Center of Mass via integration for ellipsoid I need some help with the following calculation:
I have to calculate the coordinates of the center of mass for the ellipsoid 
$$\left( \frac{x}{a} \right)^2 + \left( \frac{y}{b} \right)^2 + \left( \frac{z}{c} \right)^2 \le 1,  \quad z \ge 0$$
with mass-density $\mu(x,y,z)=z^2$.
I wanted to use:
$$
\begin{align}
x & = a r \sin\theta \cos\varphi \\
y & = b r \sin\theta \cos\varphi \\
z & = c r \cos\theta
\end{align}
$$
whereas 
$$
\begin{gather}
0 \le r \le 1, \\
0 \le \theta \le \pi, \\
 0 \le \varphi \le 2\pi
\end{gather}
$$
and
$$\frac{\partial (x,y,z)}{ \partial (r, \theta, \varphi)} = r^2 \sin\theta.$$
Did I choose the right things so far?
1) $$
\begin{align}
M & = \int\limits_E µ(x,y,z) d(x,y,z) \\
 & = \int_0^1 \hspace{-5pt} \int_0^{\pi} \hspace{-5pt} \int_0^{2\pi} c^2 r^2 \cos^2\theta \cdot r^2 \sin(\theta) d(r, \theta, \varphi) \\
 & = c^2 \int_0^1 r^4 dr \int_0^\pi \sin\theta \cdot \cos^2\theta d\theta \int_0^{2\pi} d\varphi \\
 & = \frac{4\pi c^2}{15}.
\end{align}
$$
2) $$x_s \cdot M = \ldots $$
Here I get $\int_0^{2pi} \cos\varphi \, d \varphi = 0$, so the whole product is zero, so x_s is zero too??
What am I doing wrong?
 A: This can be just a typo, but you have a wrong parametrization of the sphere, it should be
$$
\begin{align}
x & = a r \sin \theta \cos \varphi, \\
y & = b r \sin \theta \sin \varphi, \\
z & = c r \cos \theta.
\end{align}
$$
Your limits for each variable are correct though. Your Jacobian is incorrect because you forgot to take in account the factors $a,b,c$. It should be
$$\frac{\partial (x,y,z)}{\partial (r, \theta, \varphi)} = - abc r^2 \sin \theta.$$
The $-$ sign is because this parametrization of the sphere reverses orientation.
When I edited your post I made sure to clarify some things but I didn't edit a couple of mistakes, which I intend to explain now:
1) I turned $d$'s into $\partial$'s for the Jacobian to correct your notation.
2) The notations $d(x,y,z)$ and $d(r, \theta, \varphi)$ don't make sense, it is best to stick to $dx \, dy \, dz$ and $dr \, d \theta \, d \varphi$. 
Your set up for the mass is correct if you fix the Jacobian and add $abc$. The calculation seems to be too (I haven't checked that thoroughly).
I don't understand what you mean by $x_s$. If you want to compute the $x$ coordinate of the center of mass, I assume you are using
$$x_s = \frac{1}{M} \int\limits_{E} x \mu \, dV, \text{ or } x_s M = \int\limits_{E} x \mu \, dV.$$
As you have seen, this is zero, just like the others will be. This has to do with mjqxxxx's comment that the ellipsoid has symmetry about all axis, therefore its center of mass has to be at the origin.
A: The mass density is invariant under $x\rightarrow -x$ and $y\rightarrow -y$, so the center of mass must have $x=y=0$.  You do still need to find its $z$-coordinate, but since the mass density is only a function of $z$, you can reduce this to a one-dimensional integral.  At a given value of $z$, the cross-section is an ellipse with semi-major and semi-minor axes $a\sqrt{1-(z/c)^2}$ and $b\sqrt{1-(z/c)^2}$; this ellipse has area $\pi a b (1-(z/c)^2)$.  The mass of a slice of thickness $dz$ at that altitude is therefore $dm=\pi a b (z^2 - z^4/c^2)dz$.  The $z$-coordinate of the center of mass is
$$
M_z=\frac{\int_{z=0}^{z=c}zdm}{\int_{z=0}^{z=c} dm}=\frac{\int_{0}^{c}(z^3-z^5/c^2)dz}{\int_{0}^{c}(z^2-z^4/c^2)dz}=\frac{\frac{1}{4}c^4-\frac{1}{6c^2}c^6}{\frac{1}{3}c^3-\frac{1}{5c^2}c^5}=\frac{\frac{1}{4}-\frac{1}{6}}{\frac{1}{3}-\frac{1}{5}}c=\frac{5}{8}c.
$$
A: $\newcommand{\+}{^{\dagger}}%
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 \newcommand{\dd}{{\rm d}}%
 \newcommand{\down}{\downarrow}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
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 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\half}{{1 \over 2}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\ol}[1]{\overline{#1}}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}
\color{#00f}{\large\vec{R}_{\rm cm}}&\equiv\left.
{\ds{\int z^{2}\pars{x\hat{x} + y\hat{y} + z\hat{z}}\,\dd x\,\dd y\,\dd z}
\over
\ds{\int z^{2}\,\dd x\,\dd y\,\dd z}}
\right\vert_{{x^{2} \over a^{2}}\ +\ {y^{2} \over b^{2}}\ +\ {z^{2} \over c^{2}}\
<\ 1
\,,\ z\ >\ 0}
\\[1cm]&=
\left.{\ds{\int z^{3}\,\dd x\,\dd y\,\dd z}\over
\ds{\int z^{2}\,\dd x\,\dd y\,\dd z}}\,\hat{z}
\right\vert_{{x^{2} \over a^{2}}\ +\ {y^{2} \over b^{2}}\ +\ {z^{2} \over c^{2}}\
<\ 1
\,,\ z\ >\ 0}
\\[1cm]&=\left.{\ds{\verts{abc^{4}}\int z^{3}\,\dd x\,\dd y\,\dd z}
\over
\ds{\verts{abc^{3}}\int z^{2}\,\dd x\,\dd y\,\dd z}}\,\hat{z}
\right\vert_{x^{2}\ +\ y^{2}\ +\ z^{2}\ <\ 1\,,\ z\ >\ 0}
=\left.{\ds{\int z^{3}\,\dd x\,\dd y\,\dd z}
\over
\ds{\int z^{2}\,\dd x\,\dd y\,\dd z}}\ \verts{c}\,\hat{z}
\right\vert_{r\ <\ 1\,,\ z\ >\ 0}
\\[1cm]&={\ds{2\pi\int_{0}^{1}\dd r\,r^{5}\int_{0}^{\pi/2}\dd\theta\,\sin\pars{\theta}\cos^{3}\pars{\theta}}
\over \ds{2\pi\int_{0}^{1}\dd r\,r^{4}\int_{0}^{\pi/2}\dd\theta\,\sin\pars{\theta}\cos^{2}\pars{\theta}}}\
\verts{c}\,\hat{z}
={5 \over 6}\,{\ds{\int_{0}^{1}\xi^{3}\,\dd\xi}\over
\ds{\int_{0}^{1}\xi^{2}\,\dd\xi}}\ \verts{c}\,\hat{z}
=\color{#00f}{\large{5 \over 8}\,\verts{c}\,\hat{z}}
\end{align}
