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Gauss's theorem to find flux of field

  1. Feb 3, 2013 #1
    1. The problem statement, all variables and given/known data
    F(x,y,z) = x^3zi + y^3zj + xyk
    across the surface x^2 + y^2 + z^2 = 9


    2. Relevant equations
    See attached image


    3. The attempt at a solution
    See attached image.

    I'm fairly certain there's either something wrong with my initial integral, also I don't think I integrated correctly with respect to phi.. but I'm not really sure how to fix it.

    According to my exercise book solutions, the answer is 0, however if I were to continue with my working out I would get 729pi.

    Any help would be greatly appreciated, thank you.

    On an unrelated note, this forum has been a great help to me in the past, and all free of charge.. is there somewhere I can make a donation? or a paid service I can sign up for? Thanks
     

    Attached Files:

    Last edited: Feb 3, 2013
  2. jcsd
  3. Feb 3, 2013 #2

    dx

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    You did the last integral wrong.

    Use the substitution u = sin φ

    In terms of u the integral becomes

    $$ \int_0^\pi \sin^3\phi \cos \phi \ d\phi = \int_0^0 u^3 du = 0 $$


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    Last edited: Feb 3, 2013
  4. Feb 3, 2013 #3

    tiny-tim

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    hi schmiggy! :smile:

    your 3x2z + 3y2z was right,

    but then you only converted the 3x2z into polar coordinates! :rolleyes:

    (btw, that's 3r2z, which is odd in z, so isn't it obvious that its integral over that sphere will be zero? :wink:)
    there used to be an upgrade button, but now you have to click "My PF", then go down to "Paid subscriptions" near the bottom :wink:
     
  5. Feb 3, 2013 #4

    dx

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    No tiny-tim, I think he did it correctly.

    3x2z + 3y2z = 3rcosφ(x2 + y2) = 3rcosφ(r2 - z2) = 3rcosφ(r2 - r2cos2φ) = 3r3cosφsin2φ

    which is what schmiggy got
     
  6. Feb 3, 2013 #5

    tiny-tim

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    oops! :redface: i was thinking of cylindrical coordinates! :confused:

    thanks, dx :smile:

    (but my comment on "odd in z" still stands :wink:)
     
  7. Feb 4, 2013 #6
    Firstly, thanks to both of you for the help!

    I got the integral to result in
    $$ \int_0^\pi \sin^4\phi/4 = 0 $$
    (sorry, not sure how to use that code to format the integral properly, basically it's sin^4(phi) multiplied by (1/4))

    I embarrassingly sat for about 2 minutes afterwards going.. "but when I integrate with respect to theta, I'll end up with 2pi..." before realising I was integrating 0. /facepalm

    Anyway, thanks again!
     
    Last edited: Feb 4, 2013
  8. Feb 4, 2013 #7

    dx

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    Hi schmiggy,

    I think there's something wrong either with the latex or quoting. I was talking about the integral over phi at the end

    ∫ sin3φ cosφ dφ (from 0 to π)

    which can be performed with the substitution u = sin φ to give

    (1/4)sin4π - (1/4)sin40 = 0
     
  9. Feb 4, 2013 #8
    I accidentally edited your quote when I was playing around with the latex seeing how it works, my bad! I also accidentally left the d(phi) at the end. I've edited my previous post and that should be correct.. I hope.. or else I'll have a serious case of egg on my face
     
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