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Surface Integral of a helicoid.

  1. Apr 30, 2012 #1

    ElijahRockers

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    1. The problem statement, all variables and given/known data

    Evaluate

    [itex]\int\int_{S}\sqrt{1+x^2+y^2} dS[/itex]

    S is the helicoid with vector equation r(u,v) = <u cos(v), u sin(v), v>

    0<u<2, 0<v<4pi

    3. The attempt at a solution

    If I replace the term under the radical with its vector equation counterpart, and multiply that by the cross product of the partials of r(u,v) with respect to u and v, i get

    [itex]\int_{0}^{4\pi}\int_{0}^{2} \sqrt{1+u^2}u du dv[/itex]

    From there I can do a u-substitution (ill just call it a ω-sub so as not to confuse) with ω=1+u2, and dω/2 = udu.

    When I work this out, I get

    [itex]\frac{4\pi}{3}(5\sqrt{5}-1)[/itex]

    But according to the software this answer is incorrect. Anyone notice a mistake?
     
  2. jcsd
  3. Apr 30, 2012 #2

    Dick

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    You seem to have turned dS into ududv. How did you do that?
     
  4. May 1, 2012 #3

    ElijahRockers

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    S is a function of u, and v. The u du comes from the u-substitution.

    Sorry
    [itex]\frac{1}{2}\int_{0}^{4\pi}\int_{1}^5 \sqrt{w} dw dv[/itex]

    Is the after the u-sub I think.
     
  5. May 1, 2012 #4

    Dick

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    I'm asking what you got for dS using the cross product.
     
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