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Calc 3 - outward flux question

  1. May 4, 2009 #1
    The region in question is bounded by:
    the cylinder (x^2)+(y^2)=(R^2)
    the parabola x = y-((y^2)/R)
    the planes z = H, y = 0, and z = 0

    and the velocity field is:

    F = yz(i)+xz(j)+xy(k)

    and we need to calculate the outward flux of the field of the region at z = H (the top of the region).

    Ive tried doing this 2 ways:
    Doing the double integral of div(F)
    Doing the double integral of F (dot) k dA
    the solution to this problem uses the 2nd of the two; my question is this:
    why are the 2 methods in disagreement with each other? (when i take div(F) i get zero, so flux is zero)
     
  2. jcsd
  3. May 4, 2009 #2
    The divergence theorem applies to a closed volume. You are calculated the flux through a piece of the surface area and not the whole surface area of the whole volume. So you can't use the divergence theorem.
     
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