I try to calculate this surface integral, however my method does not work. Problem
Attempted sol: I tried to project the surface at the $xy$ plane using this formula and the relationship $z = \sqrt{x^2-y^2}$. Projection of a surface However, then I got some complicated expression that I do not now what to do with (and how to set up the limits).
Using a cylindrical substitution,
$$x=\rho\\y=\rho\cos\theta\\z=\rho\sin\theta$$
with bounds $2\le\rho\le3$, and $0\le\theta\le2\pi$. Letting this $(x,y,z)$ be the value of $f(\rho,\theta)$, the integral becomes $$\int_0^{2\pi}\int_2^3\rho^4\cos^2\theta\left|\left|\frac{\mathrm df}{\mathrm d\rho}\times\frac{\mathrm df}{\mathrm d\theta}\right|\right| \,\mathrm d\rho\mathrm\, d\theta$$ which is simply $$\int_0^{2\pi}\int_2^3\rho^4\cos^2\theta\sqrt{2}\rho \,\mathrm d\rho\mathrm\, d\theta$$ or $$\frac{665}{6}\sqrt{2}\int_0^{2\pi}\cos^2\theta \,\mathrm d\theta=\frac{665}{6}\sqrt{2}\pi$$