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Math Help - line integrals

  1. #1
    Junior Member
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    Chicago, IL
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    Exclamation line integrals

    ok...Compute the line integral bounded by C of ydx - xdy (i dont know if that was a typo if thats normal, ive never seen an integral like that)
    where C is the upper semicircle of radius 1 oriented from (1; 0) to (-1; 0) (i.e. counter-clockwise).

    \int xdy - ydx
    when i did the problem i got 0 as the answer, but my teachers answer page says -Pi. any ideas?

    also i used C = <cos t, sin t> from 0 to Pi.
    Last edited by Dubulus; August 6th 2008 at 04:24 PM.
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  2. #2
    Senior Member
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    Anchorage, AK
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    Is the problem the integral of y\,dx-x\,dy or x\,dy-y\,dx? In your stating words, you have the first, but for your formula you wrote the second. These two have opposite signs, and will give opposite answers. I think you mean the first, as that gives the answer you gave:

    Your parametrization gives x(t)=\cos{t}, y(t)=\sin{t}, and thus \frac{dx}{dt}=-\sin{t} and \frac{dy}{dt}=\cos{t}. Thus we have:
    \int_{C}y\,dx-x\,dy=\int_0^{\pi}\left(y\frac{dx}{dt}-x\frac{dy}{dt}\right)\,dt
    =\int_0^{\pi}\sin{t}\cdot(-\sin{t})-\cos{t}\cdot\cos{t}\,dt
    =\int_0^{\pi}-\sin^2{t}-\cos^2{t}\,dt
    =\int_0^{\pi}(-1)\,dt
    =-\int_0^{\pi}\,dt
    =-\pi

    --Kevin C.
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