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Ungula

From Wikipedia, the free encyclopedia

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

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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

[edit]

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

[edit]
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

[edit]

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

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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

where

,
is as above,
,
,
,
,

where the logarithm is natural, and

.

See also

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References

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  1. ^ Ungula at Webster Dictionary.org
  2. ^ Gregory of St. Vincent (1647) Opus Geometricum quadraturae circuli et sectionum coni
  3. ^ Blaise Pascal Lettre de Dettonville a Carcavi describes the onglet and double onglet, link from HathiTrust
  4. ^ Margaret E. Baron (1969) The Origins of the Infinitesimal Calculus, Pergamon Press, republished 2014 by Elsevier, Google Books preview
  5. ^ Solids - Volumes and Surfaces at The Engineering Toolbox
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