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.


$d\sigma_1$ 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
$$(n_n,n_y,n_z)\cdot(1,0,0).$$
