Evaluate the integral by changing to spherical coordinates. $$\int_{0}^{6}  \int_0^{\sqrt{36-x^2}} \int_{\sqrt{x^2+y^2}}^\sqrt{72-x^2-y^2} xy~ dzdydx $$
I tried converting it and I ended up with
$$\int_0^{2\pi}\int_0^{\pi}\int_0^{\sqrt{72}}\left[p\sin(\phi)\cos(\theta)p\sin(\phi)\sin(\theta)\right] p\sin^2(\phi)~ dpd\phi d\theta$$
I feel like I could be way off here. Did I get anything wrong here?
(Apologize for the formatting!)
 A: Intersect $z=\sqrt{72-x^2-y^2}$ and $z=\sqrt{x^2+y^2}$ and get $72-x^2-y^2=x^2+y^2$ and so $2x^2+2y^2=72$ and so $x^2+y^2=36$. Thus the upper-hemisphere and cone intersect along a circle of radius 6. 
Next, the outer bounds give: $0 \leq x \leq 6$ and $0 \leq y \leq \sqrt{36-x^2}$. This is a quarter of the disk $x^2+y^2 \leq 36$. 
So your region is a quarter of an ice cream cone. :)
Your bounds for $\rho$ are fine: Take a ray emanating from the origin and you first hit the upper-hemisphere (of radius $\sqrt{72}$). So $0 \leq \rho \leq \sqrt{72}$.
Since you only have a quarter of the disk (in the first quadrant), $0 \leq \theta \leq \pi/2$.
Finally, $\phi$ sweeps out from the $z$-axis. It stops when you hit the cone. The cone: $z=\sqrt{x^2+y^2}$ in spherical coordinates is $\rho \cos(\phi) = \rho \sin(\phi)$ so that $\tan(\phi)=1$ and so $\phi=\pi/4$. Thus $0 \leq \phi \leq \pi/4$.
A: $z= \sqrt{72- x^2- y^2}$ is the hemisphere above the xy-plane with center at (0, 0, 0) and radius $6\sqrt{2}$ while $z= \sqrt{1- x^2- y^2}$ is the hemisphere with center at (0, 0, 0) and radius 1.  Just on the basis of that, I would say that the $\rho$ limits would be from 1 to $6\sqrt{2}$ (or some subset of that depending on the other integrals) and the $\phi$ limits 0 to $\pi/2$.  
