# 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?

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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