Surface area of paraboloid $z=x^2+y^2$ and below $z=9$

I have attempted to convert to cylindrical coordinates and have gotten to the equation r*sqrt(4r^2+1) but I am unsure of how to set up the triple integral limits from here. If this is correct so far, how do you find the limits for each integral, if not then how would you go about solving it? Thanks.

Your $x$ and $y$ should move inside the disk of radius 3 (from $x^2+y^2=3^2$). So \begin{align} \iint_D\sqrt{1+f_x^2+f_y^2}\,dA &=\int_0^3\int_0^{2\pi} \sqrt{1+4r^2}\,r\,d\theta\,dr\\ \ \\ &=2\pi\,\int_0^3 r\sqrt{1+4r^2}\,dr\\ \ \\ &\ \ \ \ \textit{(set $u=1+4r^2$)} \\ \ \\ &=\frac\pi4\,\left.\vphantom{\int}\frac23\,u^{3/2}\right|_1^{37} =\frac\pi6\,(37^{3/2}-1) \end{align}