I understand triple integrals in cartesian and 2D/3D cylindrical polar coordinates (at least I think I do) because I can visualize their coordinate grids. Take the usual $x-y$ coordinate grid, stack them, and you have the 3D cartesian grid. In 3D cylindrical polar, you do the same thing since $z=z$. So just stack up a bunch of circles of constant $r$ that intersect with lines of constant $\theta$.
This makes understanding triple integrals in these coordinates easier since you project the shadow of the object on one of the planes, figure out the bounds in that 2D domain like you do in double integrals, and then figure out the bounds in the vertical dimension. How do you figure out the bounds in spherical coordinates. Onto what domain do you project your 3D object? So if $dV = dzdydx$, you project your object onto the $x-y$ plane, figure out the bounds in that 2D domain, then figure out the bounds in the $z$. If $dV = \rho ^2 \sin\phi d\rho d\theta d\phi$, your domain would be the "$\theta - \phi$" plane. But what plane exactly is this?
So for a sphere centered at the origin of radius $R$, the bounds corresponding to $d\rho d\theta d\phi$ from outside in are $0-\pi$, $0-2\pi$, and $0 - R$. What are the bounds if you shift that sphere up so that the sphere is centered at $z = R/2$?