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Gauss's theorem

  1. Nov 20, 2009 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

    Gauss's theorem: [tex]\int_{V}\nabla.\vec{A}dv= \oint_{s}\vec{A}.\vec{n}da[/tex]

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

    Please tell me if it is wrong. Is there a better solution?
     
  2. jcsd
  3. Nov 20, 2009 #2
    I think in this problem [tex]\vec{A}=(1,1,1)[/tex] is it correct? So [tex]\nabla.\vec{A}=0[/tex]
     
  4. Nov 20, 2009 #3

    Dick

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    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?
     
  5. Nov 21, 2009 #4

    HallsofIvy

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    What does this mean? In Gauss's theorem, below, [itex]\vec{A}\cdot\vec{n}[/itex] is a scalar function that you are integrating over the surface of the region. Here, you are integrating a vector valued function, [itex]\vec{n}[/itex], 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 "[itex]d\vec{S}[/itex]", 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 [itex]\vec{n}[/itex] itself, not a constant vector and [itex]\0\int d\vec{S}= 0[/itex] is not true!

     
  6. Nov 21, 2009 #5

    gabbagabbahey

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

    [tex]\int_{\mathcal{S}}\textbf{A}\cdot\textbf{n}da=\textbf{A}\cdot\int_{\mathcal{S}}\textbf{n}da[/tex]

    And from here, just apply Gauss' theorem.
     
  7. Nov 21, 2009 #6

    Dick

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    Right, gabbagabbahey. In the above post I meant to say "that would show that the x component of n ds is zero". Ooops.
     
  8. Nov 21, 2009 #7
    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.

    [tex]\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)[/tex]

    [tex]\int_{s}n_{x} da= \int_{s}(\vec{i}.\vec{n})da= \int_{V}\nabla.\vec{i}dv= \int_{V}0 dv= 0[/tex]

    And now I can say that [tex]\int_{s}\vec{n}da= 0[/tex]
     
    Last edited: Nov 21, 2009
  9. Nov 21, 2009 #8

    Dick

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    That's it.
     
  10. Nov 21, 2009 #9
    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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