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Math Help - Double Integral

  1. #1
    Member kjchauhan's Avatar
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    Double Integral

    Please help me to sole the following..

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

    Thank you very much..
    Last edited by kjchauhan; August 18th 2013 at 07:51 AM.
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  2. #2
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    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?
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  3. #3
    Member kjchauhan's Avatar
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    Re: Double Integral

    Thank you very much.
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  4. #4
    Member kjchauhan's Avatar
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    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?
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  5. #5
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    Re: Double Integral

    What are your bounds? Or what is your region you're integrating over?
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  6. #6
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    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.
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