= Solution
For the <scalar surface integral over an obliquely cut cylinder>, split the <scalar surface integral> into the curved <cylinder> and its two planar caps. Write $s=z-x$, so the solid has $x^2+y^2\le4$ and $-3\le s\le2$. On the curved <cylinder>, use the <parametrization>
$$
\mathbf R(\theta,s)=(2\cos\theta,2\sin\theta,2\cos\theta+s),
\qquad 0\le\theta\le2\pi,\quad -3\le s\le2.
$$
The <surface area element> follows from
$$
\mathbf R_\theta\times\mathbf R_s=(2\cos\theta,2\sin\theta,0),\qquad
dS=2\,d\theta\,ds,\qquad x-z=-s.
$$
Consequently the curved contribution is
$$
I_{\rm side}=\int_0^{2\pi}\int_{-3}^{2}(-s)\,2\,ds\,d\theta=10\pi.
$$
On either cap $z=x+s$, the <surface area element> is $\sqrt{1+z_x^2+z_y^2}\,dx\,dy=\sqrt2\,dx\,dy$ over a radius-two <disk>. The lower cap has $x-z=3$ and the upper cap has $x-z=-2$. Thus
$$
I_{\rm lower}=3\sqrt2(4\pi)=12\sqrt2\pi,\qquad
I_{\rm upper}=-2\sqrt2(4\pi)=-8\sqrt2\pi,
$$
and the complete <scalar surface integral> is
$$
\boxed{\iint_S(x-z)\,dS=\pi(10+4\sqrt2).}
$$
The <surface area element> is unsigned: opposite outward normals on the caps do not introduce extra minus signs.
\Image[/past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2021-resit-obliquely-cut-cylinder.png]
{title=Original sketch of the solid inside the cylinder between two parallel sloping caps}
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