Example. Find the flux of the vector field

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

through the part of the surface

$$
x^{2}+y^{2}+z^{2}=9
$$

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

1. The external normal at each point of the sphere $x^{2}+y^{2}+z^{2}=9$ coincides with the radius vector, i.e.,

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

2. We find the scalar product

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

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

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

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

$$
\left\{\begin{array}{c}
\varrho=3 \\
0 \leq \varphi \leq 2 \pi \\
0 \leq \theta \leq \frac{\pi}{2}
\end{array}\right.
$$

Since $d \sigma=9 \sin \theta d \varphi d \theta$, we have

$$
\Pi=\int_{0}^{2 \pi} d \varphi \int_{0}^{\pi / 2} 9 \cdot 3 \sin \theta d \theta=54 \pi
$$

Answer. $\Pi=54 \pi$ units of flux.

Conditions of the Problem. Find the flux of the vector field $\vec{a}$ through the part of the sphere cut by the plane $z=0(z \geq 0)$.

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

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

## 14.5. Calculation of Flux Using the Ostrogradsky Formula

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 closed surface $\Sigma$ (normal is external).

Plan of Solution. The flux of a vector field through a closed surface $\Sigma$ in the direction of the external normal is calculated using the Ostrogradsky formula

$$
\Pi=\iiint_{\Omega} \operatorname{div} \vec{a} d x d y d z
$$

where $\Omega$ is the region bounded by the surface $\Sigma$, and

$$
\operatorname{div} \vec{a}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}
$$

- is the divergence of the vector field $\vec{a}$.

1. Calculate the divergence $\operatorname{div} \vec{a}$.
2. Define the region $\Omega$ using inequalities.
3. Calculate the flux using formula (1) as a triple integral

$$
\Pi=\iiint_{\Omega}\left(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}\right) d x d y d z
$$

Write the answer, remembering the units.