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Total Flux through a cube

  1. Dec 9, 2008 #1
    1. The problem statement, all variables and given/known data
    Calculate the total flux of vectorF(x,y,z)=8x^2y i + 6yz^2 j + y^3z k outward through the cube whose verticies are(0,0,0), (1,0,0), (1,1,0), (0,1,0), (0,0,1), (1,0,1),(1,1,1), (0,1,1).

    2. Relevant equations

    [tex]\int[/tex][tex]\int[/tex] [tex]\widehat{}F[/tex] [tex]\bullet[/tex] (-partial z/dx i -partial z/dy j + k) dxdy

    3. The attempt at a solution
    I set up the surface S: xyz[tex]\leq[/tex] 1
    so z [tex]\leq[/tex] 1/xy

    dz/dx= 1/y lnx
    dz/dy= 1/x lny

    so F (dot) (-dz/dz i -dz/dy j + k)
    =-8x^2lnx - (6yz^2lny)/x + y^3z

    I then plugged in z

    =-8x^2lnx - 6lny/x^3y + y^2/x

    [tex]\int[/tex][tex]\int[/tex] =-8x^2lnx - 6lny/x^3y + y^2/x dxdy

    0 [tex]\leq[/tex] x [tex]\leq[/tex] 1
    0 [tex]\leq[/tex] y [tex]\leq[/tex] 1

    Does this look correct?
  2. jcsd
  3. Dec 9, 2008 #2
    Your formula there should work for the top and bottom faces of the cube. But your front/back, left/right sides might pose problems, because their normal vectors don't have a k component.

    I think you might want to consider using the general flux formula:
    [itex]\Phi=\int\int \vec{F}\circ\hat{n}dS[/itex], where [itex]\hat{n}dS[/itex] is the unit normal vector of the surface times the differential area (i.e. dx dy, dy dz, etc).
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