Evaluating an Integral by converting into polar coordinates. 
Question. Evaluate the integral by converting into polar coordinate: $$I=\int_{0}^{\sqrt 3}\int_{0}^{\sqrt {4-y^2}}\frac{dx~dy}{4+x^2+y^2}$$

My Solution. Let $f(x,y)=\frac{1}{4+x^2+y^2}$. Now the region of the integration is $S_1 \cup S_2$ as depicted in the following figure:
$xy$-plane" />
Now $I=\iint_{S_1} f(x,y)~dxdy +\iint_{S_2} f(x,y)~dxdy=I_1+I_2.$
Hence if I change the coordinate into polar co ordinate by $x=r \cos \theta;~y=r\sin \theta$ where $0<r; 0\le\theta<2 \pi$, $I_1=\int_{r=0}^{2}\int_{0}^{\pi/3}f ~rdrd\theta$
And $I_2=\int_{0}^{\sqrt 3}\int_{0}^{y}f(x,y)~dxdy$.
But I cannot figure out range of $r$ and $\theta$ when $(x,y)$ varies in $S_2$.
How can I find the range of $r$ and $\theta$ in the later case? Please help. Thank you.
 A: HINT. The region $S_2$ is a right triangle. Your start angle $\theta$ clearly is the angle formed by the line $y=\sqrt{3}x$ and the $x$-axis. There is a right triangle there in your picture, I leave it to you to find that start angle. [Think, $\tan \theta$ for that right trangle.] The end angle $\theta$ is clearly vertical at $\frac{\pi}{2}$. This only leaves $r$ to take care of. This is the tricky part.
My hint for $r$ is as follows: clearly the radius depends on the angle, since different angles will force you to go different distances $r$ from the origin to 'land' on the horizontal leg formed by the region you labeled $S_2$. Draw a radius somewhere in the middle of your upper/lower $\theta$. Notice you have a right triangle inside of $S_2$: the hypotenuse being the drawn radius, and one of the legs formed by the vertical length along the $y$-axis of length $\sqrt{3}$. You can find the angle, which we shall call $\theta'$ between these two lines (since you know $\theta$). Does basic Trig give you a formula for $r$ in terms of $\theta'$? Can you replace $\theta'$ in terms of $\theta$? Then you're done!
You can always check your answer at the end: at the start angle the radius is clearly $2$ since it lies on the circle and at angle $\frac{\pi}{2}$ the radius is clearly $1$ (the leg of the right triangle formed by the region you labeled $S_2$. Does your formula for $r$ in terms of $\theta$ agree when you plug in these $\theta$?
A: In $S_2$ clearly $\dfrac{\pi}{3}\leq\theta\leq\dfrac{\pi}{2}$ ans also $y\leq\sqrt{3}$ therefore $r\leq\dfrac{\sqrt{3}}{\sin\theta}$.
