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Surface Integrals

  1. Mar 17, 2010 #1
    1. The problem statement, all variables and given/known data
    Prove that [tex]\int\int_{S} n dS = 0 [/tex] for any closed surface S.


    2. Relevant equations



    3. The attempt at a solution

    I used divergence theorem. But i thought it is applicable only if there is another vector multiplied to that outward unit vector (n).

    [tex]\int\int_{S} F {\cdot} n dS[/tex]
     
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  3. Mar 17, 2010 #2

    Dick

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    The integral of ndS is a vector equation. Split it into components. n=(nx,ny,nz). What's a vector field F that might have the property that F.n=nx, for example?
     
  4. Mar 17, 2010 #3
    my problem is how do I show that the [tex]\int\int_{S} n dS = 0[/tex]

    for any closed surface S

    ----> I used the divergence theorem, but I don't think it will help me 'cause it is applicable for

    [tex]\int\int_{S} F{ \cdot} n dS [/tex]

    not for

    [tex]\int\int_{S} n dS [/tex]
     
  5. Mar 17, 2010 #4

    Dick

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    You basically just reposted the same thing. Look, if you show integral of nx*dS, ny*dS and nz*dS are zero, you are done right? What's an appropriate choice of F for each?
     
  6. Mar 17, 2010 #5
    Is there a value of n(outward unit vector)????
     
  7. Mar 17, 2010 #6

    Dick

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    No, n is the outward unit vector. Whatever that is. You don't have much control over that. You can choose F. Suppose you choose F=(1,0,0). What do you conclude?
     
  8. Mar 17, 2010 #7
    Yeah, I get what you are saying... So If I choose an F a vector the solution will lead to Divergence Theorem??? Am I right???
     
  9. Mar 17, 2010 #8

    Dick

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    You can use the divergence theorem to conclude something, yes. But what do you conclude?
     
  10. Mar 17, 2010 #9
    I will conclude that [tex]\int\int_{S} F \cdot n dS = 0 [/tex]
     
  11. Mar 17, 2010 #10

    Dick

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    Why do you conclude that and what is F.n and how does that help you conclude integral ndS is zero? You really aren't giving me much to go on except repeating the divergence theorem over and over again. You have to apply the divergence theorem in a specific way to solve a problem.
     
  12. Mar 17, 2010 #11

    [tex]\vec{n}=(\vec{i}\cdot\vec{n})\vec{i}+(\vec{j}\cdot \vec{n})\vec{j}+(\vec{k}\cdot\vec{n})\vec{k}[/tex]

    Then,

    [tex]\int\int_{S}\vec{n}dS=\int\int_{S}(\vec{i}\cdot\vec{n})dS\vec{i}+\int\int_{S}(\vec{j}\cdot\vec{n})dS\vec{j}+\int\int _{S}(\vec{k}\cdot\vec{n})dS\vec{k}[/tex]

    I stacked to this step...
     
  13. Mar 17, 2010 #12

    Dick

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    Now that's GOOD! You've got it. Ok, so what does the divergence theorem tell you about the first term where F=i? You may have known this all along. But I just couldn't figure out if you did from your posts.
     
    Last edited: Mar 17, 2010
  14. Mar 17, 2010 #13
    Divergence Theorem said that

    [tex]\int\int_{S} \vec{F} \cdot \vec{n} dS = \int\int\int_{V} \nabla \cdot \vec{F} dV [/tex]
     
  15. Mar 17, 2010 #14

    Dick

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    You did it again. You just quoted the divergence theorem. What DO YOU CONCLUDE from the divergence theorem? What's F and what's div(F)?
     
  16. Mar 17, 2010 #15
    [tex]div F = \left[\frac{\partial}{x}\vector{i} + \frac{\partial}{y}\vector{j}+\frac{\partial}{z}\vector{k}\right]\left(\vector{i} + \vector {j} +\vector{k}\right) [/tex]

    [tex]div F =\left[\frac{\partial}{x}\vector{0} + \frac{\partial}{y}\vector{0}+\frac{\partial}{z}\vector{0}\right][/tex]

    [tex]div F = 0 + 0 + 0[/tex]

    [tex]div F = 0 [/tex]

    Is this right??? I hope so...
     
  17. Mar 17, 2010 #16

    ideasrule

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    Yes, kind of. [tex]
    div F = \left[\frac{\partial}{x}\vector{i} + \frac{\partial}{y}\vector{j}+\frac{\partial}{z}\vector{k}\right]\left(\vector{i} + \vector {j} +\vector{k}\right)
    [/tex] should be [tex]
    div F = \left[\frac{\partial}{x}\vector{i} + \frac{\partial}{y}\vector{i}+\frac{\partial}{z}\vector{i}\right]\left(\vector{i} + \vector {j} +\vector{k}\right)
    [/tex] since F=i, and your next line is not correct since i is not equal to 0. However, it is true that div F=0. What do you conclude from that?
     
  18. Mar 17, 2010 #17
    Yeah.. That is 0...

    Thanks for helping and guiding me,Sir.

    Hope to guide me in my further studies..

    Thanks again... I LOVE YOU!!! LOL
     
  19. Mar 17, 2010 #18

    Dick

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    The argument looks a little strange. What is the F you are talking about? If F=i then Fx=1, Fy=0 and Fz=0. So sure, div(F)=d/dx(Fx)+d/dy(Fy)+d/dz(Fz)=0. All of the terms are zero.
     
  20. Mar 17, 2010 #19
    Is this for me????
     
  21. Mar 17, 2010 #20
    By the definition of divergence...see??
     
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