Center of mass of a right circular cone How can one find the center of a right circular cone with height $h$ and radius $r$? 
I've found these formulas:
$$M_{xy} = \iiint\limits_V z \rho (x,y,z) \, dx \, dy \, dz$$
$$M_{yz} = \iiint\limits_V x \rho (x,y,z) \, dx \, dy \, dz$$
$$M_{zx} = \iiint\limits_V y \rho (x,y,z) \, dx \, dy \, dz$$
$$M = \iiint\limits_V \rho (x,y,z) \, dx \, dy \, dz$$
We can assume that $\rho (x,y,z)=1$
A right circular cone with vertex pointing up, with height $h$ and base radius $r$ with base located at $z=0$ can be parameterized by:
$$x = \frac{h-u}{h} r \cos \varphi$$
$$y = \frac{h-u}{h} r \sin \varphi$$
$$z=u$$
$$u \in [0,h], \ \ \varphi \in [0, 2 \pi)$$
My problem is that I've calculated these integrals:
$$M_{xy} = \int_0 ^{2 \pi} \int_0 ^h u\,du\,d \varphi = \pi h^2$$
$$M_{yz} = \int_0 ^{2 \pi} \int_0^h \frac{h-u}{h}r \cos \varphi\, du \,d \varphi = 0$$ and similarly $M_{zx} = 0$. 
$$M = \int \int \, du  \, d \varphi = \int_0^{2 \pi}h \, d \varphi = 2h \pi.$$
There must be a mistake somewhere, because the center of mass of a right circular cone is at $\frac{3}{4}$ of its height.
Could you help me?
Thanks!
 A: I'd look at cross-sections parallel to the base.  The radius of the cross-section at height $z$ from the base is $r(h-z)/h$, so the area is $\pi r^2(h-z)^2/h^2$.  So the infinitesimal element of volume at that height is $\pi r^2(h-z)^2\,dz/h^2$.  Integrating $z$ with respect to volume, for the $z$-coordinate of the center of gravity, we get
$$
\frac{\displaystyle\iiint z\,dV}{\displaystyle\iint 1\,dV} = \frac{\displaystyle\frac{1}{h^2} \int_0^h z\Big(\pi r^2(h-z)^2\,dz\Big)}{\displaystyle\frac{1}{h^2}\int_0^h \pi r^2(h-z)^2\,dz}.
$$
The factor $\pi r^2$ does not depend on $z$, so it cancels from the numerator and the denominator.  So you have
$$
\frac{\int_0^h z(h-z)^2\,dz}{\int_0^z(h-z)^2\,dz} = \frac{\int_h^0 (h-w)w^2\,(-dw)}{\int_h^0 w^2(-dw)} = \frac{h^4/3 -h^4/4}{h^3/3} = \frac h 4.
$$
If you have a pyramid with a non-circular base, the area at height $z$ is still $(\text{constant}\cdot(h-z)^2)$, and the "constant" depends on the shape, but the "constant" again appears as a factor in both the numerator and the denominator and cancels.
A: Because of the circular symmetry, it is obvious that the center of mass is on the $z$ axis. The $z$ component of the center of mass is
$$
\frac{\int_V z \rho \;\mathrm{d}v}{\int_V \rho \;\mathrm{d}v} = \frac{M_{xy}}{M}
$$
But you mis-calculated both $M$ and $M_{xy}$. 
By far the easiest way to do these integrals is to work in cylindrical coordinates $(s, \phi, z)$ where the usual notation is to call the radial coordinate $r$ or $\rho$ but since the statement of the problem uses those two letters I use $s$ instead of $r$.  The integrals are easy but one must remember that the volume element is $s\;\mathrm{d}s\;\mathrm{d}\phi\;\mathrm{d}z$ rather than just $\mathrm{d}s\;\mathrm{d}\phi\;\mathrm{d}z$. Take $\rho = 1$ as before:
\begin{align}
M &= \int_{z=0}^{h} \int_{\phi=0}^{2\pi} \int_{s=0}^{r \frac{h-z}{h} } s\;\mathrm{d}s\;\mathrm{d}\phi\;\mathrm{d}z
=\int_{z=0}^{h} \int_{\phi=0}^{2\pi} \frac{r^2}{2} \left(\frac{h-z}{h}\right)^2\;\mathrm{d}\phi\;\mathrm{d}z \\[8pt]
&= \int_{z=0}^{h} \pi r^2\left(\frac{h-z}{h}\right)^2\;\mathrm{d}z 
= \frac{\pi r^2}{h^2} \int_{0}^{h} (z-h)^2\;\mathrm{d}z = \frac{\pi r^2}{3h^2} \left[ (z-h)^3 \right]_{0}^{h} \\
M &=  \frac{\pi r^2 h}{3}
\end{align}
\begin{align}
M_{xy} &= \int_{z=0}^{h} \int_{\phi=0}^{2\pi} \int_{s=0}^{r \frac{h-z}{h} } z s\;\mathrm{d}s\;\mathrm{d}\phi\;\mathrm{d}z
=\int_{z=0}^{h} \int_{\phi=0}^{2\pi} z \frac{r^2}{2} \left(\frac{h-z}{h}\right)^2 \;\mathrm{d}\phi\;\mathrm{d}z \\[8pt]
&= \int_{z=0}^{h} \pi z r^2\left(\frac{h-z}{h}\right)^2 \;\mathrm{d}z 
= \frac{\pi r^2}{h^2} \int_{0}^{h} z(z-h)^2 \;\mathrm{d}z 
= \frac{\pi r^2}{h^2} \int_{0}^{h} \left( z^3 - 2 z^2 + z \right) \;\mathrm{d}z \\[8pt]
&= \frac{\pi r^2}{h^2} \left[ \frac{z^4}{4} -\frac{2 z^3}{3} + \frac{z}2 \right]_{0}^{h} \\[8pt]
&=  \frac{\pi r^2 h}{3}
= \frac{\pi r^2}{h^2} h^4 \left( \frac{1}{4} - \frac{2}{3} +\frac{1}{2} \right) \\[8pt]
M_{xy} &= \frac{\pi r^2 h^2}{12}
\end{align}
$$
z_{CM} = \frac{M_{xy}}{M} = \frac{h}{4}
$$
A: The distance from the top of the cone to the center of mass is given by:
$$\frac{3}{h^{3}}\int_{0}^{h}x^{3}dx$$
A: $\newcommand{\+}{^{\dagger}}
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$\ds{x_{1} \equiv x\,,\quad x_{2} \equiv y\,\quad x_{3} \equiv z}$.

$\large\tt\mbox{With the vertex at the origin:}$
  \begin{align}
\color{#66f}{\large x_{{\rm i\ cm}}}\,\pars{{1 \over 3}\,\pi r^{2}h}
&\equiv\int_{V}x_{i}\,\dd V
=\int_{V}\nabla\cdot\pars{{1 \over 4}\,x_{i}\vec{R}}\,\dd V
=\int_{S}{1 \over 4}\,x_{i}\,\vec{R}\cdot\dd\vec{S}
\\[3mm]&=\left.{1 \over 4}
\int_{x^{2}\ +\ y^{2}\ <\ r^{2}}\ x_{i}h\,\dd x\,\dd y
\,\right\vert_{z\ =\ h}
\end{align}

By symmetry considerations, it's obvious that $\ds{x_{\rm cm} = y_{\rm cm} = 0}$.

\begin{align}
\color{#66f}{\large z_{\ cm}}\,\pars{{1 \over 3}\,\pi r^{2}h}&
={1 \over 4}\,h^{2}\
\overbrace{\left.\int_{x^{2}\ +\ y^{2}\ <\ r^{2}}\dd x\,\dd y\,
\right\vert_{z\ =\ h}}^{\ds{=\ \pi r^{2}}}
\end{align}

$$
\begin{array}{rrcl}
\mbox{Vertex at the origin:}\qquad &
\color{#66f}{\large z_{\rm cm}} & = & \color{#66f}{\large{3 \over 4}\,h}
\\
\mbox{Base at}\ xy-\mbox{surface}\qquad &
\color{#66f}{\large z_{\rm cm}} & = & h - {3 \over 4}\,h
=\color{#66f}{\large {1 \over 4}\,h}
\end{array}
$$
