[Math] How to evaluate $\nabla \frac 1r$, $\nabla^2 \frac 1r$, $\int_S \nabla \frac 1r . ndS$

divergence-operatorvector analysis

Let S be a smooth closed surface in a three-dimensional xyz-space, n, be the unit outward normal vector on S, and r be the distance between the origin and a point (x, y, z). Solve the following problems.

(1) Evaluate $\nabla (\frac 1r)$

(2) Evaluate $\nabla^2 (\frac 1r)$

(3) Evaluate the integral
$\quad \int_S \nabla (\frac 1r).ndS$

when the origin is located outside S.

(4) Evaluate the integral
$\quad \int_S \nabla (\frac 1r).ndS$

when the origin is located intside S.

First of all, I completely have no idea on what topic this problem is about.
I have tried learning divergence, differential, projection from a point to a plane, but nothing seems to be applicable to solve this problem.

Edited on 30th Oct 2018

Here is my current solution

$r = \sqrt{(x^2 + y^2 + z^2)}$

$\frac 1r = \frac 1 {\sqrt{x^2 + y^2 + z^2}}$

Answer No. 1

$\nabla (\frac 1r) = {\frac {\partial {\frac 1r}}{\partial x}} + {\frac {\partial {\frac 1r}}{\partial y}} + {\frac {\partial {\frac 1r}}{\partial z}} $

$\nabla (\frac 1r) = – \frac {x \hat i + y \hat i + y \hat k}{{(x^2 + y^2 + z^2)}^\frac 32} $

$\nabla (\frac 1r) = – \frac 1{r^3} {<x,y,z>}$

Answer No. 2

$ \nabla . \nabla \frac 1r = div \nabla \frac 1r = \nabla^2 \frac1r $

$ \nabla^2 \frac1r = 0$

This is as far as I've got
Thank you for all the helps and hints! But still I couldn't find solution to next question.

I know that on question 4 we could apply divergence theorem, but
how about question 3 when the origin is outside S?

I assume we can't apply the divergence theorem the region is not within the closed surface anymore. Is that right?

Any kind of help would be gladly appreciated!

Best Answer

Here are some hints to help you get started:

  • It seems that you think $r$ is a scalar value, but it is not in the context of this problem. $r:\mathbb{R}^3\rightarrow\mathbb{R}$ is a function that maps a point $(x,y,z)\in\mathbb{R}^3$ to a distance $\sqrt{x^2+y^2+z^2}\in\mathbb{R}$. In particular, $r$ is a hypercone in $\mathbb{R}^4$.
  • The notation $\nabla f(x)$ refers to the gradient vector of $f$, $\nabla f(x)=\langle f_x,f_y,f_z\rangle$ where $f_x$ is the partial derivative of $f$ with respect to $x$, etc...
  • $\int_sf(x,y,z)dS$ is the surface (in this case closed) integral of $f$ over $S$.
  • The divergence theorem states that $\iiint\limits_V(\nabla\cdot F)dV=\oint_S(F\cdot \hat{n})dS$
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