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2D Cartesian integral to polar integral

  1. Sep 19, 2008 #1
    Hey everybody,

    [tex] \int_{B(0,\epsilon)} log \frac{1}{r} \ dxdy \ = \ \int^{2\pi}_{0} d\theta \ \int^{\epsilon}_{0} log \frac{1}{r} \ rdr [/tex]

    when r = r(x,y)
    and B is a small ball with radius [tex] \epsilon [/tex]

    Is this right? I haven't done this in forever and I need to be sure.

  2. jcsd
  3. Sep 19, 2008 #2
    Looks right to me. If we have a small infinitesimal area in polar coordinates (in 2d) it would be roughly a box and have dimensions [tex]rd\theta \times dr[/tex] I believe, leading to the [tex]rdrd\theta[/tex] "volume" element you're using.
  4. Sep 19, 2008 #3


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    Yeah, looks ok, if by "ball" you actually meant circle.
  5. Sep 19, 2008 #4
    Yeah yeah a ball in the 2d world...

    thanks for the input ;)
  6. Sep 20, 2008 #5


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    Staff Emeritus
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    To be even more pedantic, it should be "disk", not "circle"!
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