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Green's Theorem and annulus at 0,0

  1. Jan 21, 2012 #1
    1. The problem statement, all variables and given/known data

    Use Green's Theorem to evaluate this line integral

    2. Relevant equations

    [itex]\int xe^{-2x}dx+(x^4+2x^2y^2)dy[/itex] for the annulus [itex]1 \le x^2+y^2 \le 4[/itex]

    3. The attempt at a solution

    [itex] \displaystyle \int_c f(x,y) dx + g(x,y)dy+ \int_s f(x,y) dx + g(x,y)dy = \int \int _D1 (G_x-G_y) dA=0 \implies \int_c=- \int_s= \int_{-s}[/itex]

    Let c be the out circle of radius 2 counterclockwise and s the inner radius of 1 clockwise

    x=r cos [itex]\theta[/itex], y=r sin [itex]\theta[/itex] substituting

    evaluating the last term on RHS, ie

    [itex] \displaystyle \int_{-s}= \int_0^{2 \pi} r cos \theta (e^{-2r cos \theta})(-r sin \theta d \theta) +(r^4 cos^4 \theta +2r^2 cos^2 \theta r^2 sin^2 \theta) r cos \theta d \theta[/itex]

    Is this right so far? Thanks
     
  2. jcsd
  3. Jan 21, 2012 #2

    HallsofIvy

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    If the problem is, as you say, to use Green's theorem, why are you not using Green's theorem?
     
  4. Jan 21, 2012 #3
    I guess because I was following my notes blindly which I think was just demonstrating a principle.

    For the annulus then do I use green's to calculate the inner circle and subtract it from the outer circle?

    Thanks
     
  5. Jan 21, 2012 #4
    Correction,

    I think I will split the annulus into a top and bottom and write

    [itex] \displaystyle \int \int_R (g_x-f_y) dA = \int \int_T (g_x-f_y )dA + \int \int_B (g_x-f_y) d A[/itex]...right?
     
  6. Jan 21, 2012 #5

    I calculate that the top and bottom integrals are both 0....? I thought they should be non 0 since the annulus is centred about 0?
     
  7. Jan 21, 2012 #6

    HallsofIvy

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    No, just integrate on the annulus itself. Using polar coordinates as you suggested before, take [itex]\theta[/itex] from 0 to [itex]2\pi[/itex] and r from 1 to 2.

    Now, what is [itex]g_x- f_y[/itex]?
     
  8. Jan 21, 2012 #7
    It is [itex] g_x-f_y =(4x^3 + 4xy^2)-(0)[/itex]

    If we are using green's to evaluate the above integral which does not include the origin then we expect to get a non 0 answer right? But I keep getting 0 for the above integral...
     
  9. Jan 24, 2012 #8
    Any suggestions on this one?
     
  10. Jan 24, 2012 #9

    HallsofIvy

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    Then you are doing the integral wrong!
    The integral is
    [tex]4\int_0^1 4r^4dr\int_0^{2\pi}(cos^3(\theta)+ sin(\theta)cos^2(\theta))d\theta[/tex]
    (the fourth power of r is due to the differential of area in polar coordinates being "[itex]rdrd\theta[/itex]", but it is the [itex]\theta[/itex] integral that is important.)

    The [itex]\theta[/itex] integral can be written
    [tex]\int_0^{2\pi} cos(\theta)(1- sin^2(\theta))d\theta+ \int_0^{2\pi} sin(\theta)cos^2(\theta)d\theta[/tex]

    Let [itex]u= sin(x)[/itex] in the first integral and let [itex]v= cos(\theta)[/itex] in the second equation.
     
  11. Jan 24, 2012 #10
    I calculate [itex] \displaystyle \int \int (g_x -f_y) dA = \int \int (4x^3+4xy^2) dA = \int_1^2 \int_0^{2 \pi}(4r^4 cos^3 \theta +4r^4 cos \theta sin^2 \theta) d \theta dr[/itex]....

    I am just wondering how you arrived at yours especially the limits...? Have you got the sin theta cos^2 theta switched? should it be cos theta sin^2 theta
     
  12. Jan 25, 2012 #11
    Any comments on this one folks...?
     
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