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Math Help - Green's wih polar coordinates!! :(

  1. #1
    s7b
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    Green's wih polar coordinates!! :(

    Can someone please help me with this question;

    Use Green's theorem o find the counterclockwise circulation and outward flux for the field F and curve C:

    F=(arctan(y/x))i + ln(x^2+y^2)j
    C: The boundary of the region defined by the polar coordinate inequalities
    r between 1 and 2
    theta between 0 and pi


    I know the formulas for greens, I'm just confused how to handle this with the polar coordinate boundary.
    Please help me out if you can!!!
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  2. #2
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    Quote Originally Posted by s7b View Post
    Can someone please help me with this question;

    Use Green's theorem o find the counterclockwise circulation and outward flux for the field F and curve C:

    F=(arctan(y/x))i + ln(x^2+y^2)j
    C: The boundary of the region defined by the polar coordinate inequalities
    r between 1 and 2
    theta between 0 and pi


    I know the formulas for greens, I'm just confused how to handle this with the polar coordinate boundary.
    Please help me out if you can!!!
    Note that the region of integration is the area between the top half two circles

    x^2+y^2=4 and x^2+y^2=1

    using greens theorem we should get

    \int _{D} \int \frac{x}{x^2+y^2}dA

    but this will be much easier to integrate in polar coordinates so

    \int_{0}^{\pi}\int_{1}^{2} \frac{r\cos(\theta)}{r^2}rdrd\theta=\int_{0}^{\pi}  \int_{1}^{2}\cos(\theta)drd\theta
    Last edited by TheEmptySet; March 27th 2009 at 11:37 AM. Reason: had a \theta intead of \pi
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  3. #3
    s7b
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    is this result simply 0....
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by s7b View Post
    is this result simply 0....
    That is what I got for the value of the integral
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