$\vec{E}$ is electric field
$r$ is distance between source and field points
$\hat{r}$ is a unit vector from source point to field point
$x,y,z$ are Cartesian coordinates of field point
$x',y',z'$ are Cartesian coordinates of source points
$r= \left[ (x-x')^2 + (y-y')^2 + (z-z')^2 \right]^{1/2} $
$\rho$ is uniform charge distribution density
\begin{align} \nabla \cdot \dfrac{(\hat{r})}{r^2} &=\nabla \cdot \left[ \dfrac{x-x'}{r^3} (\hat{i}) + \dfrac{y-y'}{r^3}(\hat{j}) + \dfrac{z-z'}{r^3}(\hat{k}) \right]\\ &=\dfrac{\partial}{\partial x} \left( \dfrac {x-x'}{r^3} \right) +\dfrac{\partial}{\partial y} \left( \dfrac {y-y'}{r^3} \right) +\dfrac{\partial}{\partial z} \left( \dfrac {z-z'}{r^3} \right)\\ &=0 \tag1\\ \end{align} $\text{(after some calculation, it will come out to be zero)}$
Using equation (1):
$$\nabla \cdot \vec{E} =\nabla \cdot \left[ k\ \rho \displaystyle\int_V \dfrac{dV}{r^2} (\hat{r}) \right] =k\ \rho \displaystyle\int_V \left[ \nabla \cdot \dfrac{(\hat{r})}{r^2} \right] dV =0$$
Now we do it in another way by applying Gauss divergence theorem and solid angle formula:
$\nabla \cdot \vec{E} =\nabla \cdot \left[ k\ \rho \displaystyle\int_V \dfrac{dV}{r^2} (\hat{r}) \right] =k\ \rho \displaystyle\int_V \left[ \nabla. \left( \dfrac{\hat{r}}{r^2} \right) \right] dV =k\ \rho \displaystyle \iint_S \dfrac{(\hat{r})}{r^2} \cdot \vec{dS} =4\ \pi\ k\ \rho$
Why am I getting different results?
