 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 $x^2+y^2\le 4$ and $-3\le s\le 2$. On the curved cylinder, use the parametrizationThe surface area element follows fromConsequently the curved contribution isOn either cap $z=x+s$, the surface area element is $\sqrt{1+z_x^2+z_y^2}dxdy=\sqrt{2}dxdy$ over a radius-two disk. The lower cap has $x-z=3$ and the upper cap has $x-z=-2$. Thusand the complete scalar surface integral isThe surface area element is unsigned: opposite outward normals on the caps do not introduce extra minus signs.