1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook