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Green's Th. problem

  1. Dec 6, 2007 #1
    1. The problem statement, all variables and given/known data
    Find area of indicated region: S is bounded by the curves y = 4x and y = 2x^3

    actual answer: 8/3

    note: }a:b{ is what i mean by the definite integral

    2. Relevant equations
    A(s) = 1/2 }Green's Int{ (-ydx + xdy) = 1/2 }{ }{ (dN/dx - dM/dy)dxdy

    3. The attempt at a solution

    y' = 4
    intersection: 4x = 2x^3 ---> x = sqrt(2)

    A(s) = 1/2 }0:sqrt(2){ }4x:4{ (1 + 1) dydx

    = 1/2 }0:sqrt(2){ (2(4) - 2(4x)) dx
    = 1/2 (8(sqrt(2)) - 4(2))
    = 4(sqrt(2)) - 4
    (not equal to 8/3)

    sorry bare with me i dont totally understand the theorem but i tried
    Last edited: Dec 7, 2007
  2. jcsd
  3. Dec 7, 2007 #2


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    Uh, why is there a need to use Green's theorem to solve this? Couldn't you just find the area by means of a definite integral? I thought Green's theorem only applied to line integrals of vector functions/fields?
  4. Dec 7, 2007 #3


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    I'm not quite sure what you were trying to do (the notation you used is poor, so I have no idea what's being integrated, etc.) but at the top you had the curves as 4x and 2x^2, so the intersection would occur at x=2
  5. Dec 7, 2007 #4
    yes u can just integrate the easier way, but the point is to use greens theorem, oops its y = 2x^3

    the question wants me to use this equation to solve it:
    ok ill use I(a,b) as the integral

    I(Greens) (-y/2 dx + x/2 dy) = I I (1/2 + 1/2) dA = A(s)

    i guess my real question is how do u find the intervals on which you integrate the double integral
    Last edited: Dec 7, 2007
  6. Dec 7, 2007 #5


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    Using Green's theorem, it's easy to show that the area of the region enclosed by the closed path C is [itex]\int_C x dy[/itex] or [itex]\int_C -ydx[/itex].

    I also note that y= 4x and y= 2x3 intersect at [itex](-\sqrt{2},-4\sqrt{2})[/itex], (0, 0), and [itex](-\sqrt{2},-4\sqrt{2})[/itex] so the region has two symmetric regions, one in the first quadrant and one in the third quadrant. I would recommend finding the area of the one in the first quadrant and doubling.

    [itex]\int_C -ydx[/itex] divides into two integrals: On y= 2x3, it is [itex]\int_0^{\sqrt{2}} -2x^3 dx[/itex]. On y= 4x, it is [itex]\int_{\sqrt{2}}^0 -4xdx[/itex].
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