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Flux Integral

  1. Sep 29, 2008 #1
    EDIT: NEVER MIND, I got 32/3 for the very next question, so I'm pretty sure the answer that I was comparing this to from the back of my textbook has a typo for the question number. If anyone still wants to check this, that would be OK but I'm pretty sure it's correct.

    1. The problem statement, all variables and given/known data
    Let F(x,y,z) = <2x,-3y,z> be the velocity vector (in m/s) of a fluid particle at the point (x,y,z) in steady state fluid flow.
    a) Find the net volume of fluid that passes in the upward firection through the portion of the plane x+y+z=1 in the first octant in 1 s.
    b) Assuming that the fluid has a mass density of 806 kg/m^3, find the net mass of fluid that passes in the upward direction through the surface in part (a) in 1 s.


    2. Relevant equations
    [tex]\Phi = \int\int F \cdot (r_u \times r_v) dA [/tex]


    3. The attempt at a solution
    First to parametrize the surface:
    x = u; y = v; z = 1 - u - v
    [tex]r_u = <1, 0, -1> [/tex]
    [tex]r_v = <0, 1, -1> [/tex]
    [tex]r_u \times r_v = <1,1,1>[/tex]

    [tex]\Phi = \int\int F \cdot (r_u \times r_v) dA [/tex]
    [tex]=\int\int <2u,-3v,(1-u-v)> \cdot <1,1,1> dA [/tex]
    [tex]=\int\int (2u - 3v + 1 - u - v) dA [/tex]
    [tex]=\int_0^1 \int_0^{1-u} (1 + u - 4v) dv du [/tex]
    [tex]=\int_0^1 {(v + uv - 2v^2)}_0^{1-u} du [/tex]
    [tex]=\int_0^1 ((1-u) + u(1-u) - 2(1-u)^2) du [/tex]
    [tex]=\int_0^1 (-1-3u^2+4u) du [/tex]

    which equals 0 m^3. This means that the net mass of the fluid that passes through the surface is also 0 kg. BUT the back of the textbook says the answer to part (b) is 32/3. Did I mess up somewhere?
     
  2. jcsd
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