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Consider a continuous charge distribution in volume $V'$. Draw a closed surface $S$ inside the volume $V'$.

enter image description here

Consider the following multiple integral:

$$A=\iiint_{V'} \left[ \iint_S \dfrac{\cos(\hat{R},\hat{n})}{R^2} dS \right] \rho'\ dV' =4 \pi\ m_s$$

where

$R=|\mathbf{r}-\mathbf{r'}|$

$\mathbf{r'}=(x',y',z')$ is coordinates of source points

$\mathbf{r}=(x,y,z)$ is coordinates of field points

$\cos(\hat{R},\hat{n})$ is the angle between $R$ and normal to surface element

$\rho'$ is the charge density and is continuous throughout the volume $V'$

$m_s$ is the total charge inside surface $S$


Also consider the following multiple integral:

$$B= \iint_S \left[ \iiint_{V'} \dfrac{\cos(\hat{R},\hat{n})}{R^2} \rho'\ dV' \right] dS$$

where the symbols have the meanings stated above.

\begin{align} B &= \iint_S \left[ \iiint_{V'} \rho' \dfrac{\hat{R} \cdot \hat{n}}{R^2} \ dV' \right] dS\\ &=\iint_S \left[ \iiint_{V'} \rho' \dfrac{\hat{R} }{R^2} \ dV' \right] \cdot \hat{n}\ dS\\ &=\iint_S \mathbf{E} \cdot \hat{n}\ dS \end{align}


Is $A=B\ ?$

i.e. Is interchanging the order of surface and volume integration valid? I know it is usually valid but my doubt is due to the following reasons:

  1. In the surface integral of equation $A$, when $\mathbf{r'} \in S$, we can only use spherical coordinate system with origin at point $\mathbf{r'}$ (in order to avoid improper integral with limits). So while computing $A$, we cannot use only one coordinate system. Instead, we have to use infinitely many coordinate systems.
  2. In the volume integral of equation $B$, for all $\mathbf{r}$, i.e. for all $\mathbf{r} \in S$, we can only use spherical coordinate system with origin at point $\mathbf{r}$ (in order to avoid improper integral with limits). So while computing $B$, we cannot use only one coordinate system. Instead, we have to use infinitely many coordinate systems.

Edit:

I know $\int \left[\int f(x,y)\,dx \right]dy = \int \left[\int f(x,y)\,dy \right]dx$ is true usually. Also, if in the diagram, if the volume $V'$ is contained within the surface $S$, then it is valid to change the order of integration. But here the issue is a little different. The surface $S$ is inside the volume $V'$ (please have a look at my diagram) and thus improper integral comes into play.

While computing $A$, if we need to avoid improper integrals, we have no choice except to work with infinitely many spherical coordinate systems each having their origin at points $\in V'$.

Similarly while computing $B$, if we need to avoid improper integrals, we have no choice except to work with infinitely many spherical coordinate systems each having their origin at points $\in S$.

Then how is it valid to change the order of integration in this situation? That is, how can $A=B?$

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Yes, it is valid since the once a parametrization/coordinate system is chosen for both the surface and the volume, their integration bounds don't mix, if you treat the $r$ and $r'$ coordinates as really existing in $\mathbb{R}^{2n}$. Hence Fubini's theorem applies, as long as the integrals don't diverge somewhere.

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  • $\begingroup$ $(1)$ You said "once a parametrization/coordinate system is chosen for both the surface and the volume..." Please carefully look at the two reasons for my doubt (last two para in my question). I am not using one coordinate system. In order to avoid improper integral with limits and cavities, I am forced to use infinitely many different coordinate systems while computing both $A$ and $B$. $\endgroup$ – Joe Sep 22 '19 at 15:51
  • $\begingroup$ @Joe your "infinitely many spherical coordinates" comment makes no sense since that's how divergence theorem works on a theoretical level. Really think of this integral taking place in a higher dimensional fictional space, you will see the $r$ and $r'$ coordinates do split. $\endgroup$ – Ninad Munshi Sep 24 '19 at 9:03
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    $\begingroup$ @Joe Another example: $\int_0^1 \int_0^x f(x,y)dydx$ also integrates over infinitely many coordinate systems where the origin for $x$ changes location if you were to interpret it as two integrals on $\mathbb{R}$ with one dummy variable. But there is no confusion here: as long as the function is locally integrable on $\mathbb{R}^{2n}$, everything works out. In other words, don't overthink your procedure. If there is a simple, equivalent way to think of a problem, run with it. $\endgroup$ – Ninad Munshi Sep 24 '19 at 9:06
  • $\begingroup$ Your example helped me to have an intuitive feel of your idea. In order to rigorously learn these ideas which path shall I go? I am currently taking online real analysis course. After this what course shall I take (books or videos) to rigorously clear these ideas? Please give a strong/helpful suggestion. $\endgroup$ – Joe Sep 24 '19 at 9:41
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    $\begingroup$ @Joe It might be, but I'd like to know exactly how it is. $\endgroup$ – Geremia Sep 25 '19 at 16:07

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