# Double Integral

• Aug 18th 2013, 08:14 AM
kjchauhan
Double Integral

$I=\int{\int{\ln\left(\frac{1}{\sqrt{x^2+y^2}} \right) dxdy$

Thank you very much..
• Aug 18th 2013, 12:06 PM
HallsofIvy
Re: Double Integral
The first thing you should do is use properties of logarithms to write the integrand as $ln\left(\frac{1}{\sqrt{x^2+ y^2}}\right)= -\frac{1}{2}ln(x^2+ y^2)$. then change to polar coordinates so that $x^2+ y^2= r^2$ and $dxdy= r drd\theta$. The integral become $-\frac{1}{2}\int\int ln(r^2)r dr\theta= -\frac{1}{2}\int d\theta \int ln(r^2) r dr$. Can you do that?
• Aug 18th 2013, 07:59 PM
kjchauhan
Re: Double Integral
Thank you very much.
• Aug 18th 2013, 10:14 PM
kjchauhan
Re: Double Integral
I got the answer
$I = -\frac{1}{4}\left[ tan^{-1}\left( \frac{y}{x} \right) \left[ \left(x^2+y^2 \right) ln \left(x^2+y^2 \right) - \left(x^2+y^2 \right) + c_{1} \right] + c_{2} \right]$

Is it correct?
• Aug 19th 2013, 12:38 AM
Prove It
Re: Double Integral
What are your bounds? Or what is your region you're integrating over?
• Aug 19th 2013, 07:20 AM
HallsofIvy
Re: Double Integral
ProveIt: The final integral has both x and y in it, as well as two "constants of integration" so there are no bounds- this is an "indefinite integral".

kjchauhan, I have not calculated it myself but that looks good.