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Using stokes' or divergence theorem to solve integral

  1. Oct 11, 2013 #1
    1. The problem statement, all variables and given/known data

    Use either Stokes' theorem or the divergence theorem to evaluate this integral in the easiest possible way.

    ∫∫V [itex]\cdot[/itex]ndσ over the closed surface of the tin can bounded by x2+y2=9, z = 0, z = 5, if V = 2xyi - y2j + (z + xy)k

    The bolded letters are vectors.


    2. Relevant equations

    ∫∫( xV) [itex]\cdot[/itex] ndσ = ∫V [itex]\cdot[/itex]dr Stokes' theorem

    ∫∫∫ [itex]\cdot[/itex]Vdτ = ∫∫V [itex]\cdot[/itex]ndσ Divergence theorem


    3. The attempt at a solution

    So what I did was use the divergence theorem to turn this into a volume integral. I did the ∇xV to get 2y-2y+1. But here's where I'm stuck. I can input the formula for the volume of a cylinder, right? Is that dτ? I have to integrate with respect to τ. So if 2y-2y+1 = 1, then I'm left with τ after I integrate, and is τ just ∏r2h?

    Thanks
     
  2. jcsd
  3. Oct 11, 2013 #2

    vela

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    You mean you calculated ##\nabla\cdot \vec{V}##, not the curl. ##d\tau## is the volume element. For cartesian coordinates, it's equal to ##d\tau = dx\,dy\,dz##. The fact that you're integrating over a cylindrical volume is reflected in the limits. If I were you, however, I'd switch to cylindrical coordinates, where ##d\tau = r\,dr\,d\theta\,dz##.

    EDIT: Never mind the last sentence. I misread your expression for the divergence and thought you would need to actually evaluate a non-trivial integral. Your work looks fine. If the divergence weren't simply a constant, you'd probably be better off switching to cylindrical coordinates.
     
  4. Oct 11, 2013 #3

    Zondrina

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    That all sounds reasonable. You can switch to cylindrical co-ordinates to make integrating easier I believe.
     
  5. Oct 11, 2013 #4

    Dick

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    But the divergence isn't a constant. It's 2x-2y+1. After you integrate over the cylinder then the 2x and 2y drop out. But you should say why.
     
  6. Oct 11, 2013 #5

    vela

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    That's the same mistake I made originally, but the first term is really 2y, not 2x.
     
  7. Oct 11, 2013 #6

    Dick

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    Ooo. Yes, it is.
     
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