Let $B_R\subseteq\mathbb{R}^n$ be an open ball with radius $R>0$ centered at $0$ and $f\in C^0\left(\overline{B_R}\to\mathbb{R}\right)$ be a radial function, i.e. $f(x)=f(r)$ with $r:=\left\|x\right\|_2$ being the euclidean distance of $x$.
Moreover, let $u\in C^2\left(\overline{B_R}\to\mathbb{R}\right)$ be the (unique and radial) solution of $$\left\{\begin{matrix}-\Delta u&\equiv &f&\text{in}&B_R\\u&\equiv&0&\text{in}&\partial B_R \end{matrix}\right.\tag{ED}$$ Now, let $\varphi\in C^2\left(\overline{B_R}\to\mathbb{R}\right)$ with $\varphi\equiv 0$ in $\partial B_R$. I really would like to exactly understand why $(1)$ and $(2)$ hold in
\begin{equation} \begin{split} \int_{B_R}f\varphi\;&\stackrel{(\text{ED})}{=}-\int_{B_R}\Delta u\varphi\\ &\stackrel{(1)}{=}\int_{B_R}\nabla u\nabla\varphi-\int_{\partial B_R}\frac{\partial u}{\partial \nu}\underbrace{\varphi}_{=0}\\ &\stackrel{(2)}{=}-\int_{B_R} u\Delta\varphi+\int_{\partial B_R}\underbrace{u}_{=0}\frac{\partial\varphi}{\partial\nu}\\ &=-\int_{B_R}u\Delta\varphi \end{split} \end{equation} where $\frac{\partial}{\partial\nu}$ denotes a normal derivative (and I don't understand what $\nu$ exactly is).