Rewriting triple integrals rectangular, cylindrical, and spherical coordinates
Yes your work is correct except the bounds of $\theta$. Please note that the region is in the first octant so $0 \leq \theta \leq \pi/2$.
In cartesian coordinates, note that at the intersection of the sphere and the cone,
$x^2 + y^2 = 9 - z^2 = 9 - x^2 - y^2 \implies x^2 + y^2 = 9/2$
$ \displaystyle \int_0^{3/\sqrt2} \int_0^{\sqrt{9/2-x^2}} \int_{\sqrt{x^2+y^2}}^{\sqrt{9-x^2-y^2}} (x^2 + y^2) ~ dz ~ dy ~ dx$