Proving Gauss's Theorem: Closed Surface

[rado5]

Homework Statement

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

Homework Equations

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

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?

 

[rado5]

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

 

[Dick]

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

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?

 

[HallsofIvy]

Homework Statement

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

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:
[tex]\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!

Homework Equations

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

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?

 

[gabbagabbahey]

I think the main idea here is to take note of the fact that for any constant vector $\textbf{A}$, you can say

$$\int_{\mathcal{S}}\textbf{A}\cdot\textbf{n}da=\textbf{A}\cdot\int_{\mathcal{S}}\textbf{n}da$$

And from here, just apply Gauss' theorem.

 

[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?

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

 

[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 $$\int_{s}\vec{n}da= 0$$

 

[Dick]

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 $$\int_{s}\vec{n}da= 0$$

That's it.

 

[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.