Example. Find the flux of the vector field

$$
\vec{a}=-x \vec{i}+2 y \vec{j}+z \vec{k}
$$

through the part of the plane

$$
x+2 y+3 z=1
$$

located in the first octant (the normal forms an acute angle with the $O Z$ axis).
Solution.

1. The field of unit normals to the surface defined by the equation $F(x, y, z)=0$ is given by the formula

$$
\vec{n}_{0}= \pm \frac{\operatorname{grad} F}{|\operatorname{grad} F|}
$$

In this case, $F(x, y, z)=x+2 y+3 z-1$ and, therefore,

$$
\vec{n}_{0}= \pm \frac{\{1,2,3\}}{\sqrt{14}}
$$

Considering that the normals must form an acute angle with the $O Z$ axis, i.e., $\cos \gamma= \pm 3 / \sqrt{14}>0$, we choose the plus sign in this formula. We have

$$
\overrightarrow{n_{0}}=\left\{\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}\right\}
$$

2. We find the scalar product:

$$
\left(\vec{a}, \overrightarrow{n_{0}}\right)=\frac{1}{\sqrt{14}}(-x+4 y+3 z)
$$

3. According to formula (1), the flux is determined by the surface integral:

$$
\Pi=\iint_{\Sigma} \frac{1}{\sqrt{14}}(-x+4 y+3 z) d \sigma
$$

4. We transition from the surface integral to a double integral by projecting $\Sigma$ onto the $X O Y$ plane:

$$
\begin{aligned}
\Pi=\iint_{\Sigma} \frac{1}{\sqrt{14}}(-x+4 y & +3 z) d \sigma= \\
& =\left.\iint_{D} \frac{1}{\sqrt{14}}(-x+4 y+3 z)\right|_{z=(1-x-2 y) / 3} \frac{d x d y}{|\cos \gamma|}
\end{aligned}
$$

where $D$ is the projection of $\Sigma$ onto the $X O Y$ plane and $\cos \gamma=3 / \sqrt{14}$.

The surface $\Sigma$ is defined by the conditions

$$
\Sigma=\left\{(x, y, z): \begin{array}{l}
x+2 y+3 z=1 \\
x \geq 0, y \geq 0, z \geq 0
\end{array}\right\}
$$

We find the projection $D$ of $\Sigma$ onto the $X O Y$ plane by eliminating $z$ from the conditions defining $\Sigma$:

$$
\begin{gathered}
\Sigma=\left\{(x, y, z): \begin{array}{l}
z=(1-x-2 y) / 3 \\
x \geq 0, y \geq 0, z \geq 0
\end{array}\right\} \Longrightarrow \\
\Longrightarrow D=\left\{(x, y): \begin{array}{l}
x \geq 0, y \geq 0 \\
0 \leq(1-x-2 y) / 3
\end{array}\right\}
\end{gathered}
$$

From this, we have

$$
D=\left\{(x, y): \begin{array}{l}
0 \leq y \leq 1 / 2 \\
0 \leq x \leq 1-2 y
\end{array}\right\}
$$

5. We compute the double integral, reducing it to a repeated integral:

$$
\begin{aligned}
\Pi & =\left.\iint_{D}(-x+4 y+3 z)\right|_{z=(1-x-2 y) / 3} \frac{d x d y}{|\cos \gamma|}= \\
& =\frac{1}{3} \iint_{D}(1-2 x+2 y) d x d y=\frac{1}{3} \int_{0}^{1 / 2} d y \int_{0}^{1-2 y}(1-2 x+2 y) d x=\frac{1}{18} .
\end{aligned}
$$

Answer. $\Pi=1 / 18$ units of flux.

Conditions of the problem. Find the flux of the vector field $\vec{a}$ through the part of the plane located in the first octant (the normal forms an acute angle with the $O Z$ axis).

1. $\quad \vec{a}=x \vec{i}+y \vec{j}, \quad x+y+z=1$.
2. $\quad \vec{a}=2 x \vec{i}+z \vec{k}, \quad x+y+z=1$.
3. $\quad \vec{a}=y \vec{j}+z \vec{k}, \quad 2 x+y+z=1$.
4. $\quad \vec{a}=x \vec{i}+y \vec{j}, \quad x+y+2 z=1$.
5. $\quad \vec{a}=y \vec{j}+z \vec{k}, \quad 2 x+2 y+z=1$.
6. $\quad \vec{a}=x \vec{i}+y \vec{j}, \quad 2 x+y+2 z=1$.
7. $\vec{a}=x \vec{i}+2 y \vec{j}$,
$x+y+z=1$.
8. $\vec{a}=y \vec{j}+z \vec{k}$, $2 x+2 y+z=1$.
9. $\vec{a}=x \vec{i}+2 y \vec{j}+z \vec{k}$, $x / 2+y+z / 3=1$.
10. $\vec{a}=x \vec{i}+y \vec{j}+2 z \vec{k}$,
$x / 3+y / 2+z=1$.

Answers. 1. $1 / 3$ units of flux. 2. $1 / 2$ units of flux. 3. $1 / 6$ units of flux. 4. $1 / 6$ units of flux. 5. $1 / 8$ units of flux. 6. 1/8 units of flux. 7. 1/2 units of flux. 8. $1 / 8$ units of flux. 9. 4 units of flux. 10. 4 units of flux.

## 14.3. Flux of a vector field through a part of a cylinder

Problem Statement. Find the flux of the vector field

$$
\vec{a}=P(x, y, z) \vec{i}+Q(x, y, z) \vec{j}+R(x, y, z) \vec{k}
$$

through the part of the surface

$$
x^{2}+y^{2}=r^{2}
$$

cut by the planes $z=0$ and $z=h$ (the normal is external to the closed surface formed by these surfaces).

![](https://cdn.mathpix.com/cropped/2024_05_22_5878ce417e2ca061000bg-349.jpg?height=43&width=1056&top_left_y=1191&top_left_x=121)
The surface $\Sigma$ with the field of unit normals $\vec{n}_{0}$ is defined by the formula

$$
\Pi=\iint_{\Sigma}\left(\vec{a}, \overrightarrow{n_{0}}\right) d \sigma
$$

1. The field of unit normals to the surface defined by the equation $F(x, y, z)=0$ is given by the formula

$$
\vec{n}_{0}= \pm \frac{\operatorname{grad} F}{|\operatorname{grad} F|}
$$

In this case, $F(x, y, z)=x^{2}+y^{2}-r^{2}$ and, therefore,

$$
\vec{n}_{0}= \pm \frac{\{x, y, 0\}}{\sqrt{x^{2}+y^{2}}}
$$

Considering that the normals must form an acute angle with the $O X$ axis when $x>0$, i.e., $\cos \alpha>0$ when $x>0$, and an obtuse angle with the $O X$ axis when $x<0$, i.e., $\cos \alpha<0$ when $x<0$, we choose the plus sign in this formula.

2. We find the scalar product

$$
\left(\vec{a}, \overrightarrow{n_{0}}\right)=\frac{P(x, y, z) x+Q(x, y, z) y}{\sqrt{x^{2}+y^{2}}}=f(x, y, z) .
$$

3. According to formula (1), the flux is determined by the surface integral:

$$
\Pi=\iint_{\Sigma}\left(\vec{a}, \overrightarrow{n_{0}}\right) d \sigma=\iint_{\Sigma} f(x, y, z) d \sigma
$$

4. We introduce curvilinear coordinates on the given surface (cylinder)

$$
\left\{\begin{array}{l}
x=\varrho \cos \varphi \\
y=\varrho \sin \varphi \\
z=z
\end{array}\right.
$$

In these coordinates, the surface is defined by the conditions

$$
\left\{\begin{array}{c}
\varrho=r \\
0 \leq \varphi \leq 2 \pi \\
0 \leq z \leq h
\end{array}\right.
$$

Since $d \sigma=r d \varphi d z$, we have

$$
\Pi=r \int_{0}^{2 \pi} d \varphi \int_{0}^{h} f(r \cos \varphi, r \sin \varphi, z) d z
$$

5. We compute the repeated integral and write the answer, not forgetting the dimension.