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Homework Help: Surface Integral - 2 methods give different answer

  1. Dec 9, 2012 #1
    Thanks for checking this out. Here's the problem:

    I attempted to do it by using parametrize it into spherical coordinate.

    r(r,t) = (x= cost, y= sint, z=r)
    dS=|r[itex]_{u}[/itex] x r[itex]_{v}[/itex]| dA = r[itex]\sqrt{2}[/itex] dA
    dA = rdrdt

    [itex]\int\int[/itex]x[itex]^{2}[/itex]z[itex]^{2}[/itex]dS = [itex]\int\int[/itex][itex]\sqrt{2}[/itex] cos[itex]^{2}[/itex] r[itex]^{6}[/itex] drdt

    I check my solution manual and this is how they do it. My integral has r[itex]^{6}[/itex] factor. However, solution's only has r[itex]^{5}[/itex] instead. I am very confused because these two are supposed to be from the same source...
  2. jcsd
  3. Dec 10, 2012 #2
    There is one r too many here. dS = |ru x rv| dr dθ
  4. Dec 10, 2012 #3
    Oh, so why dA not equal to rdrdt in this case?
  5. Dec 10, 2012 #4
    Why would it be? What is dA anyway?
  6. Dec 10, 2012 #5
    Based on parametric equation, if x = rcosθ, y=rsinθ, then dA = r dr dθ
    In my hw case: dS = |ru x rv| dA = |ru x rv| r dr dθ
    so that why somehow there is r^1 excess? but idk what r is excess?
  7. Dec 10, 2012 #6


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    When you are using the formula |ru x rv| dA you don't add the 'volume' part (r) to dr dθ in dA. You'll automatically get that factor from the |ru x rv|.
  8. Dec 10, 2012 #7
    Okay, thanks so much for your help :)
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