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Math Help - Evaluate the line integral

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    Evaluate the line integral

    Hello,
    please try to solve this question.
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    \oint_C x^3 dx -  2xy dy = \iint_D \left( \frac{\partial (-2xy)}{\partial x} - \frac{\partial x^3}{\partial y} \right) \ dA = -2\iint_D y \ dA =  - 2\int_0^2 \int_0^{2-x} y \ dy \ dx
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    Evaluate the line integral \int_C x^3 dx - 2xy dy where C comprises the three sides of a triangle joining O (0, 0) A (2, 0) and B (0, 2)
    From O to A we have that y does not change so dy = 0.
    \int_{O to A} x^3 dx - 2xy dy  = \int_0^2 x^3 dx

    = \frac{1}{4}x^4|_0^2 = \frac{1}{4}2^4 = 4

    From A to B the path is the line from (2, 0) to (0, 2). This is a segment of the line y = -x + 2 as x goes from 2 to 0, so
    dy = -dx

    Thus
    \int_{(2,0) to (0,2)} x^3 dx - 2xy dy

    = \int_2^0 x^3 dx + 2x(-x + 2)(-dx)

    = \int_2^0 (x^3 + 2x^2 - 4x)dx

    = \left ( \frac{1}{4}x^4 + \frac{2}{3}x^3 - 2x^2 \right ) _2^0

    = - \left ( \frac{1}{4}2^4 + \frac{2}{3}2^3 - 2 \cdot 2^2 \right ) = - \left ( 4 + \frac{16}{3} - 8 \right )

    = 4 - \frac{16}{3}

    So the overall integral will be:
    4 + 4 - \frac{16}{3} = 8 - \frac{16}{3} = \frac{8}{3}

    -Dan
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    Quote Originally Posted by topsquark View Post
    From O to A we have that y does not change so dy = 0.
    \int_{O to A} x^3 dx - 2xy dy  = \int_0^2 x^3 dx

    = \frac{1}{4}x^4|_0^2 = \frac{1}{4}2^4 = 4

    From A to B the path is the line from (2, 0) to (0, 2). This is a segment of the line y = -x + 2 as x goes from 2 to 0, so
    dy = -dx

    Thus
    \int_{(2,0) to (0,2)} x^3 dx - 2xy dy

    = \int_2^0 x^3 dx + 2x(-x + 2)(-dx)

    = \int_2^0 (x^3 + 2x^2 - 4x)dx

    = \left ( \frac{1}{4}x^4 + \frac{2}{3}x^3 - 2x^2 \right ) _2^0

    = - \left ( \frac{1}{4}2^4 + \frac{2}{3}2^3 - 2 \cdot 2^2 \right ) = - \left ( 4 + \frac{16}{3} - 8 \right )

    = 4 - \frac{16}{3}

    So the overall integral will be:
    4 + 4 - \frac{16}{3} = 8 - \frac{16}{3} = \frac{8}{3}

    -Dan
    My answer is exactly negative of your answer if you do it out by Green's theorem! Maybe you did it clockwise.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    My answer is exactly negative of your answer if you do it out by Green's theorem! Maybe you did it clockwise.
    I suppose I should have suspected this, but I didn't realize it was an integral over a closed area. I only did the integral from O to A to B. I never connected B to O.

    So from B (0, 2) to O (0, 0) dx = 0, with x = 0. Thus
    \int_{(0, 2) to (0, 0)} x^3 dx - 2xy dy = \int_2^0 (-2 \cdot 0 \cdot y) dy = \int_2^0 0 dy = 0

    So my final integral is STILL \frac{8}{3}.

    And if you look at the succession of points I did the integral in the counterclockwise sense. I can't argue with your result (though I was never very good at implementing Green's theorem anyway) but I can't find a mistake in my own work?

    -Dan
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