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Integrals 3

  1. Jan 5, 2012 #1
    1. The problem statement, all variables and given/known data

    Evlute the surface integral

    2. Relevant equations

    f(x,y,z)=x+y+z where sigma is the parallelogram with parametric equations x=u+v, y=u-v and z=1+2u+v where 0 <=u<=2 and 0<=v<=1.



    3. The attempt at a solution

    I have no idea how to tackle this. Any suggestions?
     
  2. jcsd
  3. Jan 5, 2012 #2

    LCKurtz

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    You might begin by studying the material at:

    http://tutorial.math.lamar.edu/Classes/CalcIII/SurfaceIntegrals.aspx
     
  4. Jan 6, 2012 #3
    I belive we have to use [itex] \displaystyle \int \int _R f(x(u,v),y(u,v), z(u,v) || r_u \times r_v||dA[/itex]

    I calculate the magnitude of the determinent to be \sqrt 2 hence

    The surface integral becomes

    [itex] \displaystyle \sqrt{2} \int_{0}^{1} \int_{0}^{2} (4u+v+1) du dv[/itex]....?
     
  5. Jan 6, 2012 #4

    LCKurtz

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    I agree with everything except the ##\sqrt 2##.
     
  6. Jan 7, 2012 #5
    since we have the determinant as 3i+1j-2k and the magnitude is

    [itex]\sqrt(3^2+(-2^2)+1)=\sqrt 14[/itex] cheers
     
  7. Jan 7, 2012 #6

    LCKurtz

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    The ##\sqrt {14}## is correct, but 3i + j - 2k is not a determinant; it is a vector.
     
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