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Math Help - Green's Theorem and converting to polar co-ords

  1. #1
    Senior Member chella182's Avatar
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    [SOLVED] Green's Theorem and converting to polar co-ords

    The question goes...

    Use Green's Theorem to evaluate \oint_{\Gamma}(y^3dx-x^3dy), where \Gamma, mapped counter-clockwise, is the circle x^2+y^2=4. [Hint: you may find it convenient to introduce the transformation to polars.]

    Letting x=\cos{\theta} and y=\sin{\theta} and differentiating et cetera, I end up with...
    -\int_{0}^{2\pi}\sin^{4}{\theta}+\cos^{4}{\theta} d\theta
    And I'm stuck from here. I'm guessing it's something to do with relationship \cos^{2}{\theta}+\sin^{2}{\theta}=1 somewhere, but I'm just not getting it. Can anyone help?
    Last edited by chella182; May 6th 2009 at 12:21 PM. Reason: Changing title to let people know it's been solved
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  2. #2
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    Quote Originally Posted by chella182 View Post
    The question goes...

    Use Green's Theorem to evaluate \oint_{\Gamma}(y^3dx-x^3dy), where \Gamma, mapped counter-clockwise, is the circle x^2+y^2=4. [Hint: you may find it convenient to introduce the transformation to polars.]

    Letting x=\cos{\theta} and y=\sin{\theta} and differentiating et cetera, I end up with...
    -\int_{0}^{2\pi}\sin^{4}{\theta}+\cos^{4}{\theta} d\theta
    And I'm stuck from here. I'm guessing it's something to do with relationship \cos^{2}{\theta}+\sin^{2}{\theta}=1 somewhere, but I'm just not getting it. Can anyone help?
    you didn't use Green's Theorem! first use the theorem and then change to polar. you'll end up with a very simple integral.
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  3. #3
    Senior Member chella182's Avatar
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    So doing the whole P and Q thing I end up with -3\iint_{A}x^2+y^2 dA, and NOW do I sub in \cos{\theta} and \sin{\theta} for x and y?
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    Quote Originally Posted by chella182 View Post
    Oh I thought we were meant to change to polars first o.O
    well, what you did wasn't even changing to polar. it was parametrizing the unit circle and not even the given circle. you can't use polar for line integrals. it's used for double integrals only!

    in polar you put x=r \cos \theta, \ y=r \sin \theta not x=\cos \theta, \ y= \sin \theta. (you have two variables r , \theta.)
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  5. #5
    Senior Member chella182's Avatar
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    Okay so what I edited my post you quoted to is wrong as well then?
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  6. #6
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    Quote Originally Posted by chella182 View Post
    So doing the whole P and Q thing I end up with -3\iint_{A}x^2+y^2 dA, and NOW do I sub in \cos{\theta} and \sin{\theta} for x and y?
    this is correct! now put x=r \cos \theta, \ y=r \sin \theta. the Jacobian in polar coordinates is r and x^2+y^2=r^2. thus: -3\iint_{A}x^2+y^2 dA=-3 \int_0^{2 \pi} \int_0^2 r^3 dr d \theta, which is easy to evaluate.
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  7. #7
    Senior Member chella182's Avatar
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    The r^3 comes from the r in the rdrd\theta mutliplied by the r^2 fromthe circle yes? So it'd be...

    \int_{0}^{2\pi}[\frac{r^4}{4}]_{0}^{2} d\theta

    =\int_{0}^{2\pi}4 d\theta

    =[4\theta]^{2\pi}_{0}

    =8\pi

    That right?
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  8. #8
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    Quote Originally Posted by chella182 View Post
    The r^3 comes from the r in the rdrd\theta mutliplied by the r^2 fromthe circle yes? So it'd be...

    \int_{0}^{2\pi}[\frac{r^4}{4}]_{0}^{2} d\theta

    =\int_{0}^{2\pi}4 d\theta

    =[4\theta]^{2\pi}_{0}

    =8\pi

    That right?
    yest it is correct but don't forget that you have a factor -3 outside the integral. so the final answer is -24 \pi.
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  9. #9
    Senior Member chella182's Avatar
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    Oh sugar, yeah, thanks for that
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