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.


$\let\a=\alpha \let\s=\sigma$
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?
