How would you calculate the surface of the part of the paraboloid $z=x^2+y^2$ with $1 \le z \le 4$? Do you calculate it like done below? can you calculate it in another way?
$z=x^2+y^2$
Let
$x=\sqrt{z}\cos\theta$
$y=\sqrt{z}\sin\theta$
$z=z$
where $\theta\in[0,2\pi]$ and $z\in[1,4]$
$ \dfrac{\partial{x}}{\partial{\theta}}=-\sqrt{z}\sin\theta$
$ \dfrac{\partial{y}}{\partial{\theta}}=\sqrt{z}\cos\theta$
$ \dfrac{\partial{z}}{\partial{\theta}}=0$
$ \dfrac{\partial{x}}{\partial{z}}=\dfrac{1}{2\sqrt{z}}\cos\theta$
$ \dfrac{\partial{y}}{\partial{z}}=\dfrac{1}{2\sqrt{z}}\sin\theta$
$ \dfrac{\partial{z}}{\partial{z}}=1$
$\begin{Vmatrix}\hat{i} & \hat{j} & \hat{k} \\ -\sqrt{z}\sin\theta &\sqrt{z}\cos\theta  & 0 \\ \dfrac{1}{2\sqrt{z}}\cos\theta & \dfrac{1}{2\sqrt{z}}\sin\theta & 1\end{Vmatrix}$
$=\sqrt{z}\cos\theta\hat{i}+\sqrt{z}\sin\theta\hat{j}+\dfrac{1}{2}(-\sin^2 \theta-\cos^2 \theta)\hat{k}$
$=\left<\sqrt{z}\cos\theta,\sqrt{z}\sin\theta,-\dfrac{1}{2}\right>$
And its magnitude is given by $\sqrt{z\cos^2\theta+z\sin^2\theta+\dfrac{1}{4}}=\sqrt{z+\dfrac{1}{4}}=\dfrac{1}{2}\sqrt{4z+1}$
Surface $=\displaystyle \int \int_S dS$
$=\displaystyle \int_1^4 \int_0^{2\pi} \dfrac{1}{2}\sqrt{4z+1} \,d\theta\,dz$
$=\bigg|2\pi\cdot\dfrac{1}{2}\cdot\dfrac{2}{3}(4z+1)^{\frac{3}{2}}\dfrac{1}{4}\bigg|_1^4$
$=\dfrac{\pi}{6}(17\sqrt{17}-5\sqrt{5})$

Your result is correct. By using polar coordinates, we obtain
$$\iint_S dS=\int_{\rho=1}^2 \int_0^{2\pi} \sqrt{1+f_x^2+f_y^2} \,(d\theta\,\rho d\rho)=2\pi\int_{\rho=1}^2  \sqrt{1+4\rho^2} \,\rho d\rho\\
=\dfrac{\pi}{6}\left[(1+4\rho^2)^{3/2}\right]_{\rho=1}^2
=\dfrac{\pi}{6}(17\sqrt{17}-5\sqrt{5})$$
where $f(x,y)=x^2+y^2$, $x=\rho\cos\theta$ and $y=\rho\sin\theta$.
P.S. We have that $f_x(x,y)=2x$, $f_y(x,y)=2y$ and therefore $$\sqrt{1+f_x^2+f_y^2}=\sqrt{1+4(x^2+y^2)}=\sqrt{1+4\rho^2}$$
For details take a look here: https://en.wikipedia.org/wiki/Surface_integral
