_{0}^{\frac{1}{2}}(1-2 y)^{3} d y=-\left.\frac{3}{4}(1-2 y)^{4}\right|_{0} ^{\frac{1}{2}}=\frac{3}{4}
\end{aligned}
$$