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Math Help - Green's Theorem

  1. #1
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    Green's Theorem

    Using Greenís theorem, calculate the circulation of F around the curve:
    counterclockwise around a triangle with vertices (2,0) (0,3) (-2,0)
    F = (2x^2 + 3y) i + (2x + 3y^2) j

    Is the answer (-12)?
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  2. #2
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    Re: Green's Theorem

    No, it is not. Is that all you want? Perhaps if you showed what you did we could point out errors. What did you get for the area of that triangle?
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  3. #3
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    Re: Green's Theorem

    F = (2x^2 + 3y) i + (2x + 3y^2) j
    triangle points: (2,0) (0,3) (-2,0)

    Green's theorem: ∫c F ∑ dr = ∫∫R [ (∂N/∂x) - (∂M/∂y) ] dA

    N = 2x + 3y^2
    M = 2x^2 + 3y

    ∂N/∂x = 2
    ∂M/∂y = 3

    ∫∫R (2-3) dy dx


    limits of integration:
    vertical: from 0 to 3

    horizontal:
    left side of triangle:
    y = (3/2)x + 3
    x = (2/3)y - 2

    right side of triangle:
    y = (-3/2)x + 3
    x = (-2/3)y + 2


    ∫∫R (2-3) dy dx

    ∫ (upper limit: x = (-2/3)y + 2) (lower limit: x = (2/3)y - 2) ∫ (upper limit: 3) (lower limit: 0) (2-3) dy dx

    inner integration:
    ∫ (upper limit: 3) (lower limit: 0) (2-3) dy
    ∫ (upper limit: 3) (lower limit: 0) (-1) dy
    y evaluated from 0 to 3 = -3

    outer integration:
    ∫ (upper limit: x = (-2/3)y + 2) (lower limit: x = (2/3)y - 2) -3 dx
    -3x evaluated from [x = (2/3)y - 2] to [x = (-2/3)y + 2)]
    -3(2+ 2)
    -12
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