Do the centers of mass of the whole $n$-sphere and its half coincide when $n \to \infty$? It is known that the distance between the centroid (center of mass) and the center of a unit semicircle is $\displaystyle\frac{4}{3\pi}$, whereas that of a unit hemisphere is $\displaystyle\frac{3}{8}$. I am interested in determining a closed general formula for the centroid of a unit n-hemisphere. I did some calculations and obtained $$\displaystyle\frac{1}{\sqrt{\pi}}\frac{\Gamma{(\frac{n+2}{2}})}{\Gamma{(\frac{n+3}{2}})}$$
which for $n=1, 2, 3, 4....$ gives the values $\displaystyle\frac{1}{2}, \frac{4}{3\pi},\frac{3}{8},\frac{16}{15\pi}....$. I obtained this formula using the Pappus' centroid theorem for the volume of solids of revolution, but would be happy to have confirmation of this by a proof based on integral calculations. I found a previous question on this topic, but no closed form expression was provided in the answer. 
If confirmed, an interesting and somewhat unexpected consequence of this formula would be that because $$\lim_{n \to \infty} \displaystyle\frac{1}{\sqrt{\pi}}\frac{\Gamma{(\frac{n+2}{2}})}{\Gamma{(\frac{n+3}{2}})}=0$$
then the centroid of a n-hemisphere and that of the whole n-sphere tend to coincide as $n$ increases. Is this interpretation correct? Any hint would be very appreciated. 
 A: Yes, your consequence is correct. This is a manifestation of the concentration of measure phenomenon, which in the case of the sphere means that for any $\epsilon>0$ larger and larger fraction of its volume is concentrated in the $\epsilon$-neighborhood of the sphere's equator as $n\to\infty$. More informally, asymptotically almost all of sphere's volume is located near its equatorial section, and parts of the sphere away from it contribute almost nothing to the location of the centroid. But the same is true for the hemisphere, hence the coincidence in the limit. 
This is somewhat counterintuitive but can be explained like this. The unit sphere is specified by the equation $x_1^2+x_2^2+\dots x_n^2=1$, to get the hemisphere we also constrain $x_n\geq0$. But heuristically the more variables we have the less effect constraints on any individual variable have on the collective result, in this case location of the centroid or the volume. This idea is the basis for the measure concentration in general, which has many applications, there is a whole book about it.
A: Computation of the Limit
Using Gautschi's Inequality, which is proven in this answer, and $x=\frac{n+1}{2}$, we get
$$
\sqrt{\frac{2}{\pi(n+3)}}\le\frac1{\sqrt\pi}\frac{\Gamma\left(\frac{n+2}{2}\right)}{\Gamma\left(\frac{n+3}{2}\right)}\le\sqrt{\frac{2}{\pi(n+1)}}
$$
Therefore, the centroid of the $n$-hemisphere gets arbitrarily close to the centroid of the $n$-sphere as $n\to\infty$.

Verification of the Formula
The center of mass of the upper hemisphere of the unit sphere would be the mean of the centers of the slabs weighted by the mass of the slabs. The mass of the slab at height $t$ is proportional to $\sqrt{1-t^2}^{n-1}$. Therefore,
$$
\begin{align}
\frac{\displaystyle\int_0^1t\sqrt{1-t^2}^{n-1}\,\mathrm{d}t}{\displaystyle\int_0^1\sqrt{1-t^2}^{n-1}\,\mathrm{d}t}
&=\frac{\displaystyle\int_0^1\sqrt{1-t}^{n-1}\,\mathrm{d}t}{\displaystyle\int_0^1t^{-1/2}\sqrt{1-t}^{n-1}\,\mathrm{d}t}\\
&=\left.\frac{\Gamma\left(1\right)\Gamma\left(\frac{n+1}2\right)}{\Gamma\left(\frac{n+3}2\right)}\middle/\frac{\Gamma\left(\frac12\right)\Gamma\left(\frac{n+1}2\right)}{\Gamma\left(\frac{n+2}2\right)}\right.\\
&=\frac1{\sqrt\pi}\frac{\Gamma\left(\frac{n+2}2\right)}{\Gamma\left(\frac{n+3}2\right)}
\end{align}
$$
