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[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.

Thanks!

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# 2D Cartesian integral to polar integral

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