In solid geometry, an ungula is a region of a solid of revolution, cut off by a plane oblique to its base. A common instance is the spherical wedge. The term ungula refers to the hoof of a horse, an anatomical feature that defines a class of mammals called ungulates.
The volume of an ungula of a cylinder was calculated by Grégoire de Saint Vincent. Two cylinders with equal radii and perpendicular axes intersect in four double ungulae. The bicylinder formed by the intersection had been measured by Archimedes in The Method of Mechanical Theorems, but the manuscript was lost until 1906.
A historian of calculus described the role of the ungula in integral calculus: Grégoire himself was primarily concerned to illustrate by reference to the ungula that volumetric integration could be reduced, through the ductus in planum, to a consideration of geometric relations between the lies of plane figures. The ungula, however, proved a valuable source of inspiration for those who followed him, and who saw in it a means of representing and transforming integrals in many ingenious ways.
Cylindrical ungula
A cylindrical ungula of base radius r and height h has volume
,. Its total surface area is , the surface area of its curved sidewall is , and the surface area of its top (slanted roof) is .
Proof
Consider a cylinder bounded below by plane and above by plane where k is the slope of the slanted roof: . Cutting up the volume into slices parallel to the y-axis, then a differential slice, shaped like a triangular prism, has volume
where
is the area of a right triangle whose vertices are, , , and , and whose base and height are thereby and , respectively. Then the volume of the whole cylindrical ungula is
which equals
after substituting .
A differential surface area of the curved side wall is
, which area belongs to a nearly flat rectangle bounded by vertices , , , and , and whose width and height are thereby and (close enough to) , respectively. Then the surface area of the wall is
where the integral yields , so that the area of the wall is
, and substituting yields .
The base of the cylindrical ungula has the surface area of half a circle of radius r: , and the slanted top of the said ungula is a half-ellipse with semi-minor axis of length r and semi-major axis of length , so that its area is
and substituting yields
. ∎
Note how the surface area of the side wall is related to the volume: such surface area being , multiplying it by gives the volume of a differential half-shell, whose integral is , the volume.
When the slope k equals 1 then such ungula is precisely one eighth of a bicylinder, whose volume is . One eighth of this is .
Conical ungula
A conical ungula of height h, base radius r, and upper flat surface slope k (if the semicircular base is at the bottom, on the plane z = 0) has volume
where
is the height of the cone from which the ungula has been cut out, and
.
The surface area of the curved sidewall is
.
As a consistency check, consider what happens when the height of the cone goes to infinity, so that the cone becomes a cylinder in the limit:
so that
,
, and
, which results agree with the cylindrical case.
Proof
Let a cone be described by
where r and H are constants and z and ρ are variables, with
and
.
Let the cone be cut by a plane
. Substituting this z into the cone's equation, and solving for ρ yields
which for a given value of θ is the radial coordinate of the point common to both the plane and the cone that is farthest from the cone's axis along an angle θ from the x-axis. The cylindrical height coordinate of this point is . So along the direction of angle θ, a cross-section of the conical ungula looks like the triangle . Rotating this triangle by an angle about the z-axis yields another triangle with , , substituted for , , and respectively, where and are functions of instead of . Since is infinitesimal then and also vary infinitesimally from and , so for purposes of considering the volume of the differential trapezoidal pyramid, they may be considered equal.