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Flux of a Vector field of a square on a plane x+y+z=20

  1. Nov 18, 2009 #1
    I have been trying this problem for multiple hours now, and cannot figure out what I am doing wrong.

    --Calculate the flux of the vector field F(vector)= 5i + 8j through a square of side 2 lying in the plane x + y + z = 20 oriented away from the origin.


    I realize that I need the integral of the Vector field (F) multiplied by the normal vector of the area of the square. To do this, I assigned values for the points of the triangle made, and made 2 new vectors out of it. I crossed these vectors to get the normal vector of the plane given. Once I had that, I dotted the Vector field by the normal vector, then multiplied that by the area of the square.

    The answer I got was 20800. This was the incorrect answer!

    Can anyone help me!?


    Thanks
     
  2. jcsd
  3. Nov 18, 2009 #2

    Dick

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    You are probably working too hard to find the normal, but you can do it that way. Did you find the UNIT normal or just any normal? Show the details of your solution if you are still confused.
     
  4. Nov 18, 2009 #3

    Dick

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    If you got (-1,-1,-1)/sqrt(3) for the unit normal and -52/sqrt(3) for the flux, then are you sure you've got the unit normal pointed in the right direction? The problem says it's oriented 'away from the origin'.
     
  5. Nov 19, 2009 #4
    I tried it both ways. The normal ends up being the same magnitude either way, and neither answer works.

    I cannot figure out what I am doing wrong :(
     
  6. Nov 19, 2009 #5

    Dick

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    If you tried 4*13/sqrt(3) with both signs, I'm not sure what the problem is either.
     
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