Find the center of mass of an n-dimensional hemisphere (mass is uniformly distributed) where $x_n\ge 0$

Question

I want to find the centroid of an n-dimensional hemisphere with a radius $a$. The hemisphere has uniformly distributed mass and I denote it as $B_+^n(a)=\{(x_1,\cdots,x_n):x_1^2+\cdots+x_n^2\le a^2 \ and \ x_n\ge 0\}$

Here is my attempt:

By the formula, the centroid of this n-dimensional hemisphere should be $$ \vec{P} = \dfrac{1}{M} \idotsint\limits_{B^n_+(a)} x_n \rho \ dv $$ where $\vec{P}$ is the coordinate of the centroid, $\rho$ is the density which is a constant. By the symmetry of the hemisphere, the centroid should lie in the $x_n$ axis, so only consider $\idotsint\limits_{B^n_+(a)} x_n \rho \ dv$ is enough.

The total mass is $$\rho \cdot v(B_+^n(a))=\frac{1}{2}\rho v(B^n(a)) =\frac{\rho \cdot a^n}{2} v(B^n(1)) $$ where $B^n(a)$ is the n-dimensional sphere with a radius $a$ and $v(\cdot)$ denotes the volume.We have the formula that $v(B^n(a))=a^n \cdot v(B^n(1))$

For the integration by polar coordinate transformation and Fubini theorem. $$ \begin{align*} \idotsint\limits_{B^n_+(a)} x_n \rho \ dv &= \idotsint\limits_{B^n_+(a)} r\sin\theta_1 \cdots \sin\theta_{n-2}\sin \theta_{n-1} \cdot (r^{n-1}\sin^{n-2} \theta_1 \cdot \sin^{n-3} \theta_2 \cdots \sin \theta_{n-2}) \ dr d\theta_{1} \cdots d\theta_{n-1} \\ &= \idotsint\limits_{B^n_+(a)} r^n \sin^{n-1} \theta_1 \cdot \sin^{n-2} \theta_2 \cdots \sin^2 \theta_{n-2} \sin \theta_{n-1} \ dr d\theta_{1} \cdots d\theta_{n-1} \\ &= \int_0^a r^n \ dr \int_0^{\pi/2} \sin^{n-1} \theta_1 \ d\theta_1 \cdots \int_0^{\pi/2} \sin^2 \theta_{n-2} \ d\theta_{n-2} \int_0^{2\pi} \sin \theta_{n-1} \ d\theta_{n-1} \end{align*} $$

Since the radius is $a$, the upper limit of integral of $r$ is just $a$; and since now it's a hemisphere, $\theta_1,\cdots,\theta_{n-2}$ are now in $[0,\frac{\pi}{2}]$; finally, the last angle $\theta_{n-1}$ shoudn't be affected, and it should stay in $[0,2\pi]$

But if this is true, then the $\int_{0}^{2\pi} \sin \theta_{n-1} \ d\theta_{n-1}$ will be zero, so the centroid is just the origin point. This shouldn't be right.

Thus my questions are

1. If my basic idea of calculating the centroid right?

2. Can you tell me which part of my process is wrong?

Thank you very much.

Answer 1

The problems in your approach have already been discussed in the comments.

The centroid can actually be determined in a much easier way. Consider the unit hypersphere for simplicity. The volume of the infinitesimal $(n-1)$-dimensional hyperspherical slices that make up the hemisphere along the $x_n$ direction is proportional to $r^{n-1}$, where $r=\sqrt{1-x_n^2}$ is their radius, so the centroid is at

$$ \langle x_n\rangle=\frac{\int_0^1Cx_n\left(1-x_n^2\right)^{\frac{n-1}2}\mathrm dx_n}{\int_0^1C\left(1-x_n^2\right)^{\frac{n-1}2}\mathrm dx_n}=\frac{\frac1{n+1}}{\frac{\sqrt\pi\,\Gamma\left(\frac n2+\frac12\right)}{2\Gamma\left(\frac n2+1\right)}}=\frac{\Gamma\left(\frac n2+1\right)}{\sqrt\pi\,\Gamma\left(\frac n2+\frac32\right)}\;. $$

For instance, for $n=2$, the centroid of a unit semicircle lies at $\frac{\Gamma(2)}{\sqrt\pi\Gamma\left(\frac52\right)}=\frac4{3\pi}$.

To get the centroid for arbitrary radius $a$, just multiply by $a$.