A double integral
3 Attachment(s)
Hi all.
Suppose that we want to compute the following indefinite integral: http://physicsforums.com/attachment....1&d=1330362487 The correct solution by Mathematica: http://physicsforums.com/attachment....1&d=1330362487 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)(\t heta+c_2)[/tex] If c_{1}=c_{2}=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: http://physicsforums.com/attachment....1&d=1330362487 What's wrong with this solution? Thanks in advance. 
Re: A double integral
## \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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