I have solved this problem. I wish to find out if my solution is correct.
Problem: Determine the area of the surface $A$ of the paraboloid: $$x^2+y^2-2z=0$$ where $x^2+y^2\le 8$, $y\ge x$
Solution: From the surface: $x^2+y^2-2z=0$ $$\begin{align}z=\dfrac12(x^2+y^2)\\\dfrac{\partial z}{\partial x}=\dfrac12(2x)=x\\\dfrac{\partial z}{\partial y}=\dfrac12(2y)=y\end{align}$$
Area $$A=\iint_S \,\mathrm dS$$
$$\begin{align}\mathrm dS&=\sqrt{\left(\dfrac{\mathrm dz}{\mathrm dx}\right)^2+\left(\dfrac{\mathrm dz}{\mathrm dy}\right)^2+1}\,\mathrm dA\\&=\sqrt{x^2+y^2+1}\,\mathrm dA\\&=3\,\mathrm dA\end{align}$$
$$\begin{align}\text{Area }A&=\iint_S 3\,\mathrm dA\qquad\qquad \mathrm dA=r\,\mathrm dr\,\mathrm d\theta\\&=3\int_0^{2\pi}\int_0^{2\sqrt2}r\,\mathrm dr\\&=6\pi\left[\frac{r^2}2\right]_0^{2\sqrt2}\\&=6\pi\cdot 4\\&=24\pi\end{align}$$
You've done two mistakes ;
$(1)$ : while finding $dS$ , you've used the equation of cylinder but we only use equation of that whose surface area has to be found. So,
$\begin{align}\mathrm dS=\sqrt{1+x^2+y^2}\,\mathrm dA\\&\end{align}$
$(2)$: You've changed the integral into polar coordinate system , second mistake is in limit of "$\theta$".
Look , When we take projection of the required surface of paraboloid on $XY$-plane , we get the shaded region within the circle.
Now,
$\begin{align}\text{Surface Area }&=\int_{{\frac{\pi}{4}}}^{\frac{5\pi}{4}}\int_0^{2\sqrt2}r \sqrt{1+r^2}\,\mathrm dr \mathrm d\theta \\&\end{align}$
Which I may leave to you!!