= Nested-disk subtraction for paraboloid volumes
{title2=$V=\frac\pi2(A^2-B^2)$}
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 $|\mathbf p-\mathbf q|+\sqrt B\leq\sqrt A$, 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 $A$ is
$$
\int_0^{2\pi}\int_0^{\sqrt A}(A-r^2)r\,dr\,d\theta=\frac{\pi A^2}{2}.
$$
Subtracting 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.
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