On moment of inertia calculation of a triangle I have seen multiple videos on YouTube which start off the process of calculating the moment of inertia of a triangle about it's base by writing it equal to  $\int(y^2\ dA )$ where
$y$ is the distance of the element from the axis
$dA$ is the area of the infinitesimal element 
I understand the entire approach after this. The only problem I have with this is that the way to compute moment of inertia is $\int(\pmb{dm} \cdot r^2)$. So shouldn't the above equation have a term for mass per area i.e. areal density?
 A: The videos and the formula you mention are discussing the second moment of area, or the area moment of inertia. This quantity is used extensively in engineering fields.
In physics, the term moment of inertia refers to the second moment of mass. So that's where your confusion lies.
As pointed out by @Tomek, the difference between the two uses could merely be a constant if the structure being analyzed has a constant density, or at least a constant areal density.
Engineers deal more often with shapes during their initial designing and decide on specific materials (PVC, steel, brass, titanium) later, so the use of tables with area MOIs  is convenient.  Look at an engineering mechanics text like Beer & Johnson and the flyleaf will have a table of area MOIs.
A: It is possible that for $dm$ they substitute $dm=Q\cdot dA$, where $Q$ is the density of triangle. If $Q$ is constant, then it can be putted in front of a integral, or suppressed for the sake of simplicity.
A: You are correct. There should be a term of areal density, this is also reinforced by definition of moment of inertia and the fact that if areal density term is not there then the resulting solution is dimensionally inconsistent.
It would be better if you can share a specific video where you have seen this.
