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Math Help - Calculate flux across a curve. Would like my solution checked please

  1. #1
    Senior Member
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    Calculate flux across a curve. Would like my solution checked please

    F=(x^{2}+y^{3})i+(2xy)j

    Curve C:x^{2}+y^{2}=9

    x=3cos(t), y=3sin(t); 0\leq t \leq 2\pi

    So then:

    F=(9cos^{2}(t)+27sin^{3}(t))i+(18cos(t)sin(t))j

    M=9cos^{2}(t)+27sin^{3}(t)

    N=18cos(t)sin(t)

    dx=-3sin(t)

    dy=3cos(t)

    Flux=\displaystyle \int^{2\pi}_{0}(Mdy-Ndx)dt

    Flux=\displaystyle \int^{2\pi}_{0}(27cos^{3}(t)+81sin^{3}(t)cos(t)-54sin^{2}(t)cos(t))dt

    So, then I just did the integral part by part

    \displaystyle  27 \int^{2\pi}_{0}cos^{3}(t)dt

    \displaystyle  \frac{81}{4} \int^{2\pi}_{0}(cos(t)+cos(3t))dt

    \displaystyle \frac{81}{4}[sin(t)]|^{2\pi}_{0}+\frac{81}{12}[sin(3t)]|^{2\pi}_{0}=>\frac{81}{4}(0-0)+\frac{81}{12}(0-0)=0

    \displaystyle \int^{2\pi}_{0}81sin^{3}(t)dt

    u=sin(t) du=cos(t)dt

    \displaystyle 81 \int^{2\pi}_{0}u^{3}du

    \frac{81}{4}[sin^{4}(t)]|^{2\pi}_{0}=>\frac{81}{4}(0-0)=0

    -54 \displaystyle\int^{2\pi}_{0}sin^{2}(t)cos(t))dt

    u=sin(t) du=cos(t)dt

    -54 \displaystyle\int^{2\pi}_{0}u^{2}(t)du

     \frac{-54}{3}[sin^{3}(t)]|^{2\pi}_{0}=>\frac{-54}{3}(0-0)=0

    So, that means that:

    Flux=\displaystyle \int^{2\pi}_{0}(27cos^{3}(t)+81sin^{3}(t)cos(t)-54sin^{2}(t)cos(t))dt=0

    Did, I do the problem wrong or something? I'd hate to think I did all of this work for the answer to just be 0...
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  2. #2
    Junior Member
    Joined
    May 2010
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    You are correct here. Integrals measure area across plane(s) and a curve doesn't have any area to it. Therefore it makes sense that the answer would be zero.
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