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...
 A: I think you will find it easiest if you write the integral in the form
$$\int_?^?\int_?^? \frac{1}{r^3}\,dr\,d\theta\ ,$$
that is, with $\theta$ on the "outside" integral.  Looking at your diagram, the minimum $\theta$ value in the region of integration is $\theta=0$, which occurs along the $x$ axis.  The maximum $\theta$ value occurs at the point $(1,1)$, giving $\theta=\pi/4$.  So we have
$$\int_0^{\pi/4}\int_?^? \frac{1}{r^3}\,dr\,d\theta\ .$$
For any fixed $\theta$ value, the values of $r$ in the region go from the vertical line $x=1$ to the semicircle.  At $x=1$ we have $r=\sec\theta$ as you have noted.  The easiest way to get the maximum $r$ value is to draw a line from the origin at angle $\theta$ until it hits the semicircle, and a line from there to $(2,0)$.  Then you can see in the diagram a right-angled triangle (because the angle in a semicircle is a right angle) with hypotenuse $2$, and so we get $r_{\rm max}=2\cos\theta$.  So the integral is
$$\int_0^{\pi/4}\int_{\sec\theta}^{2\cos\theta} \frac{1}{r^3}\,dr\,d\theta\ .$$
For sketching the graph,
$$y=\sqrt{2x-x^2}\quad\Rightarrow\quad x^2-2x+y^2=0
  \quad\Rightarrow\quad (x-1)^2+y^2=1\ ,$$
and not forgetting that $y\ge0$.
A: $$
\sqrt{2\,x-x^2}=\sqrt{1-(x-1)^2}.
$$
The graph is a semicircle of radius $1$ centered at $(1,0)$.
A: In polar coordinates we have 
\begin{aligned}
\int_{1}^{2}\int_{0}^{\sqrt{2x-x^2}}\frac{1}{(x^2+y^2)^2} dydx&=\int_{0}^{\frac{\pi}{4}}\int_{\sec\theta}^{2\cos\theta}\frac{r}{r^4} drd\theta \\
&=\frac{\pi}{16}
\end{aligned}
A: $\int_{1}^{2} \int_{0}^{\sqrt{2 x-x^{2}}}  \frac{1}{\left(x^{2}+y^{2}\right)^{2}} d y d x$ 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
$$
\begin{aligned}
\int_0^{\pi/4}\int_{\sec \theta}^{2 cos \theta}\frac{rdrd\theta}{r^4} & =\int_0^{\pi/4}\int_{\sec \theta}^{2 cos \theta}\frac{drd\theta}{r^3},\\
&=\int_0^{\pi/4} \frac{-1}{2r^2}|_{\sec \theta}^{2 \cos \theta} d\theta,\\
&=\int_0^{\pi/4}\frac{-1}{8\cos^2\theta}+ \frac{\cos^2\theta}{2} d\theta,\\
&=\int_0^{\pi/4}\frac{\cos^2\theta}{2}-\frac{\sec^2\theta}{8},\\
&=\frac{\theta}{4}+\frac{\sin2\theta}{8}-\frac{\tan \theta}{8}|_0^{\pi/4},\\
&=\frac{\pi}{16}.
\end{aligned}
$$
