Suppose we are in $\mathbb{R}^n$, and denote a point in the space by $x=(x_1 , \ldots , x_n$). I'm trying to calculate the intgral
$I= \int_{\partial B_1 \cap \{x_1 \geq 0 \} } x_1 d \sigma,$
where $\sigma$ is the surface measure, and $B_1$ is the unit ball.
I tried to evaluate the integral using diveregence theorem, but I'm not sure about validity of my result. My result is: $ \int _{B_1 \cap \{x_1 \geq 0 \} } \frac{x_1}{r} dV$, where $r$ is the Eucliean distance $r=(x_1^2 + \ldots x_n^2)^{1/2}$. I also not able to calculate the last integral explicitely (my result is: $I=\frac{2^{n-2}}{n(n-1)}W_0 W_1 \ldots W_{n-3}$, where $W_j= \int _0 ^{\pi /2 } \sin ^j (x) dx$ ).
First of all, by symmetry this integral must be $0$. But if you want to use the divergence theorem, you get: $$\int_{\partial B_1} x_1 \, \mathrm{d}\sigma = \int_{\partial B_1} e_1 \cdot n \, \mathrm{d}\sigma = \int_{B_1} \nabla \cdot e_1 \, \mathrm{d}V = 0,$$ where $e_1$ is the first unit vector and $n$ is the outward normal vector field, which is just $x$ in this case.