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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?

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    To supplement Anton's point with a calculation: the center of mass of a unit-radius semicircle is $2/\pi \approx 0.64$ whereas of the half disk bounded by that semicircle it is $4/(3\pi) \approx 0.42$: closer to the origin as he said.2010-08-06

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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).

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    At which coordinate does the $0$s begin in $(a,\dots,a,0,\dots,0)$?2010-08-06