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Math Help - Simple Line Integral

  1. #1
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    Simple Line Integral

    Hello! I need some help with a simple line integral. If there's someone out there who wouldn't mind explaining their work, could he/she lend a hand?

    the problem is:

    Let C be the curve which is the union of two line segments, the first going from (0, 0) to (-4, 2) and the second going from (-4, 2) to (-8, 0).
    Computer the line integral
    ∫-4dy-2dx

    i've tried solving for this by parameterizing the 2 line segments in terms of t, and calculated an incorrect answer of 32.
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  2. #2
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    I tried and got 32 too

    For the first integral:
    ∫-4dy-2dx
    r= (-4,2)t
    x=-4t
    y=2t
    dx=-4dt
    dy=2dt
    ∫-8 +8 dt=0
    For the second integral:
    ∫-4dy-2dx
    r= (-4,2) + (-8+4, 0-2)t
    x=-4 -4t
    y=2 -2t
    dx=-4dt
    dy=-2dt
    ∫8dt+8dt=32
    0<=t<=2

    or maybe it is a conservative field, and the answer is U(-8,0)-U(0,0) where U= -2x-4y
    Last edited by nathanfsu; December 2nd 2008 at 12:56 AM.
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  3. #3
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    It looks like zero to me: Start with the line going from (0,0) to (-4,2). The equation for this line is y=-1/2(x+4)+2. Parameterizing it:  C_1: \{x=t,\; y=-1/2(t+2)+2\}. That integral is, remembering to keep straight the direction we're going:

    \mathop\int\limits_{C_1} -4dy-2dx=-4\int_0^{-4} (-1/2)dt-2\int_0^{-4}dt=0

    Second one going from (-4,2) to (-8,0). Equation is y=1/2(x+8). Parameterizing: C_2:\{x=t,\; y=1/2(t+8). Keeping direction straight:

    \mathop\int\limits_{C_2} -4dy-2dx=-4\int_{-4}^{-8} (1/2)dt-2\int_{-4}^{-8}dt=0
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