Example. Find the flux of the vector field

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

through the part of the surface

$$
x^{2}+y^{2}=1
$$

cut by the planes $z=0$ and $z=2$. (The normal is outward to the closed surface formed by these surfaces).
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, $F(x, y, z)=x^{2}+y^{2}-1$ and, therefore,

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

Considering that the normals should 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

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

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

$$
\Pi=\iint_{\Sigma}\left(\vec{a}, \overrightarrow{n_{0}}\right) d \sigma=\iint_{\Sigma} \sqrt{x^{2}+y^{2}} d \sigma
$$

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

$$
\left\{\begin{array}{c}
\varrho=1 \\
0 \leq \varphi \leq 2 \pi \\
0 \leq z \leq 2
\end{array}\right.
$$

Since $d \sigma=1 \cdot d \varphi d z$ and $x^{2}+y^{2}=1$, we have

$$
\Pi=\int_{0}^{2 \pi} d \varphi \int_{0}^{2} d z
$$

## 5. We compute the double integral

$$
\Pi=\int_{0}^{2 \pi} d \varphi \int_{0}^{2} d z=4 \pi
$$

Answer. $\Pi=4 \pi$ units of flux.

Conditions of the problem. Find the flux of the vector field $\vec{a}$ through the part of the cylindrical surface $\Sigma$ cut out by the given planes.

1. $\vec{a}=x \vec{i}+y \vec{j}+2 z^{4} \vec{k}, \quad \Sigma=\left\{x^{2}+y^{2}=1\right\}, \quad z=0, z=3$.
2. $\vec{a}=x \vec{i}+y \vec{j}+x^{2} y z \vec{k}, \quad \Sigma=\left\{x^{2}+y^{2}=4\right\}, \quad z=2, z=5$.
3. $\vec{a}=(x+z y) \vec{i}-(x z-y) \vec{j}+x \vec{k}, \quad \Sigma=\left\{x^{2}+y^{2}=4\right\}, \quad z=0, z=2$.
4. $\quad \vec{a}=x \vec{i}+y \vec{j}+x z^{2} \vec{k}, \quad \Sigma=\left\{x^{2}+y^{2}=9\right\}, \quad z=0, z=3$.
5. $\quad \vec{a}=x \vec{i}+y \vec{j}-2 z \vec{k}, \quad \Sigma=\left\{x^{2}+y^{2}=4\right\}, \quad z=0, \quad z=2$.
6. $\vec{a}=x \vec{i}+y \vec{j}+y z^{3} \vec{k}, \quad \Sigma=\left\{x^{2}+y^{2}=3\right\}, \quad z=0, z=2$.
7. $\vec{a}=(x-3 y) \vec{i}+(3 x+y) \vec{j}+z^{2} \vec{k}, \quad \Sigma=\left\{x^{2}+y^{2}=5\right\}, \quad z=0, z=2$.
8. $\vec{a}=x \vec{i}+y \vec{j}+\sin ^{2} z \vec{k}, \quad \Sigma=\left\{x^{2}+y^{2}=4\right\}, \quad z=1, z=3$.
9. $\quad \vec{a}=x \vec{i}+y \vec{j}+x^{2} \vec{k}, \quad \Sigma=\left\{x^{2}+y^{2}=3\right\}, \quad z=2, z=4$.
10. $\vec{a}=3 \vec{i}+y \vec{j}+z \vec{k}, \quad \Sigma=\left\{x^{2}+y^{2}=2\right\}, \quad z=2, z=4$.

Answers. 1. $6 \pi$ units of flux. 2. $24 \pi$ units of flux. 3. $32 \pi$ units of flux. 4. $54 \pi$ units of flux. 5. $16 \pi$ units of flux. 6. $12 \pi$ units of flux. 7. $20 \pi$ units of flux. 8. $16 \pi$ units of flux. 9. $12 \pi$ units of flux. 10. $4 \pi$ units of flux.

## 14.4. Flux of a vector field through a part of a sphere

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 sphere

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

cut out by the plane $z=0 \quad(z \geq 0)$ (the normal is external to the closed surface formed by these surfaces).

Plan of Solution. By definition, the flux $\Pi$ of the vector field $\vec{a}$ through 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

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

For a sphere, obviously, the external normal at each point coincides with the radius vector of that point, i.e.,

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

2. We find the scalar product:

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

3. According to formula (1), the flux is defined 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 (sphere)

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

In these coordinates, the surface is defined by the conditions

$$
\left\{\begin{array}{l}
\varrho=r \\
0 \leq \varphi \leq 2 \pi \\
0 \leq \theta \leq \pi / 2
\end{array}\right.
$$

Since $d \sigma=r^{2} \sin \theta d \varphi d \theta$, we have

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

5. We compute the double integral and write the answer, not forgetting the units.