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Math Help - Green's theorem

  1. #1
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    Green's theorem

    The question I am attempting is;
    Verify Greens Theorem for ∫(4x-3y≤)dx+(y-4x≤)dy
    C

    Where c is the boundary of the region defined by y = x^4 and y = x,

    My problem is that I am getting two different answers.

    1
    I have put in y= x4 to get ∫ (4x-3x^8+x^4-4x^2)dx for this I get an answer of 11/30
    X=0

    0
    I then put in y=x to get ∫ (5x-7x^2)dx for this I get a value of -1/6
    X=1

    When combined these two values equal 1/5.

    Then according to Greenís thm this should equal
    1 y=x
    ∫ ∫ {d(y-4x≤)/dx Ė d(4x-3y≤)/dy}dydx
    X=0 y=x4


    = ∫∫ (6y-8x)dydx = ∫ [3y^2-8xy] between y=x and y=x4

    1
    I get this equal to ∫(3x^2-8x^2-3x^8+8x^5)dx
    X=0

    Which I get as [(-5x^3)/3-(3x^9)/9+(8x^6)/6] which equals -5/3-3/9+8/6 = -2/3 ≠ 11/30
    if you can spot where Iíve gone wrong or tell me what answer I should be looking for is it would be much appreciated thanks.
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  2. #2
    MHF Contributor

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    Quote Originally Posted by jarvis5 View Post
    The question I am attempting is;
    Verify Greens Theorem for ∫(4x-3y≤)dx+(y-4x≤)dy
    C

    Where c is the boundary of the region defined by y = x^4 and y = x,

    My problem is that I am getting two different answers.

    1
    I have put in y= x4 to get ∫ (4x-3x^8+x^4-4x^2)dx for this I get an answer of 11/30
    You have go integrate along the curve [tex]y= x^4, not just replace y by x^4. If [tex]y= x4, then dy= 4x^2 dx, not just "dx"
    [/quote] X=0

    0
    I then put in y=x to get ∫ (5x-7x^2)dx for this I get a value of -1/6
    X=1

    When combined these two values equal 1/5.

    Then according to Greenís thm this should equal
    1 y=x
    ∫ ∫ {d(y-4x≤)/dx Ė d(4x-3y≤)/dy}dydx
    X=0 y=x4


    = ∫∫ (6y-8x)dydx = ∫ [3y^2-8xy] between y=x and y=x4

    1
    I get this equal to ∫(3x^2-8x^2-3x^8+8x^5)dx
    X=0

    Which I get as [(-5x^3)/3-(3x^9)/9+(8x^6)/6] which equals -5/3-3/9+8/6 = -2/3 ≠ 11/30
    if you can spot where Iíve gone wrong or tell me what answer I should be looking for is it would be much appreciated thanks.[/QUOTE]
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