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Flux of the vector field

  1. Apr 29, 2006 #1
    Let S be the part of the plane 3x+y+z=4 which lies in the first octant, oriented upward. Find the flux of the vector field F=4i+2j+3k across the surface S.

    [tex]\int \int F\cdot dS = \int int \left( -P \frac{\partial g}{\partial x} -Q \frac{\partial g}{\partial y} +R \right) dA[/tex]

    [tex]\int \int \left( -4(-3)-2(-1)+4-3x-y \right) dA[/tex]
    [tex]\int \int \left( 18-3x-y \right) dA[/tex]

    how do I find the ends of integration?
  2. jcsd
  3. Apr 30, 2006 #2


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    UrbanXrisis, I really don't understand what you did.

    What I would have done is
    [tex]\int \int F\cdot dS = \int \int \left(F\cdot \hat{n}\right) \left(\frac{dxdy}{\hat{n}\cdot\hat{k}}\right)[/tex]

    where, [itex] \hat{n} [/itex] is the unit normal to the plane. That is, to evaluate the surface integral, I'm simply projecting the surface to the x-y plane and integrating.

    Now to find the ends of integration, just project the surface to the x-y plane. In this case, you will get a triangle. From that, you can find the limits of integration
  4. Apr 30, 2006 #3


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  5. Apr 18, 2009 #4
    I have a similar question to this, with F = 7xi+yj+zk and plane z + 4x + 2y = 12. so what I did was (7xi+ yj+ zk) . (4i+ 2j+ k), is this correct?
  6. Apr 20, 2009 #5
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