2.19. Find the flux of the vector field $\vec{r}=x \vec{i}+y \vec{j}+z \vec{k}$:

a) through the lateral surface of the cylinder $x^{2}+y^{2} \leqslant R^{2}$, $|z| \leqslant H$ (normal outward);

b) through the lateral surface of the cone $x^{2}+y^{2} \leqslant 4 z^{2}$, $z \in[0,1]$ (normal inward);

c) through the complete surface of the cube $|x| \leqslant a,|y| \leqslant a,|z| \leqslant a$ (normal outward).
Solution.

a) We use the Gauss-Ostrogradsky formula:

$$
\iint_{S} \vec{r} \cdot \vec{n} d S=\iiint_{V} \operatorname{div} \vec{r} \cdot d V
$$

Obviously, $S=S_{6 \text{ok}} \cup S_{\text {osn }}(S_{\text {bok }}, S_{\text {osn }}$ - the lateral surface and the base surface of the cylinder) and by the additivity property of the surface integral, we get:

$$
\iint_{S} \vec{r} \cdot \vec{n} d S=\iint_{S_{\infty o x}} \vec{r} \cdot \vec{n} d S+\iint_{S_{\propto \times}} \vec{r} \cdot \vec{n} d S
$$

The integrals

$$
\begin{aligned}
& \iint_{S_{\circ x n}} \dot{r} \cdot \vec{n} d S \text { and } \iiint_{V} \operatorname{div} \vec{r} \cdot d V: \\
& \iint_{S_{\mathrm{ocH}}} \vec{r} \cdot \ddot{n} d S=\iint_{x^{2}+y^{2} \leqslant R^{2}} H d x d y-\iint_{x^{2}+y^{2} \leqslant R^{2}}(-H) d x d y=2 H \pi R^{2}
\end{aligned}
$$

(see the solution of problem 2.15),

$$
\iiint_{V} \operatorname{div} \dot{r} \cdot d V=\iiint_{V} 3 d x d y d z=3 \iiint_{x^{2}+y^{2} \leqslant R^{2}} d x d y \int_{-H}^{H} d z=3 H \pi R^{2}
$$

Thus, for the sought flux through the lateral surface of the cylinder, we obtain the value

$$
\iint_{S_{(x * k}} \vec{r} \cdot \dot{n} d S=\iiint_{V} \operatorname{div} \ddot{r} \cdot d V-\iint_{S_{(<\times k}} \ddot{r} \cdot \dot{n} d S=\pi R^{2} H
$$

b) We compute the flux of the radius vector through the base of the cone $(z=1)$:

$$
\iint_{S_{\mathrm{orH}}} \dot{r} \cdot \dot{n} d S=-\iint_{x^{2}+y^{2} \leqslant 4} d x d y=-4 \pi
$$

The sign "−" indicates that $\dot{n}$ is the inward normal. Now, similarly to the previous solution, we use the Gauss-Ostrogradsky formula and the additivity of the surface integral. In the calculations, we use the outward normal $\dot{n}$, then simply change the sign to the opposite:

$$
\begin{gathered}
\iint_{S_{\text {6ox }}} \vec{r} \cdot \vec{n} d S=\iiint_{V} \operatorname{div} \ddot{r} \cdot d V-\iint_{S_{\mathrm{och}}} \vec{r} \cdot \vec{n} d S= \\
=3 \iiint_{x^{2}+y^{2} \leqslant 4} d x d y \int_{\frac{1}{2} \sqrt{x^{2}+y^{2}}}^{1} d z-4 \pi= \\
=3 \iiint_{x^{2}+y^{2} \leqslant 4} d x d y-4 \pi-\frac{3}{2} \int_{0}^{2 \pi} \int_{0}^{2} r^{2} d r d \varphi=8 \pi-\left.\frac{3}{2} 2 \pi \cdot \frac{r^{3}}{3}\right|_{0} ^{2}=0 .
\end{gathered}
$$

c) We apply the Gauss-Ostrogradsky formula. Since $\operatorname{div} \vec{r}=3$, the flux of the radius vector through the surface of the cube will be equal to three times the volume of the cube with side $2 a$. Thus,

$$
\iint_{S} \ddot{r} \cdot \ddot{n} d S=24 a^{3}
$$

## Problems related to Stokes' formula.