How to find the Surface Area?

Question

Find the surface area of the paraboloid $z = x^2+y^2$ that lies inside of the cylinder $x^2+y^2 = 4$.

I keep getting $\frac{\pi}{6} (17\sqrt{17} - 1)$

This is how I did it:

Answer 1

$dS$ is defined by the surface of the paraboloid

$dS = (-\frac {\partial z}{\partial x},-\frac {\partial z}{\partial y},1)\\ \frac {\partial z}{\partial x} = 2x, \frac {\partial z}{\partial y} = 2y $

$\iint ||dS|| \,dy \,dx\\ \iint \sqrt{4x^2 + 4y^2 + 1} \,dy \,dx$

limits of integration are defined by the cylinder lets convert to cylindrical.

$\int_0^{2\pi}\int_0^2 \sqrt{(4r^2 + 1)} r \,dr \,d\theta\\ u = 4r^2 + 1\\ du = 8r dr\\ \int_0^{2\pi}\int_1^{17} \frac 18 \sqrt{u} \,du \,d\theta\\ \int_0^{2\pi} (\frac 18)(\frac23) u^{\frac32} \,d\theta |_1^{17}\\ (2\pi)(\frac1{12})(17\sqrt{17}-1) $

I get the same answer.