Relation between surfaces of an infinitesimal tetrahedron Let $d\sigma_1,d\sigma_2, d\sigma_3$ denote the areas of the faces perpendicular to the axes $x_1,x_2,x_3$ and let $d\sigma_n$ denote the area of the inclined face with unit exterior normal n. My book says that this relation holds:
$d\sigma_i = d\sigma_n \cos(\mathbf{n},x_i)= n_id\sigma_n \quad for (i=1,2,3)$
In the limit as the tetrahedron shrinks to the point M.
I don't understand how it is derived.

 A: $\let\a=\alpha \let\s=\sigma$
You rightly tagged "geometry" your question. And no limit or
infinitesimals are needed as far as the surface is plane. Consider $\s$ (finite, not infinitesimal) and $\s_1$. They are triangles sharing a base. Can you see what's the ratio of their heights?
A: $d\sigma_1$ is the orthogonal projection of $d\sigma$ onto $x_2x_3$, so that the coefficient of proportionality must be the cosine of the angle between the normals. The latter is the dot product
$$(n_n,n_y,n_z)\cdot(1,0,0).$$
A: 

I think that working on a Figure like above you could derive the results you want. And indeed, it's valid for finite quantities and not necessarily infinitesimal.

3D version

$=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!$
EDIT


In above Figure-01
\begin{equation}
\mathbf{n}\boldsymbol{=}\left(\rm n_1,n_2,n_3\right)\boldsymbol{=}\left(\cos \theta_1,\cos \theta_2,\cos \theta_3\right)
\tag{01}\label{01}    
\end{equation}
and
\begin{equation}
\sigma_{\rm n}\boldsymbol{\equiv} \left[\rm A_1A_2A_3\right]\boldsymbol{=}
\left.
\begin{cases}
\frac12 \left(\rm A_1A_2\right)\left(\rm A_3B_3\right)\\
\frac12 \left(\rm A_2A_3\right)\left(\rm A_1B_1\right)\\
\frac12 \left(\rm A_3A_1\right)\left(\rm A_2B_2\right)
\end{cases}
\right\}
\tag{02}\label{02}    
\end{equation}
so
\begin{equation}
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\left.
\begin{cases}
\sigma_{1}\boldsymbol{\equiv} \left[\rm OA_2A_3\right]\boldsymbol{=}\frac12 \left(\rm A_2A_3\right)\left(\rm OB_1\right)\stackrel{\left(\rm OB_1\right)\boldsymbol{=}\left(\rm A_1B_1\right)\cos\theta_1}{\boldsymbol{=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!}}\frac12 \left(\rm A_2A_3\right)\left(\rm A_1B_1\right)\cos\theta_1\boldsymbol{=}\rm n_1\sigma_{\rm n}\\
\sigma_{2}\boldsymbol{\equiv} \left[\rm OA_3A_1\right]\boldsymbol{=}\frac12 \left(\rm A_3A_1\right)\left(\rm OB_2\right)\stackrel{\left(\rm OB_2\right)\boldsymbol{=}\left(\rm A_2B_2\right)\cos\theta_2}{\boldsymbol{=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!}}\frac12 \left(\rm A_3A_1\right)\left(\rm A_2B_2\right)\cos\theta_2\boldsymbol{=}\rm n_2\sigma_{\rm n}\\
\sigma_{3}\boldsymbol{\equiv} \left[\rm OA_1A_2\right]\boldsymbol{=}\frac12 \left(\rm A_1A_2\right)\left(\rm OB_3\right)\stackrel{\left(\rm OB_3\right)\boldsymbol{=}\left(\rm A_3B_3\right)\cos\theta_3}{\boldsymbol{=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!}}\frac12 \left(\rm A_1A_2\right)\left(\rm A_3B_3\right)\cos\theta_3\boldsymbol{=}\rm n_3\sigma_{\rm n}
\end{cases}
\right\}
\tag{03}\label{03}    
\end{equation}
