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?

  • $\begingroup$ Is the mass concentrated on the surface, or uniformly distributed inside the sphere? $\endgroup$ – Anton Geraschenko Aug 6 '10 at 15:14
  • $\begingroup$ @Anton: In reality, there is only the surface. But I figured, the center of mass would be the same as if it were uniformly distributed inside the sphere --- is that correct? $\endgroup$ – Jacob Aug 6 '10 at 15:19
  • $\begingroup$ Thanks for clarifying. It does make a difference whether you're cutting up a solid ball or a shell. The center of mass will be closer to the origin if it's a solid ball. You can see this intuitively by thinking about the cases $n=k=2$ or $n=k=1$. $\endgroup$ – Anton Geraschenko Aug 6 '10 at 15:31
  • $\begingroup$ @Anton: Thanks, but after checking it out, I see that an $n$-sphere actually refers to the surface of a ball so I suppose the question is still correct? $\endgroup$ – Jacob Aug 6 '10 at 15:46
  • $\begingroup$ Perhaps I'm missing something, but wouldn't k hyperplanes divide it into 2^k orthants? e.g., if n=2, 2 lines divide the plane into 4 quadrants. $\endgroup$ – Isaac Aug 6 '10 at 16:08

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).

  • $\begingroup$ Thanks for the answer! I've updated my question to add that the hyperplanes pass through the origin. How would this change $a$ and $b$? $\endgroup$ – Jacob Aug 6 '10 at 16:05
  • $\begingroup$ Thanks for the great answer! I really appreciate it and I love this place :) $\endgroup$ – Jacob Aug 6 '10 at 22:26
  • $\begingroup$ At which coordinate does the $0$s begin in $(a,\dots,a,0,\dots,0)$? $\endgroup$ – Jacob Aug 6 '10 at 22:31

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