The "position vector" of any point on the cylinder can be written as . The two derivatives:
are tangent vectors to the cylinder. Their cross product is
and the "scalar differential of surface area" is given by the length of that: .
(For this simple geometry, you could have just noted that the differential of length around the cylinder is so the differential of area is .)
Since this integral is
(In fact, just thinking about the "symmetry" (or, more correctly "asymmetry") makes this problem trivial.)