Region between a revolved surface and oblique plane
In solid geometry, an ungula is a region of a solid of revolution, cut off by a plane oblique to its base.[1] 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.[2] Two cylinders with equal radii and perpendicular axes intersect in four double ungulae.[3] 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.[4]: 146
Cylindrical ungula
Ungula of a right circular cylinder.
A cylindrical ungula of base radius r and height h has volume
,.[5]
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
Ungula of a right circular cone.
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.
The differential trapezoidal pyramid has a trapezoidal base with a length at the base (of the cone) of , a length at the top of , and altitude , so the trapezoid has area
.
An altitude from the trapezoidal base to the point has length differentially close to
.
(This is an altitude of one of the side triangles of the trapezoidal pyramid.) The volume of the pyramid is one-third its base area times its altitudinal length, so the volume of the conical ungula is the integral of that:
where
Substituting the right hand side into the integral and doing some algebraic manipulation yields the formula for volume to be proven.
For the sidewall:
and the integral on the rightmost-hand-side simplifies to . ∎
As a consistency check, consider what happens when k goes to infinity; then the conical ungula should become a semi-cone.
which is half of the volume of a cone.
which is half of the surface area of the curved wall of a cone.
Surface area of top part
When , the "top part" (i.e., the flat face that is not semicircular like the base) has a parabolic shape and its surface area is
.
When then the top part has an elliptic shape (i.e., it is less than one-half of an ellipse) and its surface area is
where
,
,
,
, and
.
When then the top part is a section of a hyperbola and its surface area is