Centroid of Right circular cone
Matthew Barrera 
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?
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$\begingroup$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?
$\endgroup$ $\begingroup$Were your reasoning right, the centroid of a triangle would be at $\dfrac h2$ as for a line segment.
The centroid ordinate is a weighted average of $y$, where the weights are the measure of the set of points with the same $y$. If you revolve the triangle, these weights change.
For a line segment, the weights are constant, giving
$$\frac{\displaystyle\int y\,dy}{\displaystyle\int dy}=\frac{\dfrac{h^2}2}{h}.$$
For a triangle, the weights are linear
$$\frac{\displaystyle\int y\cdot y\,dy}{\displaystyle\int y\,dy}=\frac{\dfrac{h^3}3}{\dfrac{h^2}2}.$$
For a cone, the weights are quadratic
$$\frac{\displaystyle\int y\cdot y^2\,dy}{\displaystyle\int y^2\,dy}=\frac{\dfrac{h^4}4}{\dfrac{h^3}3}.$$
Note that the formulas are computed for the shapes upside-down, with the vertex at the origin, so you should consider $h-\bar y$.
$\endgroup$ 3 $\begingroup$The center of length of a line segment, area for an isosceles triangle and center of mass of a right circular need not have the same constant in relation to maximum dimension measured from base. The factors are respectively $ \dfrac12 , \dfrac13 , \dfrac14. $
This is due to dimension of space $1D,2D, 3D $ and how the material considered as differential entities is populated/distributed.
For $1D$ it is at midpoint. if linear density $\rho$ per unit length, $x_c$ is found by taking moments in force equilibrium.
$$ x_c= \dfrac{\int_0^a x \rho dx }{\int_0^a\rho dx } =\dfrac{a}{2} $$
For 2D similarly we can know how $\dfrac13$ point comes about if $\rho$ stands for differential triangle areal density. Integrate between same limits.
$$ x_c= \frac{ \int x \rho dA}{\int \rho dA} =\dfrac{2 a}{3} $$
For 3D same method of taking moments is followed, $ \rho$ represents weight per unit volume aka density of differential cone elements.
$$ x_c= \frac{ \int x \rho dV}{\int \rho dV} =\dfrac{3a}{4} $$
Hope you make the sketch in each case. To find distance from base subtact each of above from $a$.
As commented by Yves Douest for simplex 4D the similar solid's center of mass is $\dfrac a5$ from its "base".
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