Why is the normal vector $\nu= \frac{y-x}{r} $ for a unit ball in the divergence theorem

Question

In Evan's textbook(Mean-value formulas for Laplace's equation) If $u\in C^2(U)$ is harmonic, then $$u(x)=\bar{\int} udS$$
where $\bar{\int}$ is the average integral.

Why
$$\bar{\int}_{\partial B(x,r)} Du(y)\frac{y-x}{r} dS(y)\\
=\bar{\int}_{\partial B(x,r)} \frac{\partial u}{\partial\nu} dS(y)\\=\frac{r}{n}\bar{\int}_{ B(x,r)} \Delta u(y)dy$$

where $Du\cdot \nu=\frac{\partial u}{\partial\nu}$ and by the divergence theorem.

But why the normal vector $\nu$ is the $\frac{y-x}{r} $ in the divergence theorem?

Answer 1

A diagram will help.

It is 2D but for any point in a ball exists a section like this. Consider too that $\left|y-x\right|=r$.