Volume of Paraboloid

Theorem

The volume of paraboloid is half the volume of its circumscribing cylinder.

Proof

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Consider acylinder of radius $r$ and height $h$, circumscribing a paraboloid $y= h \paren {\dfrac x r}^2$ whose apex is at the center of the bottom base of the cylinder and whose base is the top base of the cylinder.

Also consider the paraboloid $y = h - h \paren {\dfrac x r}^2$, with equal dimensions but with its apex and base flipped.

For every height $0 \le y \le h$, the disk-shaped cross-sectional area $\pi \paren {r \sqrt {1 - \dfrac y h} }^2$ of the flipped paraboloid is equal to the ring-shaped cross-sectional area $\pi r^2 - \pi r^2 \paren {\sqrt {\dfrac y h} }^2$ of the cylinder part outside the inscribed paraboloid.

Indeed:

\(\ds \pi \paren {r \sqrt {1 - \dfrac y h} }^2\)

\(=\)

\(\ds \pi r^2 \paren {\sqrt {1 - \dfrac y h} }^2\)

\(\ds \)

\(=\)

\(\ds \pi r^2 \paren {1 - \dfrac y h}\)

\(\ds \)

\(=\)

\(\ds \pi r^2 - \pi r^2 \dfrac y h\)

\(\ds \)

\(=\)

\(\ds \pi r^2 - \pi r^2 \paren {\sqrt {\dfrac y h} }^2\)

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Therefore, the volume of the flipped paraboloid is equal to the volume of the cylinder part outside the inscribed paraboloid.

In other words, the volume of the paraboloid is $\dfrac \pi 2 r^2 h$, half the volume of its circumscribing cylinder.

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