Why is the center of mass of a hollow cone not $\frac{h}{2}$? Please don't think this is a physics question, since my question is about the mathematics in finding center of mass of a hollow cone.
I was learning how to derive center of mass of a hollow cone,  and this is what my 'proof' went like,

Let us assume that the cone is made of several sticks (infinitesimal thickness) bunched together at a point. The COM of each stick is at it's center. Hence by similarity of triangles, the COM of the whole cone should also be at a height $\frac{h}{2}$ from the base ($h$ being the height of the cone).

Then, I learned the proof using integration and found out I was wrong, the answer is actually $\frac{h}{3}$.
This brings me to my first question,
Why is my reasoning not valid?
As for my second question (which I suspect is somehow related to my earlier one),

Usually to check if a formula is valid,  we verify for degenerate cases. For example, the formula for distance between two points in a $2D$ plane $\sqrt{(\Delta X)^2 + (\Delta Y)^2}$ correctly reduces to the formula $\Delta X$ in the degenerate case of $1D$ axis (or $Y=0)$.
So I decided to check this in case of cone. A degenrate case of cone where apex angle is zero is case of a stick. COM of stick is at center. But for the degenrate case, it is not coming at the center but at a one third distance $\frac{h}{3}$.

Why is this happening? What is the fault in my argument?
 A: Your sticks cannot have the same width along their whole length.
Infinitesimal does not mean zero. In this case it means that if you have infinitely many of them, packed side-by-side, their width is such that at the bottom they fill out the whole circumference of the base without overlap. Their other end therefore also spans the same distance. You cannot get their tops any closer together without either making them pointy (i.e. triangular), or making them overlap.
In your other argument, the degenerate case has no area so it has no COG at all.
A: Your argument would apply to the centroid of a triangle just the same.  It would have to be half the altitude up from the base, looking from each corner, but you wouldn't get  single point like you do when you use $h/3$.  The problem with the argument is the extra mass at the wide end of the triangle.  Each segment of the cone should be a long narrow triangle, not a straight stick.  The second argument is essentially the same as the first, with the same problem.
