The problem: Calculate the surface integral $\int\int_S(x+y^2)d\sigma$ where $S$ is the part of the surface $x^2+y^2=4$ that is between the planes $z=0$ and $z=3$
What I've tried: I described the region in $xz$-plane. For $0≤x≤2$ and $0≤z≤3$, $y=f(x,z)=\sqrt{4-x^2}$, and from here I get the term: $\int\int_S(x+y^2)d\sigma = \int\int_R(x+y^2)(f_x(x,z)^2+f_z(x,z)^2+1)^{1/2}dA$ After tiding up, I get the following: $\int_0^3\int_0^2{{2(-x^2+x+4)}\over{\sqrt{4-x^2}}}dxdz$
How do I approach this problem?
Your approach is fine but doesn't get the whole picture. Your parameterization of $S$ constrains it to the first octant. To fix that, you need to extend the integration range to $x\in[-2,2]$ and integrate over one half of $S$, then compute the same integral again but with $y=-\sqrt{4-x^2}$ (or exploit symmetry) to account for the other half.
For the integral itself, trigonometric substitution would be a common suggestion. Continue with $x=2\sin t$. Note that parameterizing $S$ in polar coordinates from the start would more-or-less accomplish the same thing.
$$\int \frac{4+x-x^2}{\sqrt{4-x^2}} \, dx = \int \frac{4 + 2\sin t - 4\sin^2t}{\sqrt{4-4\sin^2t}} \, 2\cos t\,dt = \int \left(4\cos^2t + 2\sin t\right) \,dt$$
You can alternatively decompose the integrand as
$$\dfrac{x}{\sqrt{4-x^2}} + \sqrt{4-x^2}$$
to employ $u=4-x^2$ for the first term, and recognize the second term's contribution as a geometric area.