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Volume integral

  1. Jan 31, 2006 #1
    I have this vector function:
    [tex]\vec V = xe^{-r}\hat i[/tex]

    I have to obtain the volume integral:
    [tex]I = \int(\vec \nabla \cdot \vec V)d^3x[/tex]

    What is that 'r' and how do I compute the volume integral?
     
  2. jcsd
  3. Jan 31, 2006 #2

    Tide

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    r is the distance from the origin: You start by evaluating the divergence of the vector. Offhand, I would recommend using spherical coordinates after evaluating the divergence.
     
  4. Jan 31, 2006 #3
    Thanks, Tide!
    So [itex]r = \sqrt{x^2 + y^2 + z^2}[/itex]. I have to substitute this in the original vector expression right?
     
  5. Jan 31, 2006 #4

    Hootenanny

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    I would say so.
     
  6. Jan 31, 2006 #5

    Meir Achuz

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    I assume the integral is over all space.
    If you know the divergence theorem, you can use it to show that the integral=0.
     
  7. Feb 16, 2006 #6
    Sorry, took me a while to get back to this problem.
    Yes, the integral is over all space.
    The divergence comes out to be:
    [tex]\vec \nabla \cdot \vec v = \frac{-x^2 e^{-r}}{r} + e^{-r}[/tex]
    I have to integrate this over a sphere of radius R. The volume integration seems too cumbersome. However, if I consider the surface integral, how do I choose the area element [itex]d\vec a[/itex]?
     
  8. Feb 16, 2006 #7

    Physics Monkey

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    A typical choice for the surface of integration would be a spherical shell of radius R in the limit as R goes to infinity (since the shell must include all space). You should be able to convince yourself that the surface integral is zero without actually calculating it.
     
  9. Feb 17, 2006 #8
    I am sorry, I don't get it :frown: . Suppose I take the surface integral as S.
    [tex]S = \int \vec V \cdot d\vec a[/tex]
    As you explained, the area element for the shell would be [itex]4\pi r^2 dr[/itex]. This means, I have to convert [itex]\vec V[/itex] into spherical coordinates, right?
     
  10. Feb 17, 2006 #9

    dextercioby

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    Note that your vector field vanishes when evaluated at infinity, so does its flux thrugh the closed surface at infinity...

    Daniel.
     
  11. Feb 17, 2006 #10
    Ok, so that means, e-r vanishes with r tends to infinity?
     
  12. Feb 17, 2006 #11

    dextercioby

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    Of course it does. As Tide said, there's no need to use Gauss's formula if u need to calculate it explicitely. Do the divergence first, then switch to spherical coordinates and then do the integral.

    Daniel.
     
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