The part of the surface within the cylinder is defined over the domain
[tex]R = \left\{(x,y) ~:~ x^2 + y^2 \le 4\right\}[/tex]
which we can write in polar coordinates as
[tex]R = \left\{(r,\theta) ~:~ 0\le r\le2 \text{ and } 0 \le\theta\le2\pi\right\}[/tex]
Then the surface area of [tex]z=f(x,y)[/tex] over the region [tex]R[/tex] is given by the integral
[tex]\displaystyle \iint_R dA = \iint_R \sqrt{1+\left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2} \, dx \, dy[/tex]
Let
[tex]z = f(x,y) = x^2 + y^2[/tex]
with partial derivatives
[tex]\dfrac{\partial f}{\partial x} = 2x[/tex]
[tex]\dfrac{\partial f}{\partial y} = 2y[/tex]
Then in the integral, we have
[tex]\displaystyle \iint_R dA = \iint_R \sqrt{1+4x^2+4y^2}\,dx\,dy \\\\ ~~~~~~~~ = \int_0^{2\pi} \int_0^2 r\sqrt{1+4r^2}\,dr\,d\theta \\\\ ~~~~~~~~ = 2\pi \int_0^2 r\sqrt{1 + 4r^2} \, dr \\\\ ~~~~~~~~ = \frac\pi4 \int_1^{17} \sqrt s\,ds = \boxed{\frac\pi6 \left(17^{3/2} - 1\right)}[/tex]
where we substituted [tex]s=1+4r^2[/tex] and [tex]ds=8r\,dr[/tex].
Alternatively, we can parameterize the surface [tex]S[/tex] by
[tex]\vec s(r,\theta) = r\cos(\theta)\,\vec\imath + r\sin(\theta)\,\vec\jmath + r^2\,\vec k[/tex]
with the same domain as [tex]R[/tex]. Then the normal vector to [tex]S[/tex] is
[tex]\vec n = \dfrac{\partial\vec s}{\partial r}\times\dfrac{\partial\vec s}{\partial \theta} = -2r^2\cos(\theta)\,\vec\imath - 2r^2\sin(\theta)\,\vec\jmath + r\,\vec k \\\\ \implies \|\vec n\| = \sqrt{r^2 + 4r^4} = r\sqrt{1+4r^2}[/tex]
Then the area is given by the surface integral
[tex]\displaystyle \iint_R dS = \int_0^{2\pi} \int_0^2 \|\vec n\| \, dr \, d\theta = \int_0^{2\pi} \int_0^2 r \sqrt{1+4r^2}\, dr\,d\theta[/tex]
just as before.
