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A double integral

  1. Feb 27, 2012 #1
    Hi all.

    Suppose that we want to compute the following indefinite integral:

    [​IMG]

    The correct solution by Mathematica:

    [​IMG]

    Now here is the (apparently) incorrect solution by using polar coordinates:
    [tex]\iint\frac{1}{\sqrt{x^2+y^2}}dxdy=\iint\frac{1}{r}rdrd\theta=(r+c_1)(\theta+c_2)[/tex]
    If c1=c2=0, then one solution is:
    [tex]r\theta=\sqrt{x^2+y^2}\tan^{-1}\left ( \frac{y}{x} \right )[/tex]
    But it isn't:

    [​IMG]

    What's wrong with this solution?

    Thanks in advance.
     

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  2. jcsd
  3. Feb 28, 2012 #2
    ## \iint\frac{1}{r}rdrd\theta ## looks good to me but that would just be ## \iint drd\theta ## with ## 0<r<\infty ## and ## 0<\theta<2\pi ## which is ## \infty ##. I was already worried about that when I saw the pole at the origin but apparently you can fix it by chopping the radius of interest.
     
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