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Math Help - Basic line integral

  1. #1
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    Basic line integral

    Hey, I must be getting something wrong about how to calculate line integrals, so I was wondering if someone here might want to give me a pointer to set me straight.

    We have a Cartesian vector field

    F = (yx^2)i - (y)j,

    where I use i, j and k as the unit vectors in the x, y and z directions. We then have a loop in the form of a triangle, with the first leg starting at the origin and going to point (1,1), the second leg going from (1,1) to (2,0) and the third from (2,0) back to the origin. I want to calculate the line integral of F around this loop.

    On the third leg y = 0, so the line integral should be 0 as well.

    For the first leg, I've got the notion that
    x = y, dx = dy, and the differential displacement should be dl = (dx)i + (dy)j.
    Taking the dot product between F and dl gives
    (yx^2)dx - (y)dy = (x^3 - x)dx
    Taking the integral of this from x = 0 to x = 1 gives the line integral along leg 1, I1 = -1/4

    For the second leg we have
    y = -x, dy = -dx and differential displacement dl = (dx)i - (dy)j.
    Taking the dot product between F and dl gives
    (yx^2)dx + (y)dy = (-x^3 + x)dx
    Taking the line integral of this from x = 1 to x = 2 gives the line integral along leg 2, I2 = -7/4

    I1 + I2 = -2, and the answer should supposedly be 7/6, so I must be getting something hilariously wrong about all this. Any help finding my mistake/misunderstanding greatly appreciated!
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  2. #2
    Super Member
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    Re: Basic line integral

    On the second leg, y = 2 - x.
    Thanks from Scurmicurv
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  3. #3
    Junior Member
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    Re: Basic line integral

    That's, uh... that's... well... that makes sense... a lot. Like, really a lot.

    *facepalm*.

    Thanks man, appreciate it.
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