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?

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
$$
M_z=\frac{\int_{z=0}^{z=c}zdm}{\int_{z=0}^{z=c} dm}=\frac{\int_{0}^{c}(z^3-z^5/c^2)dz}{\int_{0}^{c}(z^2-z^4/c^2)dz}=\frac{\frac{1}{4}c^4-\frac{1}{6c^2}c^6}{\frac{1}{3}c^3-\frac{1}{5c^2}c^5}=\frac{\frac{1}{4}-\frac{1}{6}}{\frac{1}{3}-\frac{1}{5}}c=\frac{5}{8}c.
$$
