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Let $V(r,z)$ denote the volume of a napkin ring of outer radius $r$ and height $2z$. We have that $V(r,rz) = r^3 V(1,z)$ and $V(1,az) = a^3 V(1,z)$. The former equality is trivial. To see the latter one, note that $V(1,z) = 4\pi z^3/3$.*
It follows that $V(r,z) = r^3 V(1,r^{-1}z) = r^3 r^{-3} V(1,z) = V(1,z)$.
*This might want some fleshing out: e.g. decomposing the complementary part of the sphere into two spherical cones, a cylinder, and two "negative" cones. The volumes of the cylinder and cones can be trivially computed; the volume of a spherical cone can be computed without calculus using its solid angle $\Omega$ and taking the proportion $\Omega/4\pi$.
Let $V(r,z)$ denote the volume of a napkin ring of outer radius $r$ and height $2z$. We have that $V(r,rz) = r^3 V(1,z)$ and $V(1,az) = a^3 V(1,z)$. The former equality is trivial. To see the latter one, note that $V(1,z) = 4\pi z^3/3$ (this might want some fleshing out).z^3/3$.* It follows that$V(r,z) = r^3 V(1,r^{-1}z) = r^3 r^{-3} V(1,z) = V(1,z)$. *This might want some fleshing out: e.g. decomposing the complementary part of the sphere into two spherical cones, a cylinder, and two "negative" cones. The volumes of the cylinder and cones can be trivially computed; the volume of a spherical cone can be computed without calculus using its solid angle$\Omega$and taking the proportion$\Omega/4\pi$. 1 Let$V(r,z)$denote the volume of a napkin ring of outer radius$r$and height$2z$. We have that$V(r,rz) = r^3 V(1,z)$and$V(1,az) = a^3 V(1,z)$. The former equality is trivial. To see the latter one, note that$V(1,z) = 4\pi z^3/3$(this might want some fleshing out). It follows that$V(r,z) = r^3 V(1,r^{-1}z) = r^3 r^{-3} V(1,z) = V(1,z)\$.