Example 3. Find the flux of the vector field

$$
\mathbf{a}=\frac{\mathbf{r}}{|\mathbf{r}|^{3}}
$$

through a sphere of radius $\boldsymbol{R}$ centered at the origin.
Solution. Since the normal p to the sphere is collinear with the radius-vector $\mathbf{r}$, we can take $\boldsymbol{\text { m }} \mathbf{n}^{\text {n }}=\mathbf{r}^{n}=\frac{\mathbf{r}}{|\mathbf{r}|}$. Therefore,

$$
\left(\mathbf{m}, \mathbf{n}^{\mathbf{n}}\right)=\left(\frac{\mathbf{r}}{|\boldsymbol{r}|^{1}}, \frac{\mathbf{r}}{|\mathbf{r}|}\right)=\frac{1}{|\mathbf{r}|^{\left.\right|^{2}}}(\mathbf{r}, \mathbf{r})=\frac{|\mathbf{r}|^{2}}{|\mathbf{r}|^{4}}=\frac{1}{|\mathbf{r}|^{2}}
$$

On the sphere $S$ we have $|\mathbf{r}|=\boldsymbol{R}$, therefore ( $\left.\mathbf{a}, \mathbf{n}^{\prime \prime}\right)=\frac{1}{R^{R}}$.

The sought flux IT will be equal to

$$
\mathrm{I}=\oiint_{S}\left(\mathrm{a}, \mathrm{n}^{n}\right) d S^{\prime}=\frac{1}{R^{2}} \oiint_{S} d S=4 \pi
$$

since the area of the sphere $S$ is $\phi d S=4 \pi R^{2}$.

## Problems for independent solving

104. Calculate the flux of the vector field $\mathbf{n}=3 \mathbf{j}$ through the plane, having the form of a triangle with vertices at points $M_{1}(1,2,0), M_{2}(0,2,0), M_{3}(0,2,2)$ and the side where the origin of coordinates is located.
105. Find the flux of the vector field $\mathbf{a}=\alpha \mathbf{i}+\beta \mathbf{j}+\gamma \mathbf{k}$, where $\alpha, \beta, \gamma-$ are constants, through a circle perpendicular to the $O z$ axis and having a radius of $R$, in the positive direction of the $Oz$ axis.
106. Find the flux of the vector field $\mathbf{a}=\mathbf{r}$ through the side surface of a circular cone, the vertex of which is at the origin of coordinates, the radius of the base is $R$ and the height is $h$ (the axis of the cone is along the $Oz$ axis).

![](https://cdn.mathpix.com/cropped/2024_05_22_267d965536514099194eg-22.jpg?height=26&width=745&top_left_y=1112&top_left_x=118)
at the origin of coordinates

## $2^{\circ}$. Methods for calculating the flux of a vector.

## 1. Method of projecting onto one of the coordinate planes

Let the unbounded surface $S$ be projected one-to-one onto the plane $x O_{y}$ in the region $D_{x y}$. In this case, the surface $S$ can be described by the equation $z=f(x, y)$, and since the element of area $d S$ of this surface is equal to

$$
d S=\frac{d x d y}{|\cos \gamma|}
$$

the calculation of the flux $\Pi$ through the chosen side of the surface $S$ is reduced to the calculation of a double integral by the formula

$$
\Pi=\iint_{S}\left(\mathrm{a}, \mathrm{n}^{\prime \prime}\right) d S=\left.\iint_{D_{s y}} \frac{\left(\mathbf{a}, \mathrm{n}^{0}\right)}{|\cos \gamma|}\right|_{x-f(x, y)} d x d y
$$

Here the unit vector $\mathbf{n}^{0}$ of the normal to the chosen side of the surface $S$ is found by the formula

$$
n^{0}= \pm \frac{\operatorname{grad}[z-f(x, y)]}{\mid \operatorname{grad}[z-f(x, y)||}= \pm \frac{-\left(\frac{\partial f}{\partial x}\right) i-\left(\frac{\partial f}{\partial y}\right) j+k}{\sqrt{\left(\frac{\partial f}{\partial x}\right)^{2}+\left(\frac{\partial f}{\partial y}\right)^{2}+1}}
$$

and $\cos \gamma$ is equal to the coefficient of the unit vector $\mathbf{k}$ in formula (2):

$$
\cos \gamma= \pm \frac{1}{\sqrt{\left(\frac{\partial f}{\partial x}\right)^{2}+\left(\frac{\partial f}{\partial y}\right)^{2}+1}}
$$

If the angle $\gamma$ between the $O z$ axis and the normal $\mathbf{n}^{0}$ is acute, then in formulas (2) and (3) the sign $\propto+"$ is taken, and if the angle $\gamma$ is obtuse, then the sign $-»$ is taken. The symbol

$$
\left.\frac{\left(a, n^{0}\right)}{|\cos \gamma|}\right|_{z=f(x, y)}
$$

means that in the integrand function, $z$ is replaced by $f(x, y)$.

If it is more convenient to project the surface $S$ onto the coordinate planes $y O_{z}$ or $x O_{z}$, then for calculating the flux $\Pi$ the formulas are used respectively:

$$
\Pi=\left.\iint_{D_{y x}} \frac{\left(\mathrm{a}, \mathrm{n}^{\prime}\right)}{|\cos \gamma|}\right|_{z=\varphi(y, z)} d y d z
$$

or

$$
\Pi=\left.\iint_{D_{x=}} \frac{\left(\mathbf{a}, \mathbf{n}^{0}\right)}{|\cos \gamma|}\right|_{x=\psi\left(x_{,} x\right)} d x d z
$$

Formula (4) is applied in the case when the surface $S$ is projected one-to-one onto the region $D_{y z}$ of the plane $y O z$, and hence, it can be described by the equation $x=\varphi(y, z) ; \cos \alpha$ is found as the coefficient of the unit vector $\mathbf{i}$ in the formula

$$
\mathbf{n}^{0}= \pm \frac{\operatorname{grad}[x-\varphi(y, z)]}{|\operatorname{grad}[x-\varphi(y, z)]|}= \pm \frac{1-\left(\frac{\partial \varphi}{\partial y}\right) \mathbf{j}-\left(\frac{\partial \varphi}{\partial z}\right) \mathbf{k}}{\sqrt{1+\left(\frac{\partial \varphi}{\partial y}\right)^{2}+\left(\frac{\partial \varphi}{\partial z}\right)^{2}}}
$$

i.e.,

$$
\cos \alpha= \pm \frac{1}{\sqrt{1+\left(\frac{\partial \varphi}{\partial y}\right)^{2}+\left(\frac{\partial \varphi}{\partial z}\right)^{2}}}
$$

The sign “+” is taken if the angle $\alpha$ between the $O x$ axis and the normal $\mathbf{n}^{0}$ is acute, and if $\alpha$ is an obtuse angle, then the sign $-»$ is taken.

Formula (5) is applied when the surface $S$ is projected one-to-one onto the plane $\pm O z$; in this case, $S$ can be described by the equation $y=\psi(x, z)$ and then

$$
\mathbf{n}^{0}= \pm \frac{\operatorname{grad} \mid y-\psi(x, z)]}{|\operatorname{grad}[y-\psi(x, z)]|}= \pm \frac{-\left(\frac{\partial \psi}{\partial x}\right) \mathbf{i}+\mathbf{j}-\left(\frac{\partial \psi}{\partial z}\right) \mathbf{k}}{\sqrt{1+\left(\frac{\partial \psi}{\partial x}\right)^{2}+\left(\frac{\partial \psi}{\partial z}\right)^{2}}}
$$

$\cos \beta$ is the coefficient of the unit vector $\mathbf{j}$ in the last formula, i.e.,

$$
\cos \beta= \pm \frac{1}{\sqrt{1+\left(\frac{\partial \psi}{\partial x}\right)^{2}+\left(\frac{\partial \psi}{\partial z}\right)^{2}}}
$$

If the angle $\beta$ between the $O y$ axis and the normal $\mathbf{n}^{0}$ is acute, then the sign $«+\infty$ is taken, and if the angle $\rho$ is obtuse, then the sign * - is taken.

Remark. In the case where the surface $S$ is given implicitly by the equation $\Phi(x, y, z)$, the unit normal vector

$$
\mathrm{n}^{0}=1 \cos \alpha+\mathbf{j} \cos \beta+\mathbf{k} \cos \gamma
$$

is found by the formula

$$
\mathbf{n}^{n}= \pm \frac{\operatorname{grad} \Phi(x, y, z)}{\mid g \text { grad } \Phi(x, y, z) \mid}= \pm \frac{\left(\frac{\partial \Phi}{\partial x}\right) \mathbf{i}+\left(\frac{\partial \Phi}{\partial y}\right) \mathbf{j}+\left(\frac{\partial \Phi}{\partial z}\right) \mathbf{k}}{\sqrt{\left(\frac{\partial \Phi}{\partial x}\right)^{2}+\left(\frac{\