Lots of doubts abot the surface area of a cylinder. I recently started to study parametric surfaces, and I come across this exercise that I try to solve but I have a lot of doubts reganding  the correctness of my resolution, and also I don't find similar examples on the internet.
I need to find the surface area of the cylinder $$x^{2} + y^{2} = 4x$$ bounded by z=0 and z+ x =4. The cylinder is centered at (2,0) with radius 2.
I made the parametrization $$<2+rcos(t) , rsen(t), 2-rcos(t)>$$ with r between 0 and 2, and t between 0 and 2π (First doubt : Is the parametrization right?)
Then, if everything is ok, I would proceed to do the formula of a surface area (I would not write the whole formula because I am very bad at MathJax). But you know, the double integral of the norm of the vector ("u") being "u" the cross product of the partial derivatives of the parametrization. The vector u in this case is (r,0,r) and the norm is $$\sqrt{2}r $$
Then, if everything is right, the area of the surface is the double integral
$$\int_0^{2π}\int_0^2\sqrt{2}*r^2 dθdr  $$
Is this resolution right? If not, can you help me? Thanks.
PS : I know that the cylinder bounded by the plane is half of the full cylinder. This is the main reason that I think this resolution is wrong.

Let us solve for
$$(x-2)^2+y^2=4,\\z\in[0,2-(x-2)].$$
in polar coordinates. We have
$$A=\int_0^{2\pi}\int_0^{2-2\cos\theta}\,dz\,2\,d\theta=2\int_0^{2\pi}(2-2\cos\theta)\,d\theta=8\pi$$ (the cosine term cancels out).
