Suppose I want to find the volume of a regular tetrahedron with side length $s$. Intuitively, I thought we could sum over the areas of all equilateral triangles of side length $x$, where $0 \leq x \leq s$, and that should be the answer. Since the height of any equilateral triangle is $s\sqrt{3}/{2}$ we can compute
$$\int_0^s \frac{1}{2}\cdot\frac{\sqrt{3}}{2} x\cdot x\ dx = \frac{\sqrt{3}}{12}s^3.$$
but this gives the wrong formula for the volume.
I know another way to find the solution so I am not looking for that kind of an answer, but I am looking for an explanation as to why the above is incorrect. Does it not account for something geometrically?
You are using a variable along the side of the triangle, so the thickness of your infinitesimal slices is not $dx$. You have to multiply by the cosine of the angle between the side and the altitude, which is $\sqrt{2/3}$. So your final number is multiplied by that and you should get the right formula.