= Solution
The surface is an annular band of the downward-opening paraboloid, bounded by the circles $r=1,z=4$ and $r=2,z=1$. Choose the upward orientation.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ia/paper-2-paraboloid-annulus.png]
{title=An annular surface cut from a paraboloid}
The <curl> of $\mathbf A=(3y,-xz,yz^2)$ is
$$
\nabla\times\mathbf A=(z^2+x,0,-z-3).
$$
For the graph $z=5-x^2-y^2$, the upward vector <surface integral> element is
$$
d\mathbf S=(2x,2y,1)\,dx\,dy.
$$
Over the annulus $1<r<2$, the odd term $2xz^2$ integrates to zero. Therefore
$$
\begin{aligned}
I&=\iint_{1<r<2}(2x^2-z-3)\,dx\,dy\\
&=\int_1^2\int_0^{2\pi}
(2r^2\cos^2\theta+r^2-8)r\,d\theta\,dr\\
&=\int_1^2(4\pi r^3-16\pi r)\,dr
=\boxed{-9\pi}.
\end{aligned}
$$
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