# Question regarding surface and volume integral

1. Aug 10, 2013

### yungman

I have two question:

I use $\vec r$ for $\vec x$ and $\vec r'$ for $\vec x'$ in the article.
$$\vec F(\vec r)=\frac {1}{4\pi}\nabla\left[\int_{v'}\nabla'\cdot\left(\frac{\vec F(\vec r')}{|\vec r-\vec r'|}\right)d\tau'-\int_{v'}\frac{\nabla'\cdot \vec F(\vec r')}{|\vec r-\vec r'|}d\tau'\right]+\frac {1}{4\pi}\nabla\times\left[-\int_{v'}\nabla'\times\left(\frac{\vec F(\vec r')}{|\vec r-\vec r'|}\right)d\tau'+\int_{v'}\frac{\nabla'\times \vec F(\vec r')}{|\vec r-\vec r'|}d\tau'\right]$$

(1) it said
$$\frac {1}{4\pi}\nabla\int_{v'}\nabla'\cdot\left(\frac{\vec F(\vec r')}{|\vec r-\vec r'|}\right)d\tau'=\int_{s'}\left(\frac{\vec F(\vec r')}{|\vec r-\vec r'|}\right)\cdot d\vec s'$$
where this term goes to zero as $|\vec r-\vec r'|\rightarrow\;\infty$. But not the $\int_{v'}\frac{\nabla'\cdot \vec F(\vec r')}{|\vec r-\vec r'|}d\tau'$. Why is this true?

(2)Also it said using a version of Stokes theorem to change the third term $\int_{v'}\nabla'\times\left(\frac{\vec F(\vec r')}{|\vec r-\vec r'|}\right)d\tau'$ to a surface integral and it will goes to zero as $|\vec r-\vec r'|\rightarrow\;\infty$. I don't know a version of Stokes theorem that change from a volume integral to surface integral.

Thanks

2. Aug 10, 2013

### SteamKing

Staff Emeritus
3. Aug 10, 2013

### yungman

Thanks for the reply. I really don't understand manifolds. Can you write out what $\int_{v'}\nabla'\times\left(\frac{\vec F(\vec r')}{|\vec r-\vec r'|}\right)d\tau'$ equal to and explain how that become zero?

Thanks

4. Aug 10, 2013

### SteamKing

Staff Emeritus
Forget manifolds. In the Wiki, imagine that Ʃ refers to a surface S and that ∂Ʃ refers to a closed curve C, just like what is shown in the pitcher.

I'm not going to try to explain why your integral goes to zero. To be polite about it, that's your task, and I don't pretend to understand fully your original problem.

5. Aug 10, 2013

### Djokara

I'll try to answer your questions, but have in mind that English is not my native language.
1)Volume integral doesn't have to go to zero, it could be but doesn't have to. It could be just some constant, and derivative of a const is 0.
2) I'm not completely sure about this one, but I would suggest that you try to express interger as a divergence of some other function and its easy from that point to make from it surface integral.

6. Aug 10, 2013

### yungman

Then
$$\iint_{\Sigma} \nabla \times \mathbf{F} \cdot d\mathbf{\Sigma} = \oint_{\partial\Sigma} \mathbf{F} \cdot d \mathbf{r}$$
is just
$$\int_{s'}\nabla\times\vec F\cdot d\vec s'=\oint_{c}\vec F\cdot d\vec l$$
That is just simple Stokes theorem from a closed surface to a closed line integral!!! But in my question, it is transforming from a VOLUME integral to a surface integral using a "generalized Stoke Theorem"!!! That is how to transform the following to a surface integral and become zero as the article indicates:
$$\int_{v'}\nabla'\times\left(\frac{\vec F(\vec r')}{|\vec r-\vec r'|}\right)d\tau'$$

Regarding to surface integral equal to zero as $|\vec r-\vec r'|\rightarrow\;\infty$:
$$\frac {1}{4\pi}\nabla\int_{v'}\nabla'\cdot\left(\frac{\vec F(\vec r')}{|\vec r-\vec r'|}\right)d\tau'=\int_{s'}\left(\frac{\vec F(\vec r')}{|\vec r-\vec r'|}\right)\cdot d\vec s'$$
where this term goes to zero as $|\vec r-\vec r'|\rightarrow\;\infty$.

I am thinking the reason this goes to zero because the frunction $\left(\frac{\vec F(\vec r')}{|\vec r-\vec r'|}\right)$ goes to zero as $|\vec r-\vec r'|\rightarrow\;\infty$. Then the divergence of zero is zero.

But the $\int_{v'}\frac{\nabla'\cdot \vec F(\vec r')}{|\vec r-\vec r'|}d\tau'$ is not zero because as long as the volume contain the function, no matter if the volume extends to infinity, its still contain the function. As long as the function is not zero, the volume integral is not zero no matter how big the volume extends to.

Am I correct? I really just want to follow the proof of Helmholtz theorem. I've been searching through a lot of articles and even went on youtube. There is always something very funny about the derivation here and there. Most I understand, but not agree.

Last edited: Aug 10, 2013
7. Aug 10, 2013

### Djokara

Regarding to surface integral equal to zero as $|\vec r-\vec r'|\rightarrow\;\infty$:
$$\frac {1}{4\pi}\nabla\int_{v'}\nabla'\cdot\left(\frac{\vec F(\vec r')}{|\vec r-\vec r'|}\right)d\tau'=\int_{s'}\left(\frac{\vec F(\vec r')}{|\vec r-\vec r'|}\right)\cdot d\vec s'$$
where this term goes to zero as $|\vec r-\vec r'|\rightarrow\;\infty$.

I am thinking the reason this goes to zero because the frunction $\left(\frac{\vec F(\vec r')}{|\vec r-\vec r'|}\right)$ goes to zero as $|\vec r-\vec r'|\rightarrow\;\infty$. Then the divergence of zero is zero.

I don't think this is true. This function is some vector field. And surface integral of some vector field is not 0 if force lines of that field go through that surface. And in this case this surface is in ∞ so it contains whole space so there is no interceptions between force lines of this function and surface. And that's the reason why this integral is 0.

8. Aug 10, 2013

### SteamKing

Staff Emeritus
I think the point of the Stokes transformation was a reduction in the order of the calculation, to move from evaluating the vector functions over a volume to evaluating them over a surface instead. In a way, it is analogous to using Green's Theorem in the Plane to convert a double integral evaluated over some plane region into an equivalent line integral which can be evaluated over the boundary of the plane region. It's advantageous because it is easier to describe a curve than a 2-dimensional region.

9. Aug 10, 2013

### yungman

Yes, I understand this, problem is in the article, the equation is a volume integral, how can you reduce
$$\int_{v'}\nabla'\times\left(\frac{\vec F(\vec r')}{|\vec r-\vec r'|}\right)d\tau'$$
to a surface integral?

10. Aug 10, 2013

### yungman

I don't think so either. I am just making a wild guess as I totally run out of ideas. Steamking want me to at least try, so I just try!!! $\vec F(\vec r)$ is not specified like electric field $\vec E=\frac {q}{4\pi\epsilon_0 r^2}$ that I can say something about $\vec E\propto \frac 1 {r^2}$. Not to mention even surface integral of the divergence of E is only proportion to 1/r only. So I have absolutely no idea.

Last edited: Aug 11, 2013
11. Aug 11, 2013

### yungman

I found the solution of this one:
$$\int_{v'}\nabla'\times\left(\frac{\vec F(\vec r')}{|\vec r-\vec r'|}\right)d\tau'=\int_{s'}\left(\frac{\vec F(\vec r')}{|\vec r-\vec r'|}\right)\times d\vec s'$$

It just happened it is one of the exercise problem in Griffiths and I know how to derive the equation already.

So now, the only remain question is how do I proof the two surface integral is zero when extend the surface to infinity. As I explained in the last post, I don't know $\vec F$ whether it is proportion to $\frac {1}{r^2}$ like the electric field, even $\int_{v'}\nabla\cdot\vec E \;dv'=\int_{s'}\vec E\cdot d\vec s'=\frac {\rho}{\epsilon}$ which is not zero no matter how far the surface extends to. I have no idea why the integral is zero at infinity.

Thanks

Last edited: Aug 11, 2013