Let r = (x, y, z) and $r_ε = \sqrt {x^2+y^2+z^2+ε} $ for any number $ε \ge0$ fε : R 3 → R is the scalar field fε(r) = $1\over r_ε $
Use the Divergence Theorem to prove that
$$lim_{ε→0} \int \int \int_{v(R)} ∇^2 f_ε dx dy dz = −4π$$
where V (R) is the ball of radius R > 0 centred at the origin.
so far I have proved that ∇fε(r) = $-r\over r_ε^3$
and I tried to do ∇^2 fε and got $3r^2 (r_ε)^{-5}-3r_ε^{-3} $ but I don't think that is right could someone help me out. If it is right I am not sure where to go from here
$\vec{r} = (x, y, z)$ and $r_\epsilon = \sqrt{x^2+y^2+z^2+ε}, \, \epsilon \geq 0$
$\displaystyle f_\epsilon (r) = \frac{1}{r_\epsilon} = \frac{1}{\sqrt{r^2 + \epsilon}}$
We need to prove $\lim_{\epsilon→0} \iiint_{v(R)} \nabla^2 f_\epsilon \, dx \, dy \, dz = −4π$
Please note $\nabla^2 f_\epsilon$ is the divergence of $\nabla f_\epsilon \, $ (the gradient of the scalar field $f_\epsilon)$ and the surface is a ball of radius $R$ centered at $(0, 0, 0)$.
As per divergence theorem,
$\iiint_S div (\vec{F}) \, dV = \iint_S \vec {F} \cdot \hat{n} \, dS$
So, $\iiint_S \nabla \cdot \nabla f_\epsilon \, dx \, dy \, dz = \iint_S (\nabla {f_\epsilon}) \cdot \hat{n} \, dS$
$\displaystyle \nabla f_\epsilon = - \frac{\vec{r}}{(r^2 + \epsilon)^{3/2}}$
The unit normal vector is $\displaystyle \frac{\vec{r}}{r}$
So the surface integral over a ball of radius $R (\gt 0)$ centered at origin is given by,
$I = \displaystyle \int_0^{2\pi} \int_0^{\pi} - \frac{\vec{r}}{(R^2 + \epsilon)^{3/2}} \cdot \frac{\vec{r}}{R} \, (R^2 \sin \phi) \, d\phi \, d\theta$
$\displaystyle = - \frac{R^3}{(R^2 + \epsilon)^{3/2}} \int_0^{2\pi} \int_0^{\pi} \sin \phi \, d\phi \, d\theta = - \frac{4R^3\pi}{(R^2 + \epsilon)^{3/2}}$
Now as $\epsilon$ tends to $0$, you can see what happens to your surface integral.