Find the volume between $x^2+y^2=9 , z=9-x^2-y^2, x^2+y^2+(z-16)^2=9$ So we have the surfaces$$x^2+y^2=9 $$$$ z=9-x^2-y^2$$$$ x^2+y^2+(z-16)^2=9$$
I was thinking this integral may be easier with cylindrical coordinates with the limits $0\le r\le 3$; $0\le\theta\le 2\pi$; and $9-r^2\le z\le S$.
Not sure how to define the superior function of $z$ ($S$) and then set the integral

Your approach by polar coordinates is correct. 
Let $x=r\cos \theta, y=r\sin \theta$, then the first equation gives integrating area, while the second and third equation give the lower and upper bound:
$$
\begin{align}
0&\leq r\leq3 \\
9-r^2&\leq z\\
z&\leq 16-\sqrt {9-r^2}
\end{align}
$$
so we can write the double integral as 
$$
\int^{2\pi}_0 \int^3_0 [ 16-\sqrt {9-r^2}-(9-r^2)]rdrd\theta
$$
which gives us $171\pi/2$
