Calculating a surface integral I'm working on a problem which involves calculating a surface integral. The problem is as follows:
Consider S the part of the surface with equation $x^2+y^2=13-z$ which lies above the plane $z=4$. Calculate the following integral
$$ I =\iint_S \vec{r} \cdot\vec{n} \mathrm{d}O$$
where $\vec{r} = x\vec{u_1}+y\vec{u_2}+z\vec{u_3}$ and $\vec{n}$ is the normal to $S$ pointing upwards.
In my textbook I found that I can use that $ I =\iint_S \vec{r} \cdot\vec{n} \mathrm{d}O = \iiint_G \mathrm{div} \vec{v}$ where $G$ is the area in $\mathbb{R}^3$ enclosed by $S$. $\mathrm{div} \vec{v}$ was obvious and easy to calculate and equal to $3$. I feel the problem will be easiest if I convert to cylindrical coordinates so I put
$$ \begin{cases} x = r\cos\theta \\ y=r\sin\theta \\ z = 13-r^2  \end{cases} $$
Now I my problem lies with finding out my boundaries in cylindrical coordinates. I know that $0 \leq \theta \leq 2\pi$ and that $ 4\leq z$ but further I'm not sure what the boundaries would be.
Is my working so far correct? If so what would the remaining boundaries be? If not, what would be the correct method to work this problem out? Is there any other method to working this problem out? Thanks in advance for any help.
 A: Switching to a volume integral over the divergence indeed seems like a good idea, since now our problem is to simply find the volume of the region of integration (and then multiply it by $3$). Note that for a fixed value of $z$, say $z=0$, we get a 'slice' of the surface given by the equation $x^2+y^2=13$, which you may recognize as a circle with radius $\sqrt{13}.$ As $z$ increases, the circles become smaller, and when $z=13$ the slices disappear since the equation $x^2+y^2=13-z$ has no solutions for $z \geq 13$. So the lower $z$-limit of integration should be $4$ (which is given) and the upper limit should be $13$. Since the whole problem is cylindrically symmetric, cylindrical coordinates seem like a good idea. With this, the integral becomes $$3\int_4^{13} \int_0^{2\pi} \int_0^{\sqrt{13-z}} r\,dr\,d\theta\,dz,$$ since for any fixed $z$ the radius of the circle is $\sqrt{13-z}.$ This integral is not too hard to evaluate.
EDIT: As Ted pointed out, the above would be valid if the bottom was part of the surface integral, but it isn't. The fix to this is to subtract the flux integral over the bottom part, which is $$\int_{x^2+y^2=9,\\z=4} (x,y,z) \cdot (0,0,-1) \,dx \,dy = -36\pi.$$
