# Gauss's theorem

1. Nov 20, 2009

### rado5

1. The problem statement, all variables and given/known data

Using Gauss's theorem prove that $$\int_{s}\vec{n}ds=0$$ if s is a closed surface.

2. Relevant equations

Gauss's theorem: $$\int_{V}\nabla.\vec{A}dv= \oint_{s}\vec{A}.\vec{n}da$$

3. The attempt at a solution
In this problem $$\vec{A}$$ is constant so $$\nabla.\vec{A}=0$$ so $$\int_{s}\vec{n}ds= \int_{V}\nabla.\vec{A}dv=0$$

Please tell me if it is wrong. Is there a better solution?

2. Nov 20, 2009

### rado5

I think in this problem $$\vec{A}=(1,1,1)$$ is it correct? So $$\nabla.\vec{A}=0$$

3. Nov 20, 2009

### Dick

Not quite. The integral of nds is a vector, call it N. If you use A=(1,0,0) that would show that the x component of A is zero, right?

4. Nov 21, 2009

### HallsofIvy

Staff Emeritus
What does this mean? In Gauss's theorem, below, $\vec{A}\cdot\vec{n}$ is a scalar function that you are integrating over the surface of the region. Here, you are integrating a vector valued function, $\vec{n}$, itself, over what? Below you use "da" as a "differential of surface area". What is "ds"? Do you mean "ds" to represent the "vector differential of surface area" that I would call "$d\vec{S}$", the vector perpendicular to the surface with length the differential of surface area:
$$\oint d\vec{S}= \oint \vec{n}\cdot \vec{n}dS[/itex] (and my "dS" is your "da"). Apparently not, because in that case, your "A" below is just the the normal vector $\vec{n}$ itself, not a constant vector and $\0\int d\vec{S}= 0$ is not true! 5. Nov 21, 2009 ### gabbagabbahey I think the main idea here is to take note of the fact that for any constant vector $\textbf{A}$, you can say [tex]\int_{\mathcal{S}}\textbf{A}\cdot\textbf{n}da=\textbf{A}\cdot\int_{\mathcal{S}}\textbf{n}da$$

And from here, just apply Gauss' theorem.

6. Nov 21, 2009

### Dick

Right, gabbagabbahey. In the above post I meant to say "that would show that the x component of n ds is zero". Ooops.

7. Nov 21, 2009

### rado5

Ok, first of all I would like to thank all of you for your great generosity to make me understand this problem.

Now I think I can prove it, and this is my work.

$$\int_{s}\vec{n}da= \int_{s}(n_{x},n_{y},n_{z})da= (\int_{s}n_{x} da ,\int_{s}n_{y} da ,\int_{s}n_{z} da)$$

$$\int_{s}n_{x} da= \int_{s}(\vec{i}.\vec{n})da= \int_{V}\nabla.\vec{i}dv= \int_{V}0 dv= 0$$

And now I can say that $$\int_{s}\vec{n}da= 0$$

Last edited: Nov 21, 2009
8. Nov 21, 2009

### Dick

That's it.

9. Nov 21, 2009

### rado5

Thank you very much, you are all wonderful instructors and I do respect you a lot. I would like to donate to this wonderful site, but unfortunately in my terrible country, we have no credit cards and no western union, previously I had wanted to donate to http://ocw.mit.edu but I couldn't.

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