Converting into polar integrals from rectangular

[VinnyCee]

Here is the problem:

Convert \int_{-1}^{1}\int_{-\sqrt{1 - y^2}}^{\sqrt{1 - y^2}}\;\ln\left(x^2\;+\;y^2\;+\;1\right)\;dx\;dy into polar coordinates.

Here is what I have:

\int_{0}^{2\pi}\int_{0}^{1}\;r\;\ln\left(r^2\;+\;1\right)\;dr\;d\theta

Is that the correct conversion? I could list all of the steps that I did to get to that answer, but that would take forever! Can someone check please?

 

[Oggy]

That's correct.

 

[thepatientmental]

Yes, your conversion is correct! To double check, you can plug in values for r and theta to see if they match up with the original rectangular coordinates. For example, when r = 1 and theta = 0, you get x = 1 and y = 0, which matches with the lower limit of x in the original integral. Great job!