$\displaystyle \int_{0}^{\pi/2} \int_{0}^{\pi/2} \int_{1}^{2} \rho^{2} \sin \phi \ d \rho \ d \phi \ d \theta $
I evaluated it and got $\displaystyle 0 $.
So what type of solid is this?
i did not get zero, and you should not either. this integral is giving the volume of a solid. it gives the volume of the region in the first octant bounded between a sphere of radius 1 and a sphere of radius 2, both centered at the origin.
Think of this. Everything from the outside to the grey thing. Except in this solid it is 1/2 the size of semi-earth. So divide that shape into two equal part to get your solid.