2
$\begingroup$

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 $$\int_0^{2\pi}\int_0^\infty r^3 \,\,dr\,d\theta$$?

  • 7
    Yes. $ $ $ $ $ $2011-12-27
  • 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.2011-12-27
  • 0
    @DidierPiau: Thanks!2011-12-27
  • 0
    @Srivatsan: Thanks! I have edited it now. :)2011-12-27
  • 0
    Can this kind of "Is ... correct/true?" question be improved further?2011-12-27

1 Answers 1

2

The answer to your question:

Yes.