Currently I have made the graph in geogebra and I have also made the equality of the $z$ heights to show the intersection curve in the xy plane.
So in the xy plane I have $y = 6-\sqrt{36-x^2}$ and the lines $x=6$ and $y=6$, now I have a problem with setting up the double integral since I don't know how to use the limits of integration together with the $z$ heights.
My 2 height functions are $z=6-y$ and $z=\sqrt{36-x^2}$
If $(x,y,z)$ belongs to your region, then $y\geqslant0$ and, since $y+z\leqslant6$, $y\leqslant6-z\leqslant6$. In fact, $y$ can take any value from $0$ to $6$ (every point $(0,y,0)$ with $y\in[0,6]$ belongs to your region). Now, for each $y\in[0,6]$, $z$ can take any value from $0$ to $6-y$. Finally, for each $y\in[0,6]$ and any $z\in[0,6-y]$, $x$ can take any value from $0$ to $\sqrt{36-z^2}$. So, the volume is\begin{align}\int_0^6\int_0^{6-y}\int_0^{\sqrt{36-z^2}}1\,\mathrm dx\,\mathrm dz\,\mathrm dy&=\int_0^6\int_0^{6-y}\sqrt{36-z^2}\,\mathrm dz\,\mathrm dy\\&=\int_0^618\arcsin\left(1-\frac y6\right)-\frac12(y-6)\sqrt{12y-y^2}\,\mathrm dy\\&=54\pi-72.\end{align}