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Direct Integration vs. Green's Theorem

  1. Sep 17, 2013 #1
    Problem:

    ##\int_R (x-y)dx \ dy=-2/3 ## for ##R=\{(x,y):x^2+y^2 \geq 1; y \geq 0\}## by

    a.) Direction integration,

    b.) Green's theorem.

    Attempt at a Solution:

    I'm a little confused with part a. Wouldn't the region R be defined by all the points above the y-axis that lie on, in addition to above, the circle of radius 1 centered at the origin?

    I'm confused on what the limits of integration would be for the integral ##\iint (x-y) dx \ dy##.
     
    Last edited: Sep 17, 2013
  2. jcsd
  3. Sep 17, 2013 #2

    vela

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    I'm sure that's a typo. It probably should be ##x^2+y^2 \le 1##.
     
  4. Sep 17, 2013 #3
    So by direct integration I have,

    [tex]\int_{-1}^{1} \int_{0}^{\sqrt{1-x^2}}(x-y) dy \ dx = 2/3[/tex] Is it supposed to be negative?
     
  5. Sep 17, 2013 #4

    D H

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    You made a mistake somewhere. It is -2/3 rather than +2/3.
     
  6. Sep 17, 2013 #5
    Dang. My limits are correct, yes?

    Also, how could I apply Green's theorem. I thought the idea was to change a line integral into a double integral, right?

    EDIT: Never mind, it was an error on my part. Still not sure how to apply Green's theorem.
     
    Last edited: Sep 17, 2013
  7. Sep 17, 2013 #6

    vela

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    Well, you have a double integral. You want to change it into a line integral.
     
  8. Sep 17, 2013 #7
    So it'd just be a line integral over the boundary of the region, correct?

    Since Green's theorem in a plane is given by [tex]\iint_R (\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y})dx \ dy \ =\oint_C (F_1 dx +F_2 dy)[/tex]

    for this specific problem we'd have ##\frac{\partial F_2}{\partial x}=x## and ##\frac{\partial F_1}{\partial y}=y##.

    Thus, ##F_1=\frac{1}{2}y^2## and ##F_2=\frac{1}{2}x^2##. Right?
     
    Last edited: Sep 18, 2013
  9. Sep 18, 2013 #8

    vela

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    Looks good.
     
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