Example 5. Calculate the flux P of the vector field $\vec{F}=y z \vec{i}+$ $+x z \vec{j}+x y \vec{k}$ through the portion of the plane $x+y+z=1$, located in the first octant, along the normal vector to this plane.
Solution. The flux of the vector field $\vec{F}$ is calculated using the formula $\Pi=\int_{\sigma} \int(\vec{F} \cdot \vec{n}) d \sigma$. For the segment of the plane $z=1-x-y$, we have $\cos \gamma=\frac{1}{\sqrt{3}}$ (the normal vector of the plane $x+y+z=1$ has coordinates $\vec{N}=\{1,1,1\},|\vec{N}|=\sqrt{3}$, then the unit normal vector: $\left.\vec{n}=\frac{1}{\sqrt{3}} \vec{N}=\frac{1}{\sqrt{3}}\{1,1,1\}\right)$, and $d \sigma=\sqrt{3} d x d y$.

Thus,

$$
\vec{F} \cdot \vec{n}=\frac{1}{\sqrt{3}} y z+\frac{1}{\sqrt{3}} x z+\frac{1}{\sqrt{3}} x y
$$

Therefore,

$$
\begin{aligned}
& \Pi=\iint_{\sigma} \frac{1}{\sqrt{3}}(y z+x z+x y) d \sigma= \\
& \frac{1}{\sqrt{3}} \iint_{D_{x y}}(y(1-x-y)+x(1-x-y)+x y) \sqrt{3} d x d y= \\
&=\int_{0}^{1} d x \int_{0}^{1-x}\left(x+y-x y-x^{2}-y^{2}\right) d y=\frac{1}{8}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_446938f1e4808e7d6e07g-074.jpg?height=72&width=267&top_left_y=627&top_left_x=131)