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 $171\pi/2$