Linear shape function, mass matrix, and element volume

In unsteady CFD problem, under certain time integration scheme, L2 inner product of shape functions gives what so called mass matrix, i.e. the matrix with its entries in the form $M_{ij}=\int_{\Omega}{N_i N_j}\,d{\Omega}$, where $N_i$ is shape function. When shape function is linear, there is a better way to demonstrate $M_{ij}$ is proportional to local element volume (area) than trivially plugging in and calculating or using Gaussian integral. Assume in 2D triangular element ABC whose area is $A^{ABC}$,  and $N_i(x,y), i=1,2,3$ stands for the shape function whose unit value is reached at A, B, C correspondingly.

If we interpret local mass entry $M_{13}^{ABC}=(N_1, N_3)$ over element ABC shown above as the mass of a piece of wood splinter in the shape of tetrahedron A1-A-B-C, where $N_1(x,y)$ is the thickness function on the element, while $N_3(x,y)$ is the density function, then it’s not difficult to see that $M_{13}=(N_1,N_3)=(N_2,N_3)=M_{23}$, because they are both the mass of tetrahedron with the same volume and density distribution from bottom (A1-A-B or B1-A-B) to top (C). The next step is to show that the mass of tetrahedron C1-A-B-C ($M_{33}$) is the half of the total mass of the prism A1-B1-C1-A-B-C with the same density distribution. The argument is based on linearity, which allows us to assign a single parameter for above integrals. Set this single coordinate $s$ as the line parameter from C to line A-B, i.e. $s=0$ on AB, $s=1$ at C. Then, for mass of prism and tetrahedron, we both have

mass $=\int{L(s) H(s) D(s)}\,ds$

where $L\times H$ gives the area for cross section parallel to A1-A-B-B1 at $s$ . $H$ is the thickness of ABC, while $D$ is the density distribution. We have

$L\propto (1-s),\quad D\propto s$,
and $H=1$ for prism, $H\propto s$ for tetrahedron. Then we have

$M_p\propto \int{(1-s) s}\,ds$

$M_{33}\propto \int{(1-s) s^2}\,ds$

which proves

$2 M_{33}=M_p=\frac{1}{3} A_{ABC}$

Since $M_{33}+M_{13}+M_{23}$ is nothing but the sum of the mass of three tetrahedrons, which is the mass of the prism. This sum is then $\frac{1}{3}A_{ABC}$. Using $M_{13}=M_{23}$, finally we have

$M_{11}=M_{22}=M_{33}=\frac{1}{6}A_{ABC}$

$M_{12}=M_{13}=M_{23}=\frac{1}{12}A_{ABC}$

Note: for 3D tetrahedron element, linear shape function gives similar relation, in this case, above ratio over the volume of a tetrahedron would be 1/10 and 1/20:

$M_{11}=M_{22}=M_{33}=M_{44}=\frac{1}{10}V_{ABCD}$

$M_{12}=M_{13}=M_{14}=M_{23}=M_{24}=M_{34}=\frac{1}{20}V_{ABCD}$