Let ##V'## be the volume of dipole distribution and ##S'## be the boundary.
The potential of a dipole distribution at a point ##P## is:
##\displaystyle\psi=k \int_{V'}
\dfrac{\vec{\nabla'}.\vec{M'}}{r}dV'
+k \oint_{S'}\dfrac{\vec{M'}.\hat{n}}{r}dS'##
If ##P\in V'## and ##P\in S'##, the integrand is discontinuous (infinite) at the point ##r=0##. If we use spherical and polar coordinates, the integrand is continuous everywhere:
\begin{align}
\psi &=\bbox[orange,5px]{k \int_{V'} \dfrac{\vec{\nabla'}.\vec{M'}}{r} r^2 \sin \theta\ d\theta\ d\phi\ dr}\\
&\bbox[pink,5px]{+k\oint_{S'_1} \dfrac{\vec{M'}.\hat{n}}{r} \sqrt{{f_x}^2+{f_y}^2+1}\ R\ dR\ d\theta'}
\bbox[yellow,5px]{+ k \oint_{S'_2}\dfrac{\vec{M'}.\hat{n}}{r}dS'}\\
&=\bbox[orange,5px]{k \int_{V'} \vec{\nabla'}.\vec{M'}\ r \sin \theta\ d\theta\ d\phi\ dr}\\
&\bbox[pink,5px] {+ k \oint_{S'_1} \vec{M'}.\hat{n}\ \sqrt{{f_x}^2+{f_y}^2+1}\ \dfrac{R}{\sqrt{R^2+f^2}} dR\ d\theta'}
\bbox[yellow,5px] {+ k \oint_{S'_2}\dfrac{\vec{M'}.\hat{n}}{r}dS'}
\end{align}
The field of a dipole distribution at a point ##P## is:
##\displaystyle\nabla\psi=k \int_{V'}
(\vec{\nabla'}.\vec{M'})\ \nabla \left( \dfrac{1}{r} \right) dV'
+k \oint_{S'} (\vec{M'}.\hat{n}) \nabla \left( \dfrac{1}{r} \right) dS'##
If ##P\in V'## and ##P\in S'##, the integrand is discontinuous (infinite) at the point ##r=0##. If we use spherical and polar coordinates:
\begin{align}
\nabla\psi&=\bbox[orange,5px]{k \int_{V'}
(\vec{\nabla'}.\vec{M'})\ \left( \dfrac{\hat{r}}{r^2} \right) r^2 \sin \theta\ d\theta\ d\phi\ dr}\\
&\bbox[pink,5px]{+ k \oint_{S'_1} (\vec{M'}.\hat{n}) \left( \dfrac{\hat{r}}{r} \right) \sqrt{{f_x}^2+{f_y}^2+1}\ \dfrac{R}{\sqrt{R^2+f^2}} dR\ d\theta'}
\bbox[yellow,5px] {+ k \oint_{S'_2}\dfrac{\vec{M'}.\hat{n}}{r}dS'}
\end{align}
The first term has the integrand continuous everywhere. The second term has the integrand discontinuous (infinite) at the point ##r=0##.
(i) Is it necessary to remove a small circle around ##R=0## in order to remove the singularity in the second term so that we can compute the integral?
(ii) If so what about the contribution to the surface integral from the point ##R=0##, i.e. ##r=0##? Should we use Dirac delta here because the integrand is singular at ##r=0## and is the contribution to the surface integral from the point ##r=0## significant?
The potential of a dipole distribution at a point ##P## is:
##\displaystyle\psi=k \int_{V'}
\dfrac{\vec{\nabla'}.\vec{M'}}{r}dV'
+k \oint_{S'}\dfrac{\vec{M'}.\hat{n}}{r}dS'##
If ##P\in V'## and ##P\in S'##, the integrand is discontinuous (infinite) at the point ##r=0##. If we use spherical and polar coordinates, the integrand is continuous everywhere:
\begin{align}
\psi &=\bbox[orange,5px]{k \int_{V'} \dfrac{\vec{\nabla'}.\vec{M'}}{r} r^2 \sin \theta\ d\theta\ d\phi\ dr}\\
&\bbox[pink,5px]{+k\oint_{S'_1} \dfrac{\vec{M'}.\hat{n}}{r} \sqrt{{f_x}^2+{f_y}^2+1}\ R\ dR\ d\theta'}
\bbox[yellow,5px]{+ k \oint_{S'_2}\dfrac{\vec{M'}.\hat{n}}{r}dS'}\\
&=\bbox[orange,5px]{k \int_{V'} \vec{\nabla'}.\vec{M'}\ r \sin \theta\ d\theta\ d\phi\ dr}\\
&\bbox[pink,5px] {+ k \oint_{S'_1} \vec{M'}.\hat{n}\ \sqrt{{f_x}^2+{f_y}^2+1}\ \dfrac{R}{\sqrt{R^2+f^2}} dR\ d\theta'}
\bbox[yellow,5px] {+ k \oint_{S'_2}\dfrac{\vec{M'}.\hat{n}}{r}dS'}
\end{align}
The field of a dipole distribution at a point ##P## is:
##\displaystyle\nabla\psi=k \int_{V'}
(\vec{\nabla'}.\vec{M'})\ \nabla \left( \dfrac{1}{r} \right) dV'
+k \oint_{S'} (\vec{M'}.\hat{n}) \nabla \left( \dfrac{1}{r} \right) dS'##
If ##P\in V'## and ##P\in S'##, the integrand is discontinuous (infinite) at the point ##r=0##. If we use spherical and polar coordinates:
\begin{align}
\nabla\psi&=\bbox[orange,5px]{k \int_{V'}
(\vec{\nabla'}.\vec{M'})\ \left( \dfrac{\hat{r}}{r^2} \right) r^2 \sin \theta\ d\theta\ d\phi\ dr}\\
&\bbox[pink,5px]{+ k \oint_{S'_1} (\vec{M'}.\hat{n}) \left( \dfrac{\hat{r}}{r} \right) \sqrt{{f_x}^2+{f_y}^2+1}\ \dfrac{R}{\sqrt{R^2+f^2}} dR\ d\theta'}
\bbox[yellow,5px] {+ k \oint_{S'_2}\dfrac{\vec{M'}.\hat{n}}{r}dS'}
\end{align}
The first term has the integrand continuous everywhere. The second term has the integrand discontinuous (infinite) at the point ##r=0##.
(i) Is it necessary to remove a small circle around ##R=0## in order to remove the singularity in the second term so that we can compute the integral?
(ii) If so what about the contribution to the surface integral from the point ##R=0##, i.e. ##r=0##? Should we use Dirac delta here because the integrand is singular at ##r=0## and is the contribution to the surface integral from the point ##r=0## significant?
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