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Line Integral does not match Greens Theorem?

  1. Mar 10, 2012 #1
    1. The problem statement, all variables and given/known data

    To evaluate the following line integral where the curve C is given by the boundary of the square 0 < x < 2 and 0 < y < 2 (In the anti clockwise sense):

    [itex]\oint (x+y)^2 dx + (x-y)^2 dy [/itex]

    3. The attempt at a solution

    Firstly it is noted that for a square ABDE :
    Between AB, dy = 0 ,y = 0
    Between BD, dx = 0 ,x = 2
    Between DE, dy = 0 ,y = 2
    Between EA, dx = 0 ,x =0

    Thus : [itex]\int^2_0 x^2 dx + \int^2_0 (2-y)^2 dy + \int^0_2 (x+2)^2 dx + \int^0_2 -y^2 dy [/itex]
    Giving [itex] \frac{x^3}{3} |^2_0 + \frac{-(2-y)^3}{3} |^2_0 + \frac{(x+2)^3}{3} |^0_2 + \frac{-y^3}{3} |^0_2 [/itex]

    Evaluating gives 10.6? but applying greens theorem gives -16. Can anyone spot my mistake - Probaly a negative sign?

    Thanks very much.
    P.s sorry for typos this keyboard is broken
     
  2. jcsd
  3. Mar 10, 2012 #2

    tiny-tim

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    Hi PhantomPower! :smile:

    erm :redface:

    ∫ (2 - y)2 is the same as ∫ (y - 2)2 ! :wink:

    (similary for (-y)2)
     
  4. Mar 10, 2012 #3
    Ooops.

    Thanks very much - been a long day...
     
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