Hello I'm trying to understand the following "proof" from E&M Griffiths where $V$ is the potential, for an integral over a spherical surface \begin{align} V_{\text{ave}}(R) &= \frac{1}{4 \pi R^{2}} \int V(\boldsymbol{r}) \ da \\ &= \frac{1}{4 \pi} \int V(R,\theta,\phi) \ \sin \theta \ d\theta \ d\phi \\ \frac{d V_{\text{ave}}}{d R} &= \frac{1}{4\pi } \int \frac{\partial V}{\partial R} \sin \theta \ d\theta \ d\phi \tag{1}\\ &= \frac{1}{4\pi} \int (\nabla V \cdot \boldsymbol{\hat{r}}) \ \sin \theta \ d\theta \ d\phi \tag{2}\\ &= \frac{1}{4\pi R^{2}} \int(\nabla V) \cdot (R^{2} \ \sin \theta \ d\theta \ d\phi \ \boldsymbol{\hat{r}}) \tag{3}\\ &= \frac{1}{4\pi R^{2}} \int (\nabla V ) \cdot d \boldsymbol{a} \\ &= \frac{1}{4 \pi R^{2}} \int (\nabla^{2}V) \ d\tau =0 \end{align}
Elements of confusion
In $(1)$ I'm not sure to understand what what allows me to bring the derivative inside the integral.
In $(2)$ I don't understand at all the equality $$ \frac{\partial V}{\partial R} = (\nabla V \cdot \boldsymbol{\hat{r}}),$$ I suspect this is because the parametrization is with respect to $\theta $ and $\phi$ , such that the extra variable $R$ is the normal vector with respect to the surface we're integrating over, hence the dot product ?
In $(3)$ Why does the $R^{2}$ and $\boldsymbol{\hat{r}}$ suddenly appear in the $d\boldsymbol{a}$ integral element ?
This is a direct application of the Leibinz Rule
In spherical coordinates
$$ \nabla V = {\partial V \over \partial R}\hat{\mathbf r} + {1 \over r}{\partial V \over \partial \theta}\hat{\boldsymbol \theta} + {1 \over r\sin\theta}{\partial V \over \partial \varphi}\hat{\boldsymbol \varphi} $$
if you multiply both sides by $\hat{\mathbf r}$ you will get
$$ \hat{\mathbf r} \cdot \nabla V = (\hat{\mathbf r} \cdot \hat{\mathbf r}) \frac{\partial V}{\partial R} = \frac{\partial V}{\partial R} $$