Center of gravity of an $n$-hemisphere

Connected to this question.

Keep in mind that I'm making use of the geometer definition of an $n$-sphere of distance $r$, i.e.xx _ math_3

Intend I have an $n$-sphere focused at $\bf 0$ in $\mathbb{R}^n$ with distance $r$ which has actually been separated right into $2^k$ orthants by $k$ axis-aligned hyperplanes (note, $k \le n$) in $\mathbb{R}^{n-1}$ travelling through $\bf 0$. For e.g., if $k=1$, we have $2$ $n$-hemispheres.

Below's the inquiry: just how do I find the center of gravity of such an orthant? Or (given that I'm still working with it), just how would certainly you find the center of gravity of an $n$-hemisphere?

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2019-05-05 01:27:38
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Answers: 1

One locates the centre of mass for these bodies equally as one. provides for any kind of body, by assimilation.

The $x_i$-coordinate of the centre of mass of an area $A$ in $\mathbb{R}^n$. is. $$\frac{\int_A x_i\ dx_1\cdots dx_n}{\int_A\ dx_1\cdots dx_n}.$$. Below the area could too be that in between the aircrafts $x_n=a$ and also $x_n=b$. in the round of distance $r$ centred at the beginning. Making use of the proportion. of $A$ this proportion of integrals amounts to. $$\frac{\gamma_{n-1}\int_a^b x_n(r^2-x_n^2)^{(n-1)/2} dx_n} {\gamma_{n-1}\int_a^b (r^2-x_n^2)^{(n-1)/2} dx_n}$$. ( in the $x_n$ instructions) where $\gamma_{n-1}$ is the quantity of the $(n-1)$-dimensional. device round (and also obligingly terminates). These integrals can be struck by. trig replacements.

Edited It's currently clear that my initial analysis of your inquiry. was incorrect. Nonetheless it's still unclear whether your centre of mass. is for a strong orthant or its bent surface area. Regardless if your orthant. is specified by the problems $x_1,\ldots,x_k\ge0$ after that its centre. of mass has the kind $(a,\ldots,a,0\ldots,0)$ where $a$ relies on $r$. and also $n$ yet not on $k$. One sees this from the proportion of the trouble. Hence the trouble lowers to the hemispheric instance. In the strong instance the. solution 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 "covering" instance 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 amounts right).

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2019-05-08 04:36:00
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