If I have an integral:(adsbygoogle = window.adsbygoogle || []).push({});

[itex]\int\int_{R} x^{2} + y^{2} dy dx [/itex]

Where the region R is the area enclosed by a circle centered on the origin of any given radius, is it possible to just convert x^2 + y^2 to r^2 and integrate from 0 to r over dr and 0 to 2 pi over d[itex]\theta[/itex]?

So it would become:

[itex]\int^{2\pi}_{0}\int^{r}_{0} r^{2} dr d\theta[/itex]

I have seen this done before, but for some reason there was an extra r, with the dr. I'm not sure if this was a mistake or not, but I can't see any reason for it to be there. Unfortunately, when I saw it before the radius of the circular region was 1, and so I could not check through another method, since both gave the same answer. I was also unsure if the r in the boundary should be the same as the r in the integrand. Could somebody explain if there is supposed to be an extra r there, or if anything is wrong with the transformation? So far as I can tell, it is as simple as just replacing x with rcos([itex]\theta[/itex]) and y with rsin([itex]\theta[/itex]), but the way I saw it done before suggests otherwise.

Thank you for any help.

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# Converting cartesian surface integral to polar

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