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Math Help - [SOLVED] Check out my result : line integral

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Check out my result : line integral

    I=\int_C \frac{dx+dy}{x^2+y^2} where C is described by r(t)=(\cos t, \sin t), 0\leq x \leq 2\pi.
    My attempt : |r'(t)|=1 so I=\int_0^{2\pi} \frac{\cos t dt+\sin tdt}{\cos^2t + \sin^2t}=0.
    I realize that C is a circle of radius 1 and that if my result is good then F(x,y)=\frac{dx+dy}{x^2+y^2} is a conservative field. But I'm not sure I've done it right.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by arbolis View Post
    I=\int_C \frac{dx+dy}{x^2+y^2} where C is described by r(t)=(\cos t, \sin t), 0\leq x \leq 2\pi.
    My attempt : |r'(t)|=1 so I=\int_0^{2\pi} \frac{\cos t dt+\sin tdt}{\cos^2t + \sin^2t}=0.
    I realize that C is a circle of radius 1 and that if my result is good then F(x,y)=\frac{dx+dy}{x^2+y^2} is a conservative field. But I'm not sure I've done it right.
    you didn't do it right.

    here we have x = \cos t \implies \boxed{dx = - \sin t~dt} and

    y = \sin t \implies \boxed{dy = \cos t~dt}

    (the answer is still 0 though )
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  3. #3
    MHF Contributor arbolis's Avatar
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    Thanks for the correction. I do still get 0 as you pointed out.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by arbolis View Post
    Thanks for the correction. I do still get 0 as you pointed out.
    indeed. by now you should have it memorized that integrating sine or cosine over any multiple of their period yields zero. so for instances, \int_0^{2 \pi} \sin t~dt = \int_\pi^{3 \pi} \sin t~dt = \int_{\pi /2}^{5 \pi / 2} \cos t~dt = \int_\pi^{9 \pi} \sin t~dt = \int_a^{a + 2k \pi} \sin t ~dt = \int_k^{2n \pi} \cos t~dt = 0 etc etc etc
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