2D Cartesian integral to polar integral

[Somefantastik]

Hey everybody,

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

when r = r(x,y)
and B is a small ball with radius $$\epsilon$$

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

Thanks!

 

[yaychemistry]

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 $$rd\theta \times dr$$ I believe, leading to the $$rdrd\theta$$ "volume" element you're using.

 

[Defennder]

Yeah, looks ok, if by "ball" you actually meant circle.

 

[Somefantastik]

Yeah yeah a ball in the 2d world...

thanks for the input ;)

 

[HallsofIvy]

Yeah, looks ok, if by "ball" you actually meant circle.

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