Centroid of Right circular cone

Question

We know that the centroid of a right angled triangular area is located at $Y=h/3$ and $X=b/3$ from the right angled vertex, where $h$ is height and $b$ is base length. So a right circular cone is just a rotation of this planar triangular. So now to calculate the $y$ coordinate of the centroid, it is just: $$ y_{cm} = \frac{\int Y.dm}{\int dm}$$ Now since Y is a constant it should turnout $y_{cm}= h/3$ but it is actually $h/4$. What is the mistake in what I have done?

Answer 1

As noted in the comment of André Nicolas: ''Two dimensions and three dimensions are different''.

For a correct calculus of the centroid, note that the radius at height $z$ (your $Y$) of the cone is $$ r(z)=(h-z)\frac{R}{h}=R\left(1-\frac{z}{h} \right) $$ where $h$ is the height of the cone and $R$ the radius of the basis. So, the volume of the cone can be calculated (by horizontal slices) as:

$$ \pi R^2\int_0^h\left(1-\frac{z}{h} \right)^2dz $$ and the height $z_c$ of the centroid is: $$ z_c=\frac{\pi R^2\int_0^h z\left(1-\frac{z}{h} \right)^2dz}{\pi R^2\int_0^h\left(1-\frac{z}{h} \right)^2dz} $$

can you do from this?