### 2.14. Find the centers of gravity:

a) of the hemisphere $z=\sqrt{R^{2}-x^{2}-y^{2}}$, with surface density $p=k\left(x^{2}+y^{2}\right), k=$ const;

b) of the cone $a z^{2}=b^{2}\left(x^{2}+y^{2}\right), 0 \leqslant z \leqslant b, p=p_{0}=$ const.
Solution.

a) From symmetry considerations, it is clear that $x_{c}=y_{c}=0$. We will determine $z_{c}$ using a surface integral of the first kind:

$$
z_{c}=\frac{\iint_{\sigma} z \cdot p d S}{\iint_{\sigma} p d S}
$$

The area element of the hemisphere is

$$
d S=\sqrt{1+z_{x}^{2}+z_{y}^{2}} d x d y=\frac{R d x d y}{\sqrt{R^{2}-x^{2}-y^{2}}}
$$

## Calculate the mass of the hemisphere:

$$
\iint_{\sigma} p d S=k R \iint_{x^{2}+y^{2} \leqslant R^{2}} \frac{\left(x^{2}+y^{2}\right) d x d y}{\sqrt{R^{2}-x^{2}-y^{2}}}
$$

We introduce polar coordinates $x=r \cos \varphi, y=r \sin \varphi$ with the range of variables $0 \leqslant z \leqslant R, \varphi \in[0,2 \pi]$ :

$$
\begin{gathered}
k R \iiint_{\left.x^{2}\right\lrcorner y^{2} \leqslant R^{2}} \frac{\left(x^{2}+y^{2}\right) d x d y}{\sqrt{R^{2}-x^{2}-y^{2}}}=k R \int_{0}^{2 \pi} d \varphi \int_{0}^{R} \frac{r^{3} d r}{\sqrt{R^{2}-r^{2}}}= \\
=\pi k R \int_{0}^{R^{2}} \frac{d t}{\sqrt{R^{2}-t}}=-\left.2 \pi k R \sqrt{R^{2}-t} \cdot t\right|_{0} ^{R^{2}}+2 \pi k R \int_{0}^{R} \sqrt{R^{2}-t} d t= \\
=-\left.\frac{4}{3} \pi k R\left(R^{2}-t\right)^{\frac{3}{2}}\right|_{0} ^{R^{2}}=\frac{4}{3} \pi k R^{4}
\end{gathered}
$$

We find the $z_{c}$ coordinate of the center of mass of the hemisphere:

$$
\begin{gathered}
z_{c}=\frac{3}{4 \pi k R^{4}} \iint_{\sigma} z \cdot \rho d S=\frac{3}{4 \pi R^{3}} \iint_{x^{2}+y^{2} \leqslant R^{2}}\left(x^{2}+y^{2}\right) d x d y= \\
=\frac{3}{4 \pi R^{3}} \int_{0}^{2 \pi} d \varphi \int_{0}^{R} r^{3} d r=\frac{3}{8} R .
\end{gathered}
$$

b) In the case of the cone, just like for the hemisphere, $x_{c}=$ $=y_{c}=0$.

First, we determine the mass of the cone using a surface integral:

$$
M=\iint_{\sigma} p d S=\frac{b}{a} p_{0} \iiint_{x^{2}+y^{2} \leqslant a^{2}} \sqrt{2} d x d y=\sqrt{2} p_{0} \pi a^{2} \cdot \frac{b}{a}=\sqrt{2} p_{0} \pi a b .
$$

(The area element of the cone is calculated in problem 2.12.b.)

We calculate the $z_{c}$ coordinate of the center of mass of the homogeneous cone $a z^{2}=b^{2}\left(x^{2}+y^{2}\right) ; z \in[0, b]:$

$$
\begin{gathered}
M \cdot z_{c}=\iint_{\sigma} z \cdot p d S=p_{0} \quad \iint_{x^{2}+y^{2} \leqslant a^{2}} \frac{b}{a} \sqrt{x^{2}+y^{2}} \cdot \sqrt{2} \cdot \frac{b}{a} d x d y= \\
=\sqrt{2} \frac{b^{2}}{a^{2}} p_{0} \int_{0}^{2 \pi} d \varphi \int_{0}^{a} r^{2} d r=\frac{2 \sqrt{2} \pi p_{0} b^{2} a}{3}
\end{gathered}
$$

Therefore,

$$
z_{c}=\frac{2}{3} b
$$

Problems on calculating surface integrals of the second kind.