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Surface Integral of Vector Field

  1. Nov 24, 2008 #1
    1. The problem statement, all variables and given/known data
    Find [tex]\int\int_{S}[/tex] F dS where S is determined by z=0, 0[tex]\leq[/tex]x[tex]\leq[/tex]1, 0[tex]\leq[/tex]y[tex]\leq[/tex]1 and F (x,y,z) = xi+x2j-yzk.


    2. Relevant equations
    Tu=[tex]\frac{\partial(x)}{\partial(u)}[/tex](u,v)i+[tex]\frac{\partial(y)}{\partial(u)}[/tex](u,v)j+[tex]\frac{\partial(z)}{\partial(u)}[/tex](u,v)k

    Tv=[tex]\frac{\partial(x)}{\partial(v)}[/tex](u,v)i+[tex]\frac{\partial(y)}{\partial(v)}[/tex](u,v)j+[tex]\frac{\partial(z)}{\partial(v)}[/tex](u,v)k

    [tex]\int\int_{\Phi}[/tex] F dS = [tex]\int\int_{D}[/tex] F * (TuxTv) du dv

    3. The attempt at a solution
    To start off, I'm not sure how to parametrize the surface S. Any help is appreciated.
     
  2. jcsd
  3. Nov 24, 2008 #2

    HallsofIvy

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    Since you are just talking about a portion of the xy-plane, x= u, y= v, z= 0. Oh, and the order of multiplication in [itex]T_u\times T_v[/itex] is important. What is the orientation of the surface? (Which way is the normal vector pointing?)

    (Actually that last point doesn't matter because this integral is so trivial.)
     
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