<s>[INST] Calculate $\iint_D \frac{\sqrt{x^2 + y^2}}{1+x^2+y^2} dx\,dy$ I am attempting to solve the integral: $$\iint_D \frac{\sqrt{x^2 + y^2}}{1+x^2+y^2} dx\,dy\,,$$ where $D$ is bounded by $1<x^2+y^2<9$ and within the sector bounded by lines $$-\frac{y\sqrt{2}}{2}++\frac{x\sqrt{2}}{2}=0$$ and $$\frac{y\sqrt{3}}{2}+\frac{x}{2}=0$$ The illustration of the sector is shown below:

I tried by first calculating the angles between which the sector is bounded and switching to polar coordinates. However that didn't seem to give me the correct answer. I'm not very used to working with polar coordinates so I might have missed something simple along the way. My calculations are shown below:

I'm grateful for any help, thanks in advance!
/Nick
 [/INST] Your upper bound of $\theta=-\frac{\pi}{3}$ should instead be $\theta=-\frac{\pi}{6}$. Look how the red ray in the graph you provided makes an angle of $\theta=-30^\circ$ with the positive $x-$ axis. Therefore $$\iint_{D}\frac{\sqrt{x^2+y^2}}{1+x^2+y^2}dA=\int_{-3\pi/4}^{-\pi/6} \int_1^3\frac{r^2}{1+r^2}drd\theta=\frac{7\pi}{12}\Big[2-\arctan(3)+\frac{\pi}{4}\Big]$$
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