Example. Find the flux of the vector field

$$
\vec{a}=\left(y^{2}+z^{2}\right) \vec{i}+\left(x y+y^{2}\right) \vec{j}+(x z+z) \vec{k}
$$

through the closed surface $\Sigma$, which is the complete surface of the cylinder

$$
x^{2}+y^{2}=1, \quad z=0, \quad z=1
$$

(normal outward).
## Solution.

1. We compute the divergence of the vector field:

$$
\operatorname{div} \vec{a}=\frac{\partial\left(y^{2}+z^{2}\right)}{\partial x}+\frac{\partial\left(x y+y^{2}\right)}{\partial y}+\frac{\partial(x z+z)}{\partial z}=2 x+2 y+1
$$

2. We define the region $\Omega$ using inequalities.

The surface $\Sigma$, which bounds the region $\Omega$, consists of three surfaces and can be written as

$$
\left\{\begin{array}{c}
x^{2}+y^{2}=1 \\
0 \leq z \leq 1
\end{array}\right\} \cup\left\{\begin{array}{c}
z=0 \\
x^{2}+y^{2} \leq 1
\end{array}\right\} \bigcup\left\{\begin{array}{c}
z=1 \\
x^{2}+y^{2} \leq 1
\end{array}\right\}
$$

From these conditions, we derive the system of inequalities that define the region $\Omega$:

$$
\Omega=\left\{(x, y, z): \begin{array}{l}
x^{2}+y^{2} \leq 1 \\
0 \leq z \leq 1
\end{array}\right\}
$$

The shape of the region $\Omega$ suggests it is convenient to switch to cylindrical coordinates. We have

$$
\Omega=\left\{\begin{array}{ll}
0 \leq \varphi \leq 2 \pi \\
(\varrho, \varphi, z): & 0 \leq \varrho \leq 1 \\
& 0 \leq z \leq 1
\end{array}\right\}
$$

3. We compute the flux using formula (1) as a triple integral:

$$
\Pi=\iiint_{\Omega}(2 x+2 y+1) d x d y d z
$$

Switching to cylindrical coordinates, we get

$$
\Pi=\int_{0}^{1} d z \int_{0}^{2 \pi} d \varphi \int_{0}^{1} \varrho(2 \varrho \cos \varphi+2 \varrho \sin \varphi+1) d \varrho=2 \pi
$$

Answer. $\Pi=2 \pi$ units of flux.

### Problem Conditions. Find the flux of the vector field $\vec{a}$ through the closed surface formed by the given surfaces (normal is external).

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

Answers. 1. $2 \pi / 5$ units of flux. 2. $\pi$ units of flux. 3. $\pi$ units of flux. 4. $64 \pi$ units of flux. 5. $-\pi / 2$ units of flux. 6. $1 / 12$ units of flux. 7. $\pi$ units of flux. 8. $3 \pi / 8$ units of flux. 9. $-13 \pi / 12$ units of flux. 10. $2 \pi$ units of flux.

## 14.6. Work of a Force

### Problem Statement. Find the work of the force

$$
\vec{F}=P(x, y) \vec{i}+Q(x, y) \vec{j}
$$

when moving along the curve $L$ from point $M\left(x_{1}, y_{1}\right)$ to point $N\left(x_{2}, y_{2}\right)$.

### Plan of Solution.

1. The work $A$ of the force field is equal to the line integral of the second kind along the curve $L$:

$$
A=\int_{L}(\vec{F}, d \vec{r})=\int_{L} P(x, y) d x+Q(x, y) d y
$$

2. We compute the line integral.

We write the answer, not forgetting the units.