Let $ \vec{A}$ be a vector field
$$ \int \limits_{V} \vec{\nabla} \cdot \Phi(\vec{r}) \vec{A}(\vec{r}) d^{3} \vec{r}= \int \limits_{V}[\Phi(\vec{r}) \vec{\nabla} \cdot \vec{A}(\vec{r})+\vec{A}(\vec{r}) \cdot \vec{\nabla} \Phi(\vec{r})] d^{3} \vec{r} \underset{\substack{\uparrow \\ \text { Gauss }}}{=} \oint \limits_{S(V)} \Phi(\vec{r}) \vec{A}(\vec{r}) \cdot d \vec{S}$$
Let $ \vec{A}(\vec{r})=\vec{\nabla} \Psi(\vec{r})$ with the scalar field $\Psi(\vec{r})$.
$$\int \limits_{V}[\Phi \Delta \Psi+(\vec{\nabla} \Phi) \cdot(\vec{\nabla} \Psi)] d V=\oint_{S(V)} \Phi(\vec{\nabla} \Psi) \cdot d \vec{S} \quad $$ (1. Green’s formula)
Why does this equality hold:
$\oint \limits_{S(V)} \Phi(\vec{r}) \vec{A}(\vec{r}) \cdot d \vec{S}=\oint_{S(V)} \Phi(\vec{\nabla} \Psi) \cdot d \vec{S} \quad $ ???
I would have thought:
$\oint \limits_{S(V)} \Phi(\vec{r}) \vec{A}(\vec{r}) \cdot d \vec{S}=\oint \limits_{S(V)} \Phi(\vec{r}) \vec{\nabla} \Psi(\vec{r}) \cdot d \vec{S}$
and that $\Phi(\vec{r}) \vec{\nabla} \Psi(\vec{r})\neq \Phi(\vec{\nabla} \Psi)$