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The Reuleaux tetrahedron, sometimes also called the spherical tetrahedron, is the intersection of four closed balls of equal radius placed so that the center of each ball lies on the boundary of the other three. The centers of the balls are therefore located at the vertices of a regular tetrahedron, and the solid consists of an "inflated" tetrahedron with six curved edges.
Note that the name, coined here for the first time, is based on the fact that the geometric shape is the three-dimensional analog of the Reuleaux triangle, not the fact that it has constant width. In fact, the Reuleaux tetrahedron is not a solid of constant width. However, Meißner (1911) showed how to modify the Reuleaux tetrahedron to form a surface of constant width by replacing three of its edge arcs by curved patches formed as the surfaces of rotation of a circular arc. Depending on which three edge arcs are replaced (three that have a common vertex or three that form a triangle), one of two noncongruent shapes can be produced that are called Meissner tetrahedra (Lachand-Robert and Oudet 2007).
To analyze the Reuleaux tetrahedron, fix a tetrahedron of unit edge length with its vertices at the points 
, 
, 
, and 
. Simultaneously solving the equations
of the two spheres centered at 
and 
for 
and 
as a function of 
gives an edge arc as
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(1)
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(2)
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Half an arc is traced out as 
passes from 
to 
, and
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(3)
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(4)
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so the arc length of the curves connecting the vertices is given by
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(5)
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 |  | ![6sqrt(2)int_(sqrt(6)/12)^(sqrt(6)(sqrt(6)-1)/12)[5-4sqrt(6)z(1+sqrt(6)z)]^(-1/2)dz.](/images/equations/ReuleauxTetrahedron/Inline30.svg) |
(6)
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Making the change of coordinates 
,
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(7)
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(8)
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(9)
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Using the Gauss-Bonnet formula gives surface area as
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(10)
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(11)
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(OEIS A202473; Harbourne 2010, Hynd 2025).
The volume is significantly trickier to calculate analytically. Set up spherical coordinates from the centroid of the tetrahedron, so that the distance from the bottom vertex to the radius vector is 1, i.e.,
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(12)
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which simplifies to
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(13)
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giving
![r(theta,phi)=1/4[sqrt(3cos(2phi)+13)-sqrt(6)cosphi].](/images/equations/ReuleauxTetrahedron/NumberedEquation3.svg) |
(14)
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By symmetry, the volume of the Reuleaux tetrahedron is given by
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(15)
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The integral over 
can be done immediately,
![V=1/8int_0^(pi/3)int_0^(phi(theta))[sqrt(3cos(2phi)+13)-sqrt(6)cosphi]^3sinphidphidtheta.](/images/equations/ReuleauxTetrahedron/NumberedEquation5.svg) |
(16)
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To determine the limiting directions, parameterize the corresponding straight edge of the original tetrahedron as a function of the
azimuthal coordinate 
as
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(17)
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(18)
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(19)
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The polar angle 
can then be solved for as a function of 
as
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(20)
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(21)
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The integral over 
can be done by making the change of coordinates
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(22)
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giving
![V=int_0^(pi/3)1/(32)[256-45sqrt(6)+42sqrt(6)cos(2tan^(-1)u)-(58sqrt(13+3cos(2tan^(-1)u)))/(sqrt(1+u^2))-6sqrt(13+3cos(2tan^(-1)u))cos(3tan^(-1)u)+3sqrt(6)cos(4tan^(-1)u)]dtheta.](/images/equations/ReuleauxTetrahedron/NumberedEquation7.svg) |
(23)
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Making the change of variables 
, with 
, gives
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(24)
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so the volume becomes
![V=int_0^1[(8sqrt(3))/(1+3t^2)-(16sqrt(2)(3t+1)(4t^2+t+1)^(3/2))/((3t^2+1)(11t^2+2t+3)^2)-(sqrt(2)(249t^2+54t+65))/((11t^2+2t+3)^2)]dt.](/images/equations/ReuleauxTetrahedron/NumberedEquation9.svg) |
(25)
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This integral can be done analytically using computer algebra, but the analytic form contains inverse trigonometric and logarithmic terms that do not express the result in simplest possible form. However, a much simpler approach using the surface area combined with some straightforward geometry gives the fully simplified form almost immediately as
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(26)
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(27)
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(OEIS A102888; Harbourne 2010, Hynd 2025, Hynd 2026).
