Surface integral - missing limits? 
$f(x,y,z)=2y(x^2+1)^{-1}(1+4z)^{-1/2}$, $S=\{z=x^2+y^2, |y|<1\}$. Find $\int_S f\; dA.$

I got a solution for the integral, but I'm not sure about its limits.
$$\int_S f\;dA=\int\int f(x,y,g(x,y))\sqrt{\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2+1}\;dA\\=\int\int 2y(x^2+1)^{-1}(1+4x^2+4y^2)^{-1/2}\sqrt{4x^2+4y^2+1}\;dxdy.$$
I changed to polar coordinates
$$\int\int 2r\sin t (r^2\cos^2 t+1)^{-1}(1+4r^2)^{-1/2}\sqrt{4r^2+1}(r)\;drdt\\=\int \int 2\sin t(\cos^2 t+1)^{-1}\;drdt$$
Then with a substitution $w=\cos t$
$$\int 2\sin t(\cos^2 t+1)^{-1}\;dt=-2\int (w^2+1)^{-1}\;dt=-2\arctan w.$$
So I think the I should put the limits in $-2\arctan \left(\frac{x}{r}\right)$. I have three problems: (1) Limits for $x$ are not given, and $z$ increases as $x$ increases so by a look at the graph it looks like the integral tends to +infinity, (2) $y$ bounded by $-1$ and $1$, but this limits seem to vanish the integral, (3) If there were limits that do not vanish the integral my solution is negative, isn't that strange?
 A: $y = r \sin \theta\\
|y| = 1\\
r |\sin \theta| = 1\\
r = |\csc \theta|$
But only for some values of $r, \theta$ are the planes |y| = 1 in play.
$r\le 1$ the planes are not in play and $\theta$ can range from $0$ to $2\pi$
But, for reasons I will detail below, I am going to suggest integrating from $-\pi$ to $\pi$
$r> 1$
$\theta \in [-\pi, -\pi + \arcsin \frac 1r]\cup[ - \arcsin \frac 1r,  + \arcsin \frac 1r]\cup[\pi - \arcsin \frac 1r, \pi] $ the planes are out of play
And otherwise the planes are in play.
This means we have several integrals to evaluate.
$\int_{-\pi}^{\pi}\int_0^1 f(r,\theta) \;dr\;d\theta + 
\int_{-\pi}^{\theta_1}\int_1^\infty f(r,\theta) \;dr\;d\theta + 
\int_{\theta_1}^{\theta_2}\pi\int_1^{\csc\theta} f(r,\theta) \;dr\;d\theta$, etc. 
Now I am going to suggest you look at it in this order
$\int_{-\pi}^{\pi}\int_0^1 f(r,\theta) \;dr\;d\theta \\+ 
\int_{-\pi}^{-\pi + \arcsin \frac 1r}\int_1^\infty f(r,\theta) \;dr\;d\theta + 
\int_{\pi - \arcsin \frac 1r}^{\pi}\int_1^{\infty} f(r,\theta) \;dr\;d\theta\\
+\int_{-\pi + \arcsin \frac 1r}^{-\arcsin \frac 1r}\int_1^{\csc\theta} f(r,\theta) \;dr\;d\theta+\int_{\arcsin \frac 1r}^{\pi-\arcsin \frac 1r}\int_1^{\csc\theta} f(r,\theta) \;dr\;d\theta\\+
\int_{- \arcsin \frac 1r}^{\arcsin \frac 1r}\int_1^{\infty} f(r,\theta) \;dr\;d\theta$ 
And if I have set this up right, each row should equal 0!
Now, to save some work....
$f(x,y,z) = -f(x,-y,z)$ evaluated over an interval that is symmetric across the x axis.
That what does that imply?
$\iint f \;dy \;dx = 0$
