I'm seeking clarification on the following identity involving surface integrals and partial derivatives:
$B_{\rho} = B(y, \rho) \subset \mathbb{R}^n$ represents the ball centered at $y$ with radius $\rho$, and $\nu$ denotes the unit normal vector to $B_{\rho}$. In this identity, we have $r = |x-y|$, $\omega = \frac{x-y}{r}$, and $u(x) = u(y + r\omega)$.
$$ \int_{\partial B_{\rho}} \frac{\partial u}{\partial \nu} \, ds = \int_{\partial B_{\rho}} \frac{\partial u}{\partial r} (y + \rho\omega) \, ds = \rho^{n-1} \int_{|\omega|=1} \frac{\partial u}{\partial r} (y + \rho\omega) \, d\omega $$
I believe the first equality can be obtained through the following reasoning:
$$ \frac{\partial u}{\partial r} = \frac{\partial x}{\partial r} \cdot \frac{\partial u}{\partial x} = \omega \cdot \frac{\partial u}{\partial x} = \frac{\partial u}{\partial \nu} $$
since $\nu = \omega$ due to both being unit normal vectors to the surface.
However, I'm confused about the second equality, which seems to involve a change of variable. I would appreciate clarification on why $(y + \rho\omega)$ suddenly appears in the integral.