Nested-disk subtraction for paraboloid volumes ($V=\frac{\pi }{2}(A^2-B^2)$)

Suppose a solid is above the elliptic paraboloid $z=x^2+y^2$ and below a plane whose vertical separation from it is $A-|\mathbf{x}-\mathbf{p}{|}^2$, where $A>0$. A second lower plane exceeds the elliptic paraboloid by $B-|\mathbf{x}-\mathbf{q}{|}^2$, with $B>0$. If $|\mathbf{p}-\mathbf{q}|+\sqrt{B}\le \sqrt{A}$, the disk where the lower plane matters lies within the upper projection. Raising the bottom to this lower plane therefore removes exactly its excess volume, without changing the outer projection.

Translate polar coordinates separately to the two centres. The double integral over a disk of squared radius $A$ isSubtracting the analogous integral with $B$ proves the displayed volume formula. The containment condition is essential: if the two disks overlap only partly, subtract only over their intersection.