<s>[INST] Integrating $\int_1^2 \int_0^ \sqrt{2x-x^2} \frac{1}{((x^2+y^2)^2} dydx $ in polar coordinates I'm having a problem converting $\int\limits_1^2  \int\limits_0^ \sqrt{2x-x^2}   \frac{1}{(x^2+y^2)^2}  dy dx $ to polar coordinates.
I drew the graph using my calculator, which looked like half a circle on the x axis. 
I know that $\frac{1}{(x^2+y^2)^2}  dydx$ turns to $ \frac{r}{(r^2)^2}drd\theta$, which would be $ \frac{1}{r^3}$
The region of integration in  $\theta $ is I guess $0 \leq  \theta  \leq  \frac{1}{4}  \pi $, but I'm stuck for r. The lower boundary is $x=1=rcos\theta \rightarrow r=1/cos\theta=sec\theta$, but what is the upper boundary?? Is it  $\sqrt{2rcos\theta-r^2cos^2\theta}$? I tried this but I couldn't integrate my equation...
Could someone please help me out?
Also, how can I draw the graph of $y= \sqrt{2x-x^2} $? I could only know how it looked like by using my calculator...
 [/INST] $\int_{1}^{2} \int_{0}^{\sqrt{2 x-x^{2}}}  \frac{1}{\left(x^{2}+y^{2}\right)^{2}} d y d x$ is an integration over region $y=0$ to $y=\sqrt{2 x-x^{2}}$ and $x=1$ to $x=2$ as shaded region in below graph. Now we will change $xy$ to polar co-ordinate. Using $x=r \cos \theta$, $y=r \sin \theta$ and $dxdy=r dr d\theta$. After changing into polar co-ordinate shaded region in polar co-ordinate will be $r=\sec \theta$ to $r=2 \cos \theta$ and $\theta=0$ to $\theta=\pi/4$.
Graph for the region
Now our integration is
$$
\begin{aligned}
\int_0^{\pi/4}\int_{\sec \theta}^{2 cos \theta}\frac{rdrd\theta}{r^4} & =\int_0^{\pi/4}\int_{\sec \theta}^{2 cos \theta}\frac{drd\theta}{r^3},\\
&=\int_0^{\pi/4} \frac{-1}{2r^2}|_{\sec \theta}^{2 \cos \theta} d\theta,\\
&=\int_0^{\pi/4}\frac{-1}{8\cos^2\theta}+ \frac{\cos^2\theta}{2} d\theta,\\
&=\int_0^{\pi/4}\frac{\cos^2\theta}{2}-\frac{\sec^2\theta}{8},\\
&=\frac{\theta}{4}+\frac{\sin2\theta}{8}-\frac{\tan \theta}{8}|_0^{\pi/4},\\
&=\frac{\pi}{16}.
\end{aligned}
$$
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