Example 13. Find the flux of the vector field $\mathbf{v}=(x-2 y+1) \mathbf{i}+(2 x+ y-3 z) \mathbf{j}+(3 y+z) \mathbf{k}$ through the part of the sphere $x^{2}+y^{2}+z^{2}$, located in the first octant, into the region where $x^{2}+y^{2}+z^{2}>1$.
Solution. In this case, we have

$$
\begin{gathered}
R=1, \quad \varphi_{1}=0, \quad \varphi_{2}=\frac{\pi}{2} \\
\theta_{1}=0, \quad \theta_{2}=\frac{\pi}{2}, \quad n^{n}=x 1+y J+z k, \quad\left(\mathrm{a}, n^{0}\right)=x^{2}+y^{2}+z^{2}+x
\end{gathered}
$$

We proceed on the sphere $x^{2}+y^{2}+z^{2}=1$ with coordinates $\varphi$ and $\theta$ such that

$$
x=\cos \varphi \sin \theta, \quad y=\sin \varphi \sin \theta, \quad z=\cos \theta
$$

Then we will have

$$
\left(a, n^{n}\right)=1+\cos \varphi \sin \theta
$$

Applying formula (14), we get

$=\int_{0}^{\pi / 2} d \varphi \int_{0}^{\pi / 2}(1+\cos \varphi \sin \theta) \sin \theta d \theta=$

$$
=\int_{0}^{\pi / 2} d \varphi \int_{0}^{\pi / 2} \sin \theta d \theta+\int_{0}^{\pi / 2} \cos \varphi d \varphi \int_{0}^{\pi / 2} \sin ^{2} \theta d \theta=\frac{3}{4} \pi
$$

## Problems for Independent Solution

124. Find the flux of the vector field $=x^{3} \mathbf{i}-y^{3} \mathbf{i}+z \mathbf{k}$ through the outer side of the part of the sphere $x^{2}+y^{2}+z^{2}=1$, cut out by the conical surface $z^{2}=x^{2}+y^{2}$ $\left(z \geqslant \sqrt{x^{2}+y^{2}}\right)$.
125. Find the flux of the vector field $\mathbf{a}=y z i+x z j+x y \mathbf{k}$ through the outer side of the part of the sphere $x^{2}+y^{2}+z^{2}=\boldsymbol{R}^{2}$, located in the first octant.
126. Find the flux of the radius vector $r=x i+y \boldsymbol{j}+\boldsymbol{z} \mathbf{k}$ through the outer side of the part of the sphere $x^{2}+y^{2}+z^{2}=2$, bounded by the planes $z=0, x=y$.
127. Find the flux of the vector field $\mathbf{a}=x z \mathbf{i}+y z \mathbf{j}+z^{2} \mathbf{k}$ through the outer part of the sphere $x^{2}+y^{2}+z^{2}=9$, cut off by the plane $z=0(z \geqslant 2)$.

## § 12. Flux of a Vector through a Closed Surface. Theorem of Gauss-Ostrogradsky

Theorem. If in some region G of space the vector

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

is continuous and has continuous partial derivatives $\frac{\partial P}{\partial x}, \frac{\partial Q}{\partial y}, \frac{\partial R}{\partial z}$, then the flux of the vector field a through any closed piecewise smooth surface $\Sigma$, located in the region $G$, is equal to the triple integral of $\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}$ over the region $V$ bounded by the surface $\Sigma$:

$$
\Pi=\oiint_{\Sigma}\left(\mathrm{a}, \mathrm{n}^{0}\right) d \sigma=\iiint_{V}\left(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}\right) d v
$$

(Gauss-Ostrogradsky formula).

The normal $\mathbf{n}$ to the surface $\Sigma$ is taken as the outer normal.