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Math Help - Converting to a Polar Integral

  1. #1
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    Converting to a Polar Integral

    How do you integrate f(x,y )=\frac{ln(x^{2}+y^{2})}{x^{2}+y^{2}}\ over the region 1{\le}x^{2}+y^{2}{\le}e^{2}?
    Last edited by mr fantastic; April 5th 2009 at 05:49 PM. Reason: Fixed the y^2
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    Quote Originally Posted by antman View Post
    How do you integrate f(x,y )=\frac{ln(x^{2}+y{2})}{x^{2}+y{2}}\ over the region 1{\le}x^{2}+y^{2}{\le}e^{2}?
     x = r \cos(\theta)

     y = r \sin(\theta) .

    Hence  x^2 + y^2 = r^2 \cos^2(\theta) + r^2 \sin^2(\theta) = r^2(\sin^2(\theta) + \cos^2(\theta)) = r^2
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    Quote Originally Posted by antman View Post
    How do you integrate f(x,y )=\frac{ln(x^{2}+y{2})}{x^{2}+y{2}}\ over the region 1{\le}x^{2}+y^{2}{\le}e^{2}?
    Switch to polar coordinates.
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    Quote Originally Posted by antman View Post
    How do you integrate f(x,y )=\frac{ln(x^{2}+y{2})}{x^{2}+y{2}}\ over the region 1{\le}x^{2}+y^{2}{\le}e^{2}?
    following your lead, and with the aid of Mush's post we convert to polar coordinates. we obtain the integral \int_0^{2 \pi} \int_1^{e^2} \frac {\ln r^2}{r^2} \cdot r ~dr~d \theta = 2 \int_0^{2 \pi} \int_1^{e^2} \frac {\ln r}r ~dr~d \theta

    you should be able to handle that
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    Quote Originally Posted by Jhevon View Post
    following your lead, and with the aid of Mush's post we convert to polar coordinates. we obtain the integral \int_0^{2 \pi} \int_1^{e^2} \frac {\ln r^2}{r^2} \cdot r ~dr~d \theta = 2 \int_0^{2 \pi} \int_1^{e^2} \frac {\ln r}r ~dr~d \theta

    you should be able to handle that
    Should still be  r^2 on the bottom?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mush View Post
    Should still be  r^2 on the bottom?
    no, i don't think so. the r from the r~dr ~d \theta cancels one of the r's at the bottom
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    Quote Originally Posted by Jhevon View Post
    no, i don't think so. the r from the r~dr ~d \theta cancels one of the r's at the bottom
    Oh, indeed.
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    Would is be  2ln(e^{2})^{2}\pi?
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    Quote Originally Posted by antman View Post
    Would is be  2ln(e^{2})^{2}\pi?
    i suppose you mean 2 (\ln e^2)^2 \pi. then yes. but you left the answer in a "clumsy" form. of course, \ln e^2 = 2...
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    You're right. So then it would simply be 8\pi.. correct?
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    Quote Originally Posted by antman View Post
    You're right. So then it would simply be 8\pi.. correct?
    yes
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