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Proving a theorem in line integrals

  1. Feb 13, 2016 #1
    At the bottom of the picture, I couldn't understand why differentiating with respect to x gives the first integral at the right-hand side 0. Thanks for reading.
     

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  2. jcsd
  3. Feb 13, 2016 #2

    HallsofIvy

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    In my opinion, that is simply wrong. Rather than taking [itex]C_1[/itex] to be "any path from (a, b) to [itex](x_1, y)[/itex]" we must choose [itex]C_1[/itex] to be the vertical line from (a, b) to [itex](a, y)[/itex] then take [itex]C_2[/itex] to be the horizontal line from [itex](a, y)[/itex] to [itex](x, y)[/itex].
     
  4. Feb 13, 2016 #3
    I'm no math pro, but my guess would be that since you have the hypothesis that your integral is path independent, then
    [tex]
    \int_{C_1} F\cdot dr = \int^{(x1,y)}_{(a,b)} F\cdot dr = f(x1,y) - f(a,b)
    [/tex]
    which differentiated w.r.t. x gives 0 and w.r.t. y does not (since the point (x,y) is arbitrary, y is arbitrary but x1,a,b are fixed)
     
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