I have to evaluate
$$\iint_{S}\vec{F}\cdot d\vec{S}$$
with $$\vec{F}(x,y,z)=(y,-x,z)$$ and $S$ is bounded by $z=9-x^2-y^2$ and $z=5$. I can't use Gauss Theorem to do it.
I thought about using the following "formula" $$\int_{S}\vec{F}\cdot d\vec{S}=\iint_{D}\left(-P\cdot\frac{\partial g}{\partial x}-Q\cdot\frac{\partial g}{\partial y}+R\right)dA$$ with $\vec{F}=(P,Q,R)$ and $S$ is the graph of $z=g(x,y)$.
However, as $z\geq5$, do I have to take $$g(x,y)=9-x^2-y^2-5?$$
Parameterize the paraboloid with
$$\vec s_P = (u \cos v, u \sin v, 9-u^2), \quad u\in[0,2] \land v \in[0,2\pi]$$
and normal vector
$$\vec n_P = \frac{\partial \vec s_P}{\partial u} \times \frac{\partial \vec s_P}{\partial v} = \left(2u^2\cos v,2u^2\sin v,u\right)$$
and the disk with
$$\vec s_D = (u \cos v, u \sin v, 5), \quad u\in[0,2] \land v\in[0,2\pi]$$
and normal (easy to determine by inspection) $\vec n_D = (0,0,-1)$.
Integrating along the respective surfaces yields
$$\begin{align*} \iint\limits_{\rm paraboloid} \vec F \cdot d\vec S &= \int_0^{2\pi} \int_0^2 \left(9u-u^3\right) \, du \, dv \\[2ex] \iint\limits_{\rm disk} \vec F \cdot d\vec S &= \int_0^{2\pi} \int_0^2 -5u \, du\,dv \\[2ex] \implies \iint_S \vec F\cdot d\vec S &= \int_0^{2\pi} \int_0^2 \left(4u-u^3\right) \, du \, dv \end{align*}$$
which agrees with your solution via divergence theorem.