- #1
Mike400
- 59
- 6
##\mathbf{M'}## is a vector field in volume ##V'## and ##P## be any point on the surface of ##V'## with position vector ##\mathbf {r}##
Now by Gauss divergence theorem:
\begin{align}
\iiint_{V'} \left[ \nabla' . \left( \dfrac{\mathbf{M'}}{\left| \mathbf{r}-\mathbf{r'} \right|} \right) \right] dV'
&=\unicode{x222F}_{S'} \left[ \left( \dfrac{\mathbf{M'}}{\left| \mathbf{r}-\mathbf{r'} \right|} \right) . \hat{n} \right] dS' \tag{1}\\
\end{align}
Both of these expressions, ##LHS## and ##RHS## of equation ##(1)##, have their respective integrands singular at ##\mathbf{r'}=\mathbf{r}##. Now how shall we show that both the volume and surface integration is unaffected by this singularity at ##\mathbf{r'}=\mathbf{r}##?
Now by Gauss divergence theorem:
\begin{align}
\iiint_{V'} \left[ \nabla' . \left( \dfrac{\mathbf{M'}}{\left| \mathbf{r}-\mathbf{r'} \right|} \right) \right] dV'
&=\unicode{x222F}_{S'} \left[ \left( \dfrac{\mathbf{M'}}{\left| \mathbf{r}-\mathbf{r'} \right|} \right) . \hat{n} \right] dS' \tag{1}\\
\end{align}
Both of these expressions, ##LHS## and ##RHS## of equation ##(1)##, have their respective integrands singular at ##\mathbf{r'}=\mathbf{r}##. Now how shall we show that both the volume and surface integration is unaffected by this singularity at ##\mathbf{r'}=\mathbf{r}##?