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Surface Integrals

  1. Feb 7, 2007 #1
    I am wondering if someone could help me evaluate the following:
    I am asked to find the surface integral ∫∫ydS where S is part of the paraboloid y = x^2+z^2 that lies inside the cylinder x^2+z^2 = 4.
    The double integral could be rewritten as ∫∫y*√(4(x^2+z^2)+1)dS, or ∫∫(x^2+z^2)*√(4(x^2+z^2))dxdz. But this seems very difficult to integrate, so if I convert to polar coordinates, I should have ∫∫r^2*√(4r^2+1)rdrdθ, where r is between 0 and 2 and θ is between 0 and 2π. But I’m not really sure how to integrate r^3*√(4r^2+1)? Or did I set this up incorrectly? Thanks.
     
  2. jcsd
  3. Feb 7, 2007 #2

    HallsofIvy

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    Write [itex]r^3\sqrt{4r^2+ 1}dr[/itex] as [itex] r^2\sqrt{4r^2+ 1}(rdr)[/itex] and let [itex]u= 4r^2+1[/itex].
     
  4. Feb 7, 2007 #3
    Many thanks.
     
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