Without computations, one could recognize this integral as the volume of one half of circular cone with height $9$ and base radius $3$. The volume is $\frac12 \frac{\pi 3^2\cdot 9}{3} = \frac{27\pi}{2} $
To set up integrals, begin with the inner variable, $z$: $ \int_{0}^{9-3\sqrt{x^2+y^2}}\,dz = \int_0^{9-3r}\,r\,dz$ The $x,y$ limits describe the upper half of disk of radius $3$ (because $y\ge 0$). This means $0\le r\le 3$ and $0\le \theta \le \pi$. So, the result is $\int\limits_{0}^{\pi}\int\limits_{0}^{3}\int\limits_{0}^{9-3r}r\,dz\, dr\,d\theta =\int\limits_{0}^{\pi}\int\limits_{0}^{3} (9r-3r^2)\,dr\,d\theta = \pi (9\cdot 3^2/2-3^3) = \frac{27 \pi}{2} $
Negative values of $r$ are sometimes useful in drawing polar curves, but they should never be used in integration.