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The Torricelli solid/Gabriel's Horn is defined as the rotation-invariant solid delimited by a hyperbola. It appears in De solido hyperbolico acuto where Torricelli proves that it has a finite volume, despite the fact that it is infinite.

Nowadays, that paradox appears in a different guise, as the fact that the area of that surface that bounds the Torricelli solid is infinite. In simple terms, the Torricelli trumpet cannot be painted but it can be filled with paint. This is for example how the story is told in Wikipedia.

However, I do not know who made this observation regarding the area, and when.

Using integral calculus, these facts can be given relatively straightforward proofs.

Torricelli's proof that the solid has finite volume uses an approximation of the solid as a stack of “infinitesimally” small concentric cylinders. This is indeed enough to bound that volume.

However, to evaluate the area, that would not be enough and to write a proof in Torricelli's style (or in the style of the early inventors of the infinitesimal calculus), it would be necessary to consider the solid as a stack of infinitesimall truncated cones.

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  • $\begingroup$ It is a good question who converted Torricelli's infinite solid of finite volume into the painter's paradox. But that the surface area is infinite follows from Pappus's centroid theorem known to Kepler in 1615 and proved by Guldin (Pappus's proof has been lost) in a 1640 book that Torricelli was familiar with due to its critique of indivisibles. However, that was expected of an infinite body, to Torricelli and his contemporaries it was the finite volume that was the astonishing part. $\endgroup$
    – Conifold
    Commented Oct 29, 2023 at 11:42
  • $\begingroup$ @Conifold, Guldin didn't prove the rule but merely checked it in a few cases, and declared that it must therefore patently be true. Both Cavalieri and Torricelli took him up on it; see the recent thesis by Héctor Manuel Delgado. $\endgroup$
    – Mikhail Katz
    Commented Oct 29, 2023 at 12:50
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    $\begingroup$ Antoine, the truncated cones are not a problem because we are only looking for a lower bound to show that the total area is infinite. The "horizonal" part of the cones (i.e., parallel to the axis) already gives infinite area. $\endgroup$
    – Mikhail Katz
    Commented Oct 29, 2023 at 12:52

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