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:
\oint d\vec{S}= \oint \vec{n}\cdot \vec{n}dS[/itex] (and my &quot;dS&quot; is your &quot;da&quot;). Apparently not, because in that case, your &quot;A&quot; below is just the the normal vector \vec{n} itself, not a constant vector and \0\int d\vec{S}= 0 is not true!<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> <h2>Homework Equations</h2><br /> <br /> Gauss&#039;s theorem: \int_{V}\nabla.\vec{A}dv= \oint_{s}\vec{A}.\vec{n}da<br /> <br /> <h2>The Attempt at a Solution</h2><br /> 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<br /> <br /> Please tell me if it is wrong. Is there a better solution? </div> </div> </blockquote>

 

[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.