Cartesian into polar integral. I have set up a double integral in course of proving Gauss theorem in physics. I am considering a gaussian cube of edge $a$ and I supposed that mid point of cube is at origin and a charge is placed at this origin. 
The integral is as follow: 
$$\dfrac{qa}{2(4\pi\epsilon_0)}\int_{-a/2}^{a/2}\int_{-a/2}^{a/2}\dfrac{dx\;dy}{\left(\frac{a^2}{4}+x^2+y^2\right)^{\frac{3}{2}}}$$ 
In this integral, $z$ coordinate is fixed on the top surface and is at a height of $\dfrac{a}{2}$ away from origin.
 
Sorry for extremely poor quality of the image, I am on a bad laptop at the moment.
Basically, I could integrated this in Cartesian coordinates and it do yield the correct answer. But I want to obtain the result in spherical 3D coordinates. And I am really having trouble changing the limits. Is there anyway to do it in polar? 
I took the following substitution
$x=r\cos\theta \sin\phi$  ,  $y=r\sin\theta \sin\phi$  ,  $\dfrac{a}{2}\text{(z=constant)}=r\cos\phi$ 
But it didn't yield anything good. 
And please try to keep the answer simple, as i am high school student, thank you.
 A: If I ignore you $z$ coordinate, and just look at the double integral 
$$A= C \int_{-a/2}^{a/2} \int_{-a/2}^{a/2} \frac{dx dy}{(\frac{a^2}{4}+ x^2+y^2)^{3/2}}$$
I would see a square bounded by the points $(-a/2,a/2)\;(a/2,a/2) \;(a/2,-a/2) \; (-a/2,-a/2)$.
Thus I would write this double integral using polar coordinates( and not spherical coordinates because I have just double integral and not triple), so I will divide this square into four parts using its diagonals, thenthe new four bounded domains are specified by $\theta $ and $r$, where $r$  will be a function of $\theta$
$$ A=C .\int_{-\pi/4}^{\pi/4} \int_{0}^{a/(2*\cos \theta )} f(r,\theta) drd\theta + \int_{\pi/4}^{3\pi/4} \int_{0}^{a/(2*\sin \theta )} f(r,\theta) drd\theta+ \int_{3\pi/4}^{5\pi/4} \int_{0}^{-a/(2*\cos \theta )} f(r,\theta) drd\theta +\int_{5\pi/4}^{7\pi/4} \int_{0}^{-a/(2*\sin \theta )} f(r,\theta) drd\theta $$
where $f(r,\theta)=\frac{r}{(\frac{a^2}{4}+ r^2)^{3/2}} $>
Just to inform how I get the boundaries: The first domain of the four new domains is a triangle bounded by the three points $(0,0) \;(a/2,-a/2) \; (a/2,a/2)$. In this domain clearly $\theta $ is ranging between  $-\pi /4$ and $\pi /4$. On the other hand, any ray issued from the origin outward, it cross the domain until the line $x=a/2$ i.e. $r \cos \theta = a/2$  which implies $ r= a/(2* \cos \theta)$. Similarly for the remaining domains.
I hope this will help you.
