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Green's Thm: How am I not applying it correctly?

  1. Jan 14, 2010 #1
    Curve consists of part of a circle x2 + y2 = 4
    theta from 0 to pi/4

    Vector field F = <y2-yx2 , yx2>

    (let the i component be P, and the j component be Q)

    I used the following formula: double integral of [(partial derivative of Q with respect to x) - (partial derivative of P with respect to y) dA]

    So I got: double integral of [ (2xy) - (2y - x2) dy dx ] y from 0 to x, and x from 0 to 2

    I got answer = 16/3 which is wrong.

    My question is: What is wrong about the way I applied Green's thm?
     
    Last edited: Jan 14, 2010
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  3. Jan 14, 2010 #2

    nicksauce

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    Well, for one thing, your bounds of integration do not describe a circle.
     
  4. Jan 14, 2010 #3
    the curve doesn't consist of the whole circle. it is only part of the circle where theta is from 0 to pi/4
     
  5. Jan 14, 2010 #4

    nicksauce

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    Still, your bounds of integration describe a triangular region, not a circular region.
     
  6. Jan 15, 2010 #5

    HallsofIvy

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    That's the whole point! Green's theorem says that
    [tex]\int Mdx+ Ldy= \int\int \left(\frac{\partial L}{\partial x}- \frac{\partial M}{\partial y}\right)dxdy[/tex]
    Where the path integral is over a closed path and the double integral is over the region bounded by that path. The arc of a circle from 0 to [itex]\pi/4[/itex] is not a closed path. And your double integral was over the triangle bounded by the lines y= 0, x= 2, and y= x which doesn't have anything to do with the given path! The lines y= 0 and y= x are okay but the curved side is NOT the straight line x= 2.

    If you really want to use Green's theorem to find that integral you will have to add the line y= 0, from x=0 to 2 and the line y= x. To integrate over the area bounded by those and the arc of the circle, I would recommend changing to polar coordinates, taking [itex]\theta[/itex] from 0 to [itex]\pi/4[/itex] and r from 0 to 2. If you really want to use rectangular coordinates, you will need to do two integrals, the first with y from 0 to x and then x from 0 to [itex]\sqrt{2}[/itex], since y= x intersects [itex]x^2+ y^2= 4[/itex] at [itex](\sqrt{2}, \sqrt{2})[/itex]. The second integral would be with y from 0 to [itex]\sqrt{4- x^2}[/itex] and x from [itex]\sqrt{2}[/itex] to 2.

    And, of course, the integral on that region will NOT be the path integral on the arc, it will be the path integral on the arc plus the path integrals on the two straight lines. To find the path integral on the arc, you would need to calculate the path integrals on the two straight lines separately and subtract them from the area integral.
     
  7. Jan 15, 2010 #6
    The question did say that the curve is bounded by x = 0 and y = 0. I didn't mention it because I mistakenly thought it was obvious.

    Anyway, so to convert [y2 - 2y + x2] dx dy into polar coordinates I did: x = 2cos(theta), y = 2sin(theta), then got: [r2-4sin(theta)]r dr d(theta) Is that incorrect? Because I'm not getting the right answer even though I integrated r from 0 to 2, and theta from 0 to pi/4
     
  8. Jan 15, 2010 #7

    HallsofIvy

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    No. [itex]x= r cos(\theta)[/itex] and [itex]y= r sin(\theta)[/itex], not "[itex]2cos(\theta)[/itex] and [itex]2sin(\theta)[/itex]. [itex]y^2- 2y+ ^2[/itex] becomes [itex]r^2- 2r sin(\theta)[/itex]. Your integral should be
    [tex]\int_{\theta= 0}^{\pi/4}\int_{r= 0}^2 (r^2- 2r sin(\theta))r drd\theta[/tex]

    And that double integral will NOT be the path integral over the curve because you have not taken into acount the two straight line boundaries of the region. That will be equal to the integral on the x-axis, from x= 0 to x= 2 plus the integral over the curve, plus the integral on y= x from x= [itex]\sqrt{2}[/itex] to 0.
     
  9. Jan 16, 2010 #8
    i can't figure out how to take into account the line segments from (0,0) to (2,0), and from (square root of 2, square root of 2) to (0, 0)

    what's the most efficient way to set up the path integral using green's thm?
     
  10. Jan 17, 2010 #9

    HallsofIvy

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    I don't know what you mean by "set up the path integral using green's thm". You don't use Green's theorem to set up a path integral! You use it to give a hopefully simpler area integral equal to the path integral on a closed path. One difficulty I have with this is that you have never said what the original problem was!

    If the problem is just to find the path integral you cite- the integral of [itex]<y^2-yx^2 , yx^2>[/itex] over the unit circle from [itex]\theta= 0[/itex] to [itex]2\pi[/itex], I see no reason to use Green's theorem at all.

    If you must use Green's theorem then, like I said, you must take into account the straight line sides. On the x-axis, y= 0 so you can take x= t, y= 0, dx= dt, dy= 0. The vector function [itex]<y^2-yx^2 , yx^2>= < 0, 0>[/itex] so the integral is just 0.

    On the line y= x, take x= t, y= t so dx= dy= dt. [itex]<y^2-yx^2 , yx^2>= <t^2- t^3, t^3>[/itex] so the integral is

    [tex]\int_{t=\sqrt{2}}^0 (t^2- t^3)dt+ t^3dt[/tex]

    [tex]= \int_{t= \sqrt{2}}^0[/tex][tex] t^2 dt[/tex]
     
    Last edited by a moderator: Jan 18, 2010
  11. Jan 26, 2010 #10
    thank you so much :)
     
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