Center of gravity 1/4 sphere I am solving problem, where i have to find the center of gravity of homogenus 1/4 of a sphere ($x^2+y^2+z^2 \leq 1 ; x,y > 0)$. I Know that the center is in $xy$ plane on $y=x$. So I just tried to calculate the $r_T$. So I am trying to use integral in spherical coordinates:
$$
r_T= \frac{\int \int \int_V \rho r dV}{\int \int \int_V \rho dV}
$$
Is this the valid way, because I don't know how to use Jacobian matrix in this formula. What I am asking is if we can determine position of centre of gravity  with that equation only for Cartesian coordinates ($x,y$ and $z$) or for position of mass centre for any coordinate system, eg. for any new coordinates $q$:
$$
q_T= \frac{\int \int \int_V \rho q dV}{\int \int \int_V \rho dV}
$$
 A: The $x$-coordinate of the center can be integrated in spherical coordinates as
\begin{align}
x_c & = \frac1{\frac14\cdot \frac{4\pi}3} \int_{V} xdV\\
&=\frac3{\pi}\int_0^{\pi/2}\int_0^\pi\int_0^1
(r\cos\phi\sin\theta)r^2\sin\theta drd\theta d\phi\\
&= \frac3{\pi}\int_0^{\pi/2} \cos\phi d\phi
\int_0^\pi\sin^2\theta d\theta \int_0^1 r^3dr \\
&= \frac38\\
\end{align}
and $y_c=x_c$. Thus, the center of gravity is $(\frac38, \frac38, 0)$.
A: I'll go about this in the usual way. I'll use $\mathbf{r}$ as a generic position vector and $\mathbf{r}_{\text{C.O.M}}$ to denote the position vector of the center of mass. We expect that the answer should be something like $$\mathbf{r}_{\text{C.O.M}}=\frac{1}{M}\iiint\limits _{\Omega } \rho \mathbf{r}\mathrm{d} V$$
Where $\Omega$ is our region and $M$ is our total mass. Since this region has constant density and is a quarter-sphere, $M$ is easy to calculate - it is simply $\frac{4\pi\rho}{3}$. We can now cancel out the $\rho$ in the numerator and denominator to be left with $$\mathbf{r}_{\text{C.O.M}}=\frac{3}{4\pi }\iiint\limits _{\Omega }\mathbf{r}\mathrm{d} V.$$
Now let's focus on that nasty looking triple integral. I'm going to convert to spherical polar coordinates and use the $(r,\theta,\phi)$ convention. We know that in polar spherical coordinates, $$\mathrm{d}V=r^2\sin(\phi)\mathrm{d}r\mathrm{d}\theta\mathrm{d}\phi.$$
And our position vector is simply $\mathbf{r}=r\hat{\mathbf{r}}$.
What about the bounds of our integral? Well, it is fairly obvious that $r$ will range from $0$ to $1$, $\theta$ will range from $0$ to $\pi/2$, and $\phi$ will range from $0$ to $\pi$. Converting $\hat{\mathbf{r}}$ into the standard basis, we have:
$$\mathbf{r}_{\text{C.O.M}}=\frac{3}{4\pi}\int\limits ^{\pi }_{0}\int\limits ^{\pi /2}_{0}\int\limits ^{1}_{0}r(\cos(\theta)\sin(\phi)\hat{\mathbf{i}}+\sin(\theta)\sin(\phi)\hat{\mathbf{j}}+\cos(\phi)\hat{\mathbf{k}})r^{2}\sin( \phi )\mathrm{d} r\mathrm{d} \theta \mathrm{d} \phi .$$ 
Split the integral into $\hat{\mathbf{i}}$,$\hat{\mathbf{j}}$ and $\hat{\mathbf{k}}$ components and take it from here.
EDIT: I acknowledge this is a rather long way of doing it. I tried to be as formal as possible to really show the spirit of volume integrals in different coordinate systems.
