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Greens Theorem

  1. Apr 28, 2012 #1
    1. The problem statement, all variables and given/known data
    Let C be the boundary of the region bounded by the curves y=[itex]x^{2}[/itex] and y=x. Assuming C is oriented counter clockwise, Use green's theorem to evaluate the following line integrals (a) [itex]\oint(6xy-y^2)dx[/itex] and (b) [itex]\oint(6xy-y^2)dy[/itex]

    2. Relevant equations

    3. The attempt at a solution

    [itex]\int^{0}_{1} 6x^2 - x^2[/itex]
    [itex]\int^{0}_{1} 5x^2 [/itex] = -[itex]\frac{5}{3}[/itex]
    [itex]\int^{1}_{0} 6x^3 - x^4[/itex] = [itex]\frac{6}{4} - \frac{1}{5} = \frac{13}{10}[/itex]

    so [itex]\oint[/itex] = -[itex]\frac{11}{30}[/itex]

    [itex]\int\int_{R}[/itex] [itex]\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}[/itex])dxdy
    M=6xy-[itex]y^{2}[/itex] and N=0
    [itex]\frac{\partial M}{\partial y} 6x-2y[/itex]


    [itex]\int^{1}_{0} [ \int^{x}_{y=x^2} (6x-2y)dy] dx [/itex]

    [itex]\int^{1}_{0} 5x^2 - 6x^3 - x^4 dx [/itex]

    = [itex]\frac{-1}{30}[/itex]
    anyone got any idea what im doing wrong here!stumped
  2. jcsd
  3. Apr 28, 2012 #2


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    Gold Member

    Poorly written, but assuming you are doing the circuit integral -11/30 is correct.

    Check your sign on that ##x^4## term in the second to last line. And don't you want ##-M_y## for Green's theorem?
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