Example 1. Calculate the integral $I=\iint_{\sigma} \frac{d \sigma}{(1+x+z)^{2}}$, where $\sigma-$ is the part of the plane $x+y+z=1$, located in the first octant.
Solution. The surface $\sigma$ can be written explicitly: $z=1-x-y$. From this, $z_{x}^{\prime}=-1, z_{y}^{\prime}=-1, d \sigma=\sqrt{1+{z_{x}^{\prime}}^{2}+{z_{y}^{\prime}}^{2}} d x d y=\sqrt{3} d x d y$. The projection of the surface $\sigma$ onto the plane $O x y$ is a triangle $0 \leqslant x \leqslant 1,0 \leqslant y \leqslant 1-x$. Thus, the surface integral reduces to a double integral:

$$
\begin{aligned}
& \iint_{\sigma} \frac{d \sigma}{(1+x+z)^{2}}=\int_{D} \frac{\sqrt{3} d x d y}{(1+x+1-x-y)^{2}}=\sqrt{3} \int_{0}^{1} d x \int_{0}^{1-x} \frac{d y}{(2-y)^{2}}= \\
& \quad=\left.\sqrt{3} \int_{0}^{\mathrm{I}} d x \cdot \frac{1}{2-y}\right|_{0} ^{1-x}=\left.\sqrt{3}\left(\ln (1+x)-\frac{x}{2}\right)\right|_{0} ^{1}=\sqrt{3}\left(\ln 2-\frac{1}{2}\right) . \\
& \text { Answer. } \quad I=\frac{\sqrt{3}}{2}(2 \ln 2-1) .
\end{aligned}
$$