Center of mass of an $n$-hemisphere Related to this question.
Note that I'm using the geometer definition of an $n$-sphere of radius $r$, i.e.$ \\{ x \in \mathbb{R}^n : \|x\|_2 = r  \\} $
Suppose I have an $n$-sphere centered at $\bf 0$ in $\mathbb{R}^n$ with radius $r$ which has been divided into $2^k$ orthants by $k$ axis-aligned hyperplanes (note, $k \le n$) in $\mathbb{R}^{n-1}$ passing through $\bf 0$. For e.g., if $k=1$, we have $2$ $n$-hemispheres.
Here's the question: how do I find the center of mass of such an orthant? Or (since I'm still working on it), how would you find the center of mass of an $n$-hemisphere?
 A: One finds the centre of mass for these bodies just as one
does for any body, by integration.
The $x_i$-coordinate of the centre of mass of a region $A$ in $\mathbb{R}^n$
is
$$\frac{\int_A x_i\ dx_1\cdots dx_n}{\int_A\ dx_1\cdots dx_n}.$$
Here the region might as well be that between the planes $x_n=a$ and $x_n=b$
in the sphere of  radius $r$ centred at the origin. Using the symmetry
of $A$ this ratio of integrals equals
$$\frac{\gamma_{n-1}\int_a^b x_n(r^2-x_n^2)^{(n-1)/2} dx_n}
{\gamma_{n-1}\int_a^b (r^2-x_n^2)^{(n-1)/2} dx_n}$$
(in the $x_n$ direction) where $\gamma_{n-1}$ is the volume of the $(n-1)$-dimensional
unit ball (and obligingly cancels). These integrals can be attacked by
trig substitutions.
Edited
It's now clear that my original interpretation of your question
was wrong. However it's still not clear whether your centre of mass
is for a solid orthant or its curved surface. In any case if your orthant
is defined by the conditions $x_1,\ldots,x_k\ge0$ then its centre
of mass has the form $(a,\ldots,a,0\ldots,0)$ where $a$ depends on $r$
and $n$ but not on $k$. One sees this from the symmetry of the problem.
Thus the problem reduces to the hemispheric case. In the solid case the
answer is
$$a=\frac{\int_0^r x(r^2-x^2)^{(n-1)/2} dx}
{\int_0^r (r^2-x^2)^{(n-1)/2} dx}$$
while in the "shell" case it is
$$a=\frac{\int_0^r x(r^2-x^2)^{(n-3)/2} dx}
{\int_0^r (r^2-x^2)^{(n-3)/2} dx}$$
(if I've done my sums right).
