Ungula

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

A cylindrical ungula of base radius r and height h has volume

V={2 \over 3}r^{2}h

,.^[5]

Its total surface area is

A={1 \over 2}\pi r^{2}+{1 \over 2}\pi r{\sqrt {r^{2}+h^{2}}}+2rh

,

the surface area of its curved sidewall is

A_{s}=2rh

,

and the surface area of its top (slanted roof) is

A_{t}={1 \over 2}\pi r{\sqrt {r^{2}+h^{2}}}

.

Proof

Consider a cylinder $x^{2}+y^{2}=r^{2}$ bounded below by plane $z=0$ and above by plane $z=ky$ where k is the slope of the slanted roof:

k={h \over r}

.

Cutting up the volume into slices parallel to the y-axis, then a differential slice, shaped like a triangular prism, has volume

A(x)\,dx

where

A(x)={1 \over 2}{\sqrt {r^{2}-x^{2}}}\cdot k{\sqrt {r^{2}-x^{2}}}={1 \over 2}k(r^{2}-x^{2})

is the area of a right triangle whose vertices are, $(x,0,0)$ , $(x,{\sqrt {r^{2}-x^{2}}},0)$ , and $(x,{\sqrt {r^{2}-x^{2}}},k{\sqrt {r^{2}-x^{2}}})$ , and whose base and height are thereby ${\sqrt {r^{2}-x^{2}}}$ and $k{\sqrt {r^{2}-x^{2}}}$ , respectively. Then the volume of the whole cylindrical ungula is

V=\int _{-r}^{r}A(x)\,dx=\int _{-r}^{r}{1 \over 2}k(r^{2}-x^{2})\,dx

\qquad ={1 \over 2}k{\Big (}[r^{2}x]_{-r}^{r}-{\Big [}{1 \over 3}x^{3}{\Big ]}_{-r}^{r}{\Big )}={1 \over 2}k(2r^{3}-{2 \over 3}r^{3})={2 \over 3}kr^{3}

which equals

V={2 \over 3}r^{2}h

after substituting $rk=h$ .

A differential surface area of the curved side wall is

dA_{s}=kr(\sin \theta )\cdot r\,d\theta =kr^{2}(\sin \theta )\,d\theta

,

which area belongs to a nearly flat rectangle bounded by vertices $(r\cos \theta ,r\sin \theta ,0)$ , $(r\cos \theta ,r\sin \theta ,kr\sin \theta )$ , $(r\cos(\theta +d\theta ),r\sin(\theta +d\theta ),0)$ , and $(r\cos(\theta +d\theta ),r\sin(\theta +d\theta ),kr\sin(\theta +d\theta ))$ , and whose width and height are thereby $r\,d\theta$ and (close enough to) $kr\sin \theta$ , respectively. Then the surface area of the wall is

A_{s}=\int _{0}^{\pi }dA_{s}=\int _{0}^{\pi }kr^{2}(\sin \theta )\,d\theta =kr^{2}\int _{0}^{\pi }\sin \theta \,d\theta

where the integral yields $-[\cos \theta ]_{0}^{\pi }=-[-1-1]=2$ , so that the area of the wall is

A_{s}=2kr^{2}

,

and substituting $rk=h$ yields

A_{s}=2rh

.

The base of the cylindrical ungula has the surface area of half a circle of radius r: ${1 \over 2}\pi r^{2}$ , 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 $r{\sqrt {1+k^{2}}}$ , so that its area is

A_{t}={1 \over 2}\pi r\cdot r{\sqrt {1+k^{2}}}={1 \over 2}\pi r{\sqrt {r^{2}+(kr)^{2}}}

and substituting $kr=h$ yields

A_{t}={1 \over 2}\pi r{\sqrt {r^{2}+h^{2}}}

. ∎

Note how the surface area of the side wall is related to the volume: such surface area being $2kr^{2}$ , multiplying it by $dr$ gives the volume of a differential half-shell, whose integral is ${2 \over 3}kr^{3}$ , the volume.

When the slope k equals 1 then such ungula is precisely one eighth of a bicylinder, whose volume is ${16 \over 3}r^{3}$ . One eighth of this is ${2 \over 3}r^{3}$ .

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

V={r^{3}kHI \over 6}

where

H={1 \over {1 \over h}-{1 \over rk}}

is the height of the cone from which the ungula has been cut out, and

I=\int _{0}^{\pi }{2H+kr\sin \theta  \over (H+kr\sin \theta )^{2}}\sin \theta \,d\theta

.

The surface area of the curved sidewall is

A_{s}={kr^{2}{\sqrt {r^{2}+H^{2}}} \over 2}I

.

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:

\lim _{H\rightarrow \infty }{\Big (}I-{4 \over H}{\Big )}=\lim _{H\rightarrow \infty }{\Big (}{2H \over H^{2}}\int _{0}^{\pi }\sin \theta \,d\theta -{4 \over H}{\Big )}=0

so that

\lim _{H\rightarrow \infty }V={r^{3}kH \over 6}\cdot {4 \over H}={2 \over 3}kr^{3}

,

\lim _{H\rightarrow \infty }A_{s}={kr^{2}H \over 2}\cdot {4 \over H}=2kr^{2}

, and

\lim _{H\rightarrow \infty }A_{t}={1 \over 2}\pi r^{2}{{\sqrt {1+k^{2}}} \over 1+0}={1 \over 2}\pi r^{2}{\sqrt {1+k^{2}}}={1 \over 2}\pi r{\sqrt {r^{2}+(rk)^{2}}}

,

which results agree with the cylindrical case.

Proof

Let a cone be described by

1-{\rho  \over r}={z \over H}

where r and H are constants and z and ρ are variables, with

\rho ={\sqrt {x^{2}+y^{2}}},\qquad 0\leq \rho \leq r

and

x=\rho \cos \theta ,\qquad y=\rho \sin \theta

.

Let the cone be cut by a plane

z=ky=k\rho \sin \theta

.

Substituting this z into the cone's equation, and solving for ρ yields

\rho _{0}={1 \over {1 \over r}+{k\sin \theta  \over H}}

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

z_{0}=H{\Big (}1-{\rho _{0} \over r}{\Big )}

.

So along the direction of angle θ, a cross-section of the conical ungula looks like the triangle

(0,0,0)-(\rho _{0}\cos \theta ,\rho _{0}\sin \theta ,z_{0})-(r\cos \theta ,r\sin \theta ,0)

.

Rotating this triangle by an angle $d\theta$ about the z-axis yields another triangle with $\theta +d\theta$ , $\rho _{1}$ , $z_{1}$ substituted for $\theta$ , $\rho _{0}$ , and $z_{0}$ respectively, where $\rho _{1}$ and $z_{1}$ are functions of $\theta +d\theta$ instead of $\theta$ . Since $d\theta$ is infinitesimal then $\rho _{1}$ and $z_{1}$ also vary infinitesimally from $\rho _{0}$ and $z_{0}$ , 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 $rd\theta$ , a length at the top of ${\Big (}{H-z_{0} \over H}{\Big )}rd\theta$ , and altitude ${z_{0} \over H}{\sqrt {r^{2}+H^{2}}}$ , so the trapezoid has area

A_{T}={r\,d\theta +{\Big (}{H-z_{0} \over H}{\Big )}r\,d\theta  \over 2}{z_{0} \over H}{\sqrt {r^{2}+H^{2}}}=r\,d\theta {(2H-z_{0})z_{0} \over 2H^{2}}{\sqrt {r^{2}+H^{2}}}

.

An altitude from the trapezoidal base to the point $(0,0,0)$ has length differentially close to

{rH \over {\sqrt {r^{2}+H^{2}}}}

.

(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: