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Evaluating line integrals versus Green's Theorem

  1. Apr 26, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the simple closed integral of (x+xy-y)(dx+dy) counterclockwise around the path of straight line segments from the origin to (0,1) to (1,0) to the origin...
    a)as a line integral
    b)using green's theorem

    2. Relevant equations

    Eq of line segment r(t)=(1-t)r0+tr1

    Greens Theorem states that Integral of Pdx+Qdy on a closed simple path will equal the double integral or area of dQ/dx-dP/dy (see the work I did)

    3. The attempt at a solution

    See the pictures. The two were supposed to be equal, I just needed to show how to arrive at the same answer using both methods. The triangle shows the graph of the path, horizontal is x axis, vertical is y axis. The C2 on the left of the triangle should say C3, sorry. Also the first integral should read (1+y) not (1-y)
     

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    Last edited: Apr 27, 2010
  2. jcsd
  3. Apr 27, 2010 #2

    HallsofIvy

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    One obvious error is in saying that
    [tex]\frac{\partial Q}{\partial x}= 1- y[/tex]
    Q= x+ xy- y so
    [tex]\frac{\partial Q}{\partial x}= 1+ y[/tex]

    but it is written correctly in the next line so that's not the problem.

    I notice you then integrate, in integrating over the area,
    [tex]\int_{y=0}^{-x}[/tex]
    Why is that? The upper boundary is y= x, not y= -x. It looks to me like you should be integrating [itex]2x+ x^2/2- x^2= 2x- x^2/2[/itex] with respect to x. That would give [itex]\left[x^2- x^3/6\right]_0^1= 5/6[/itex] rather than -1/2.
     
    Last edited: Apr 27, 2010
  4. Apr 27, 2010 #3
    But that wouldn't give you the same answer for part a and b...
    Why would that give y=x for the boundary. I looked it over again with my calculator, that upper boundary line (hypotenuse) is y=-x+1
    however integrating that you end up with 2, which is still not the same as part a.
     
  5. Apr 27, 2010 #4

    HallsofIvy

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    Yes, you are right about that- I misread your problem.

    But, still, the integral should be from 0 to 1- x, not to just -x.
     
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