Divergence of a radial $1/r^2$ vector field How to obtain the divergence of the function $F(r,\varphi,\theta)=\hat{r}/r^2$ where $\hat r$ is the unit vector in radial direction? Is there a solution without computing the surface integral for definition of divergence?
The issue is: in this divergence, the delta function will be present. But if you obtain divergence from formula $\nabla\cdot F$, that is equal to zero. At the  point $r=0$,  this formula cannot be used.
 A: First off, you mean to say $f(r) = \hat r / r^2$ since as you've noted, you know the divergence is zero when $r \neq 0$ since $$\nabla \cdot f(r) \hat r =  \frac{1}{r^2} \frac{\partial  }{\partial r} ( r^2 f(r)  )$$
but let's try to under stand what happens when $r \to 0$. Since the function doesn't make sense here, let's try to understand this in a weak sense (or the sense of distributions, a surface integral will come about only when we start trying to make sense of this). Personally, I like to notice that $ -\nabla \frac{1}{r} = f(r)$, i.e. $\Delta \frac{1}{r}$ is the same as $\nabla \cdot f(r)$ (where $\Delta$ is the Laplacian operator). Let $\phi \in C^\infty$ be a nice test function, and consider the integral over the ball $B = \{ x \in \mathbb{R}^3 : ||x|| < \epsilon \}$.
$$ \int _B  \phi(x) \Delta \left (  \frac{1}{r} \right ) dx $$
Using Green's identity ( i.e . integration by parts), you may check that
$$ \int _B  \left [ \phi(x) \Delta \left (  \frac{1}{r} \right )  - \frac{1}{r} \Delta \phi \right ]dx = \oint_{\partial B} \left [ \phi \frac{\partial }{\partial r} \frac{1}{r} - \frac{1}{r} \frac{\partial \phi }{ \partial r} \right ] dS$$
Note that $dx = r^2 \sin \theta dr  \, d\theta  \, d \varphi$, thus
$$ \int _B  \phi(x) \Delta \left (  \frac{1}{r} \right ) dx= \underbrace{ \int_B \frac{1}{r} \Delta \phi dx }_{ \mathcal{O}(\epsilon)}+ \oint_{\partial B} \left [ \underbrace{\phi \frac{\partial }{\partial r} \frac{1}{r} }_{\mathcal{O}(1)}- \underbrace{\frac{1}{r} \frac{\partial \phi }{ \partial r}}_{\mathcal{O}(\epsilon)} \right ] dS$$
Now if we take the limit as the ball shrinks to the origin (the bad point) $\epsilon \to 0$, we see terms of order $\epsilon$ die, and we're left with
$$ \lim_{\epsilon \to 0} \int _B  \phi(x) \Delta \left (  \frac{1}{r} \right ) dx = \lim_{ \epsilon \to 0} \oint_{\partial B} \phi \frac{\partial }{\partial r} \frac{1}{r} dS = \lim_{ \epsilon \to 0} -\int_0^\pi \int_0^{2 \pi } \phi(x) \sin \theta d \theta d \varphi = - 4 \pi \phi(0)$$
Thus some people like to write
$$ \Delta \frac{1}{r} = - 4 \pi \delta (r) $$
