Center of mass of $x^2+y^2 \leq z \leq h$ I'm trying to find the center of mass of this shape $x^2+y^2 \leq z \leq h$but im having difficulties founding the limits of integration.
using cylindrical coordinates, $x=rcos \theta$, $y=r \sin \theta$
I would suspect that since $0\leq x^2+y^2 \leq z \leq h$, then $0 \leq r \leq \sqrt{h}$, and $r^2 \leq z \leq h$ and the jacobian is $r$.
But the integral $$M=\int_{0}^{2\pi} \int_{0}^{\sqrt{h}} \int_{r^2}^{h}r dzdrd\theta = \frac{\pi h^2}{2}$$
That is the mass of this shape.
Now let's find the moment in the $xy$ direction:
$$M_{xy} = \int_{0}^{2\pi} \int_{0}^{\sqrt{h}} \int_{r^2}^{h}rz dzdrd\theta =\frac{\pi h^3}{3}$$
So the $z$ coordinate at the center of mass should be:
$$z_{cm}=\frac{M_{xy}}{M}=\frac{2}{3}h$$
but the answer given in the worksheet says $z_{cm}=\frac{3}{4}h$
Where am I mistaken? I'm sure it's the integration domain. But why?
 A: Everything is fine with the calculation. An alternative parametrization is 
$M_{xy}=\int_{0}^{2\pi} \int_{0}^{h} \int_{0}^{\sqrt{z}}r~\mathrm dr~\mathrm dz~\mathrm d\theta =2\pi\frac{h^2}{2}$ and 
$M_{xy}=\int_{0}^{2\pi} \int_{0}^{h} \int_{0}^{\sqrt{z}}rz~\mathrm dr~\mathrm dz~\mathrm d\theta =2\pi\frac{h^3}{6}$. 
A: I would say that you have counted correctly. Z coordinate of the center of mass is $\frac{2}{3}h$:
$dM=\pi(x^2+y^2)\,dz=\pi z\,dz\,\Rightarrow \,M=\pi\int_0^hdM=\pi\int_0^hz\,dz=\pi[\frac{z^2}{2}]_0^h=\pi\frac{h^2}{2}$
$dM_{xy}=z\pi(x^2+y^2)\,dz=\pi z^2\,dz\,\Rightarrow \, M_{xy}=\pi\int_0^hz^2\,dz=\pi[\frac{z^3}{3}]_0^h=\pi\frac{h^3}{3}$
$\Rightarrow \,z_{cm}=\frac{2}{3}h$
Result in the worksheet $\left(\frac{3}{4}h \right)$ is not correct. To find also in a collection of mathematical formulas.
A: Given
$$
x^2 + y^2 \le z \le h
$$
Then consider cylinder co-ordinates
$$
\begin{eqnarray}
x &=& r \cos(\phi)\\
y &=& r \sin(\phi)\\
z &=& z
\end{eqnarray}
$$
Then you obtain
$$
\iiint_V dx dy dz \Psi = \int_0^h dz \int_0^\sqrt{z} dr \int_0^{2\pi} d\phi r \Psi
$$
So the volume $(\Psi=1)$ is simply given by
$$
\begin{eqnarray}
V &=& \int_0^h dz \int_0^\sqrt{z} dr \int_0^{2\pi} d\phi r\\
&=& 2 \pi \int_0^h dz \int_0^\sqrt{z} dr r\\
&=& \pi \int_0^h dz z\\
&=& \frac{\pi h^2}{2}
\end{eqnarray}
$$
Uniform density implies that we have
$$
\langle \Psi \rangle = \frac{2}{\pi h^2} \int_0^h dz \int_0^\sqrt{z} dr \int_0^{2\pi} d\phi r \Psi
$$
and for center of mass we obtain
$$
\begin{eqnarray}
\langle \Psi_x \rangle &=& \frac{2}{\pi h^2} \int_0^h dz \int_0^\sqrt{z} dr \int_0^{2\pi} d\phi r r \cos(\phi)\\
\langle \Psi_y \rangle &=& \frac{2}{\pi h^2} \int_0^h dz \int_0^\sqrt{z} dr \int_0^{2\pi} d\phi r r \sin(\phi)\\
\langle \Psi_z \rangle &=& \frac{2}{\pi h^2} \int_0^h dz \int_0^\sqrt{z} dr \int_0^{2\pi} d\phi r z
\end{eqnarray}
$$
So it is clear that
$$
\begin{eqnarray}
\langle \Psi_x \rangle &=& 0\\
\langle \Psi_y \rangle &=& 0
\end{eqnarray}
$$
and
$$
\begin{eqnarray}
\langle \Psi_x \rangle &=& \frac{2}{\pi h^2} \int_0^h dz \int_0^\sqrt{z} dr r z\\
&=& \frac{2}{h^2} \int_0^h dz z^2\\
&=& \frac{2}{3} h
\end{eqnarray}
$$
A: I managed to show its $\frac{2}{3}h$ in another way.
Realize that $x^2+y^2 <z$ in a 2d plane represents a circle with $\sqrt{z}$ radius.
Let's call that circle $A$.
We are asked to calculate $\int_{0}^{h} \iint_A dxdy dz$ (Fubini shows us that order of integration does not matter)
however, $\iint_A dxdy$ is merely the area of $A$, which is according to formula for area of circle, $\pi z$.
So the integral for the mass is $\int_{0}^{h} \pi zdz=\frac{\pi h^2}{2}$
With a similar argument, you can see that $M_{xy}=\frac{\pi h^3}{3}$ and as such, $z_{cm}=\frac{2}{3}h$
