I saw the following definition for the barycentre of a set $\Omega \subseteq \mathbb{R}^d$:$$\mathrm{bc}^\Omega=\frac{1}{\mathrm{vol}(\Omega)}\int_\Omega x dx \in\mathbb{R}^n.$$So I wanted to compute the barycenter for a ball of radius $r$ with center $x_0, \quad \Omega= B_r(x_0)$, and from my understanding it should be $\mathrm{bc}^\Omega=x_0.$In the Original post my computation was wrong here is the corrected version:\begin{equation}\begin{aligned}\mathrm{bc}^\Omega_i&=\frac{1}{\omega_n r^n}\int_0^r\int_{\partial B_\rho(x_0)}\xi_i dS\xi d\rho\\&=\frac{1}{\omega_n r^n}\int_0^r\int_{\partial B_\rho(x_0)}|\xi-x_0|\left(\left\langle e_i, \frac{\xi-x_0}{|\xi-x_0|} \right\rangle + \left\langle e_i, \frac{x_0}{|\xi-x_0|} \right\rangle \right) dS\xi d\rho\\&=\frac{1}{\omega_n r^n}\int_0^r\rho\int_{B_\rho(x_0)}div(e_i) dx + \int_{\partial B_\rho(x_0)}\rho\left\langle e_i, \frac{x_0}{|\xi-x_0|} \right\rangle dS\xi d\rho\\&=\frac{x_{0, i} \cdot n}{r^n}\int_{0}^{r}\rho^{n-1}d\rho=x_{0, i}\end{aligned}\end{equation}
Where I first transformed the integral domain to integrate over spheres, then reformulated the integrand to include the unit normal to those spheres. Applied the divergence theorem to the constant $e_i$ and integrated as usual in the second summand.I also used the formulae:$$\int_{B_r(x_0)}1 dx= r^n\omega_n,\quad \int_{\partial B_r(x_0)}1 dS\xi= r^{n-1}\omega_n\cdot n$$where $\omega_n=\frac{\pi^{n/2}}{\Gamma(\frac n2+1)}$ is the volume of the unit ball.