Show that $\iint_S (x^2+y^2) dA = 9 \pi /4$ In exam it was asked to show that 

$$\iint_S x^2+y^2 dA = 9 \pi /4$$ for $$S = {\{(x, y, z) | x>0, y>0,3>z>0, z^2 = 3(x^2 + y^2)}\}$$

I have tried many times but I don't get the $9 \pi /4$. 
$$\begin{align}
\iint_S\sqrt{1+f_x^2+f_y^2}\,dA &=\int_0^{\sqrt3}\int_0^{2\pi} r^2\sqrt{1+36r^2}\,r\,d\theta\,dr\\
&=2\pi\,\int_0^\sqrt3 r^3\sqrt{1+36r^2}\,dr = 2 \pi \dfrac{\left(36r^2+1\right)^\frac{3}{2}\left(54r^2-1\right)}{9720}\Bigg|_0^{\sqrt{3}} \\&= 2 \pi\dfrac{161{\cdot}109^\frac{3}{2}+1}{9720} \ne 9 \pi/4 \end{align}$$ 
Where I did it wrong?

Your problem arises from using the fact that $f_x = 6x$ and $f_y = 6y$. Indeed this isn't the case, as $z^2 = 3(x^2+y^2)$, not $z$. Thus you would have:
$$z=f(x,y) = \sqrt{3(x^2+y^2)}$$
$$f_x = \frac{\sqrt{3}x}{\sqrt{x^2+y^2}} \quad \quad f_y = \frac{\sqrt{3}y}{\sqrt{x^2+y^2}}$$ Then we have:
$$\iint_S (x^2+y^2)\sqrt{1+f_x^2 + f_y^2}dxdy = \iint_S (x^2+y^2)\sqrt{1+\frac{3x^2+3y^2}{x^2+y^2}} dxdy = \iint_S 2(x^2+y^2) dxdy$$
Now use polar coordinates and compute the integral. But be careful that $\theta \in \left(0,\frac \pi2\right)$, as $x,y$ are positive.
