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If I want to change the following integral from Cartesians to Polars:

$$\int_{-\infty}^\infty\int_{-\infty}^\infty (x-a)^2+(y-b)^2\,\,dx\,dy$$

in a way such that we are centered at $(a,b)$, so $(x-a)^2+(y-b)^2=r^2$,

Is the polar form simply $$\int_0^{2\pi}\int_0^\infty r^3 \,\,dr\,d\theta$$?

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Yes. $ $ $ $ $ $ –  Did Dec 27 '11 at 9:40
2  
There's a small typo in the equation of the circle: it should be $(x-a)^2+(y-b)^2 = r^2$. Nevertheless you got that right in the integral. –  Srivatsan Dec 27 '11 at 9:48
    
@DidierPiau: Thanks! –  Georgie Dec 27 '11 at 10:01
    
@Srivatsan: Thanks! I have edited it now. :) –  Georgie Dec 27 '11 at 10:02
    
Can this kind of "Is ... correct/true?" question be improved further? –  Jack Dec 27 '11 at 13:19

1 Answer 1

up vote 2 down vote accepted

The answer to your question:

Yes.

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