2D Cartesian integral to polar integral

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Discussion Overview

The discussion revolves around the conversion of a 2D Cartesian integral to a polar integral, specifically evaluating the integral of the logarithm function over a small ball (or disk) in the plane. The scope includes mathematical reasoning and technical explanation related to integration in different coordinate systems.

Discussion Character

  • Technical explanation, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant presents an integral in Cartesian coordinates and seeks confirmation on its correctness when transformed into polar coordinates.
  • Another participant agrees with the transformation, noting the area element in polar coordinates as approximately a box with dimensions rdθ × dr, leading to the volume element of rdrdθ.
  • A third participant questions the terminology used, suggesting that "ball" should be replaced with "circle" for clarity.
  • A later reply acknowledges the previous comment but humorously affirms the use of "ball" in a 2D context.
  • Another participant further refines the terminology, suggesting that "disk" is the more precise term instead of "circle."

Areas of Agreement / Disagreement

Participants generally agree on the correctness of the integral transformation, but there is a minor disagreement regarding the terminology used to describe the geometric shape involved.

Contextual Notes

There is some ambiguity regarding the definitions of "ball," "circle," and "disk," which may affect clarity in communication. The discussion does not resolve these terminological issues.

Somefantastik
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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.

Thanks!
 
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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.
 
Yeah, looks ok, if by "ball" you actually meant circle.
 
Yeah yeah a ball in the 2d world...

thanks for the input ;)
 
Defennder said:
Yeah, looks ok, if by "ball" you actually meant circle.

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

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