Let $B(x,r)\subset \mathbb R^n$ the ball $\{y\in\mathbb R^n\mid \|y-x\|<r\}$, and $$\partial B(x,r)=\{r\sigma \mid \sigma \in \partial B(x,1)\},$$ it's boundary.
1) How do you get $$\int_{\partial B(x,r)}u(\sigma r)\mathrm d \sigma =r^{n-1}\int_{B(x,1)}u(y)\mathrm d y.$$ In other word, why $$\mathrm d \sigma =r^{n-1}\mathrm d y\ ?$$ There is something I don't get here.
2) Why an integration by part give us $$\int_{\partial B(r,x)}\partial _r u(y)\mathrm d y=\int_{B(r,x)}\Delta u(y)\mathrm d y\ \ ?$$ The only thing I see is that $$\int_{B(r,x)}\Delta u(y)\mathrm d y=\int_{B(r,x)}\text{div}(\nabla u(y))\mathrm d y=\int_{\partial B(r,x)}\nabla u(y)\cdot \nu \mathrm d \sigma .$$ I imagine that $$\nabla u(y)\cdot \nu =\partial _ru(y),$$ but I don't see why.
3) Same, why using an integration by part gives $$\int_{B(x,r)}\partial _i u(y)\mathrm d y=\int_{\partial B(x,r)}u\nu_i\mathrm d y\ \ ?$$
By the way, the notation $u\nu_i$ looks confuse to me, what is it exactly ?