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asmani Feb27-12 11:12 AM

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 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:

http://physicsforums.com/attachment....1&d=1330362487

What's wrong with this solution?

Thanks in advance.

AdrianMay Feb28-12 07:02 AM

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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