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I would like to apologise for the question that I am going to pose; and for my curiosity.

Q: How to evaluate the integral below of the function $E(x, y, z) $over $\Omega$ which is the surface of a ball or sphere. Suppose that the radius of sphere is 1; and that it is set at the origin. Define:

$$\vec{E} = \frac{1}{4\pi \varepsilon_0 \varepsilon} \frac{Q} {r^3} \frac{\vec{r}}{r} $$

Gauss law in physics is based on the divergence theorem, stating that $$\iint_\Omega \vec{E} d\vec{S} = \frac{Q}{\varepsilon_0}$$ where $\vec{E} $ is the intensity of the electrical field defined by the Coulomb law as $\vec{F} = q\vec{E} $. This question is all about math. From this, we get the definition of DIV by the limit operation: $$\operatorname{div} \vec{F} = \lim_{\varepsilon \to 0} \frac{1}{V_\varepsilon}\iint_{\Omega_\varepsilon} \vec{F} d\vec{S} $$

sergei ivanov
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    Don't apologize for being curious! However I was always wondered but avoided the question as to how to evaluate the flux by ordinary measures? Is this impossible due to having to take into account all directions and distances? I feel that is tough but possible. Any hint? seems more a physics question than a math one. – mathcounterexamples.net Jan 12 '21 at 10:16
  • @mathcounterexamples.net Do you think? My question is all about integration. In physics, they would say that there exists the DIV theorem, and skip all mathematics to investigate physics phenomenon in which they are interested. In contrast to them, I would like not skip math and consider this tricky integration process. Hopefully, you comment might mean that I must restate Q as a typical problem integration. Then attach my attempts. Is that right? – sergei ivanov Jan 12 '21 at 10:25
  • When you say Is this impossible due to having to take into account all directions and distances?, it looks like to me a question about the possibility / impossibility in term of doing practical measures. No? – mathcounterexamples.net Jan 12 '21 at 10:27
  • @mathcounterexamples.net no, just direct integration. I would ask in engineering stack exchange about numerical computation - questions like how do I measure the thermal resistivity of a wall at home. I will keep your thinking way in mind when asking further questions if they arise. Regarding your comment, I meant the pure computation of the flux of the field E through sphere of radius R to get the result stated in the Gauss theorem. I feel I must restate Q to make it more mathematical! Thank you again. – sergei ivanov Jan 12 '21 at 10:35
  • Yeah, maybe writing the question as a pure math question and adding the context (from physics) at the end would be better. – mathcounterexamples.net Jan 12 '21 at 10:36
  • @mathcounterexamples.net unfortunately, I need extra time to work on the question that I am trying to ask because I must define all things as functions. So, I should delete Q and edit it later. Now it is a bad Q :( I see that my Q is very trivial and makes no sense at this moment... Differentiang the domain of Integration wrt time by means of Reynolds transport theorem I might get mathematical equation of charge conservation and so equation of continuity. – sergei ivanov Jan 12 '21 at 10:57
  • Is the working not given in the book? I mean how the divergence theorem is applied to $\vec{E}$ to get to the flux over unit sphere? – Math Lover Jan 12 '21 at 11:00
  • @MathLover I do not read books and the question has been in my mind for a long ime since learning the physics about Gauss law. The lecturer said that the integration directly is a tough math problem. Hence, they apply DIV theorem and skip evaluation of surface integral – sergei ivanov Jan 12 '21 at 11:09

2 Answers2

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As per divergence theorem,

$\displaystyle \iint_\Omega \vec{E} \cdot d\vec{S} = \iiint_V div(\vec{E}) \ dV$

If we take a single point charge at the origin, $\displaystyle \vec{E} = \frac{1}{4\pi \varepsilon_0} \frac{q} {r^2} \hat{r} \ $ where $\hat{r} = \frac{\vec{r}}{r}$ is a unit normal vector.

Now the outward unit normal vector for a sphere of radius $1$ centered at the origin is $\hat{r}$.

So the surface integral $\iint_\Omega \vec{E} \cdot d\vec{S} = \displaystyle \int_0^{2\pi} \int_0^{\pi} \frac{q}{4\pi \varepsilon_0} \hat{r} \cdot \hat{r} \sin \phi \ d\phi \ d\theta$

$ = \displaystyle \int_0^{2\pi} \int_0^{\pi} \frac{q}{4\pi \varepsilon_0} \sin \phi \ d\phi \ d\theta = \frac{q}{\varepsilon_0}$

If you are applying divergence theorem, in spherical coordinates,

$\displaystyle \iint_\Omega \vec{E} \cdot d\vec{S} = \frac{q}{4 \pi \varepsilon_0} \iiint_V \nabla \cdot \big(\frac{\vec{r}}{r^2}\big) \ dV = \frac{q}{4 \pi \varepsilon_0} \int_0^{2 \pi} \int_0^{\pi} \int_0^1 \nabla \cdot \big(\frac{\vec{r}}{r^2}\big) \ r^2 \ \sin \phi \ dr \ d\phi \ d\theta$

$\displaystyle = \frac{q}{\varepsilon_0}$

EDIT: one of the things I forgot to add in my original answer as to how the volume integral comes to that when the vector field has singularity at the origin. In fact the whole contribution to the flux is from the origin and it is zero everywhere else. For that you have to refer to Dirac Delta function. Let me know if you need more details on that.

Math Lover
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  • Please see my edit, something I missed to mention in my answer yesterday. – Math Lover Jan 13 '21 at 08:58
  • would you be so kind as to provide me with more details on your edit, in which you talk about the need to use Delta function? – sergei ivanov Jan 13 '21 at 10:35
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    When we did the surface integral, we got $\frac{q}{\varepsilon_0}$. But when we do the volume integral using divergence, can you see that a vector field of type $\frac{\vec{r}}{r^2}$ is not defined at the origin $(r = 0)$? There is divergence at the origin but we cannot calculate it by taking derivative as we usually do. This is derived using Dirac Delta function. $\nabla \cdot \big(\frac{\vec{r}}{r^2}\big) = 4 \pi \ \delta^3(r)$ and the way it is defined, volume integral of $\delta^3(r) = 1$. That is how we get to the result using divergence. – Math Lover Jan 13 '21 at 11:47
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    Splendid! I have understood your answer and comment ; nothing remains unclear. So, much obliged! Be happy! – sergei ivanov Jan 13 '21 at 13:34
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Your formula for $E$ has one too many factors of $1/r$: you can either write $\frac{Q}{4\pi\varepsilon_0r^2}\frac{\vec{r}}{r}$ or $\frac{Q}{4\pi\varepsilon_0r^3}\vec{r}$. In terms of infinitesimal solid angle $d\Omega$, for a radius-$r$ circle centred on the origin$$d\vec{S}=r\vec{r}d\Omega\implies\vec{E}\cdot d\vec{S}=\frac{Q}{4\pi\varepsilon_0}d\Omega\implies\int\vec{E}\cdot d\vec{S}=\frac{Q}{\varepsilon_0}.$$

J.G.
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  • That is helpful. Is the idea applicable to the general case where the integration is not over sphere but over another smooth surfaces that encloses a certain volume? – sergei ivanov Jan 12 '21 at 11:06
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    @sergeiivanov Since $\vec{nabla}\cdot\vec{E}=\rho/\varepsilon_0$, a closed surface $\partial X$ satisfies $\int_{\partial X}\vec{E}\cdot d\vec{S}=\varepsilon_0^{-1}\int_X\rho d^3x$, the last integral being the enclosed charge. So $\int\vec{E}\cdot d\vec{S}=Q/\varepsilon_0$ works there too. – J.G. Jan 12 '21 at 11:10