Vrftsjtryk

I'm struggling with surface integrals, and I still do not have much confidence with the parameterization of functions. This is the exercise I would like to solve:

Calculate the surface integral of the function

$$f = (x-1)^2 + (y-2)^2$$

extended to the surface $P equiv (1+rho cos vartheta, 2 + rho sin vartheta, 4 - rho^2)$ with $vartheta in [0, 2pi], rho in [sqrt{2}, sqrt{3}]$

without going into detail in the steps and in the final result, how should I proceed? The first thing I should do is find the parameterization of the surface, right? And can I retrieve it?

Thanks in advances

Your surface $S$ is already given in a parametrized form, i.e. your parametrization is
$$ P(u,v) = (1+ucos v, 2+usin v, 4-v^2), quad uin [sqrt 2, sqrt 3], quad v in [0,2pi]. $$
Now you should compute your surface area element $mathrm dA$, which is defined as
$$ mathrm dA = |partial_u P times partial_v P| , mathrm du , mathrm dv, $$
and the sought integral then becomes
$$ int_S f , mathrm dA = int_{0}^{2pi} int_{sqrt 2}^{sqrt 3} f(P(u,v)) |partial_u P times partial_v P| , mathrm du ,mathrm dv. $$
Can you proceed?