Ungula Explained

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]

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 isA = { 1 \over 2 } \pi r ^{2} + { 1 \over 2 } \pi r \sqrt{r ^{2} + h ^{2}} + 2 r h ,the surface area of its curved sidewall isA _{s} = 2 r h ,and the surface area of its top (slanted roof) isA _{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 = k y 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 volumeA(x) dx whereA(x) = { 1 \over 2 } \sqrt{r ^{2} - x ^{2}} ⋅ 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 isV = \int ^{r} _{- r} A(x) dx = \int ^{r} _{- r} { 1 \over 2 } k (r ^{2} - x ^{2)} dx = { 1 \over 2 } k ( [r ^{2} x] ^{r} _{- r} - [ { 1 \over 3 } x ^{r} _{- r} ) = { 1 \over 2 } k ( 2 r ^{3} - { 2 \over 3 } r ^{3)} = { 2 \over 3 } k r ^{3} which equalsV = { 2 \over 3 } r ^{2} h after substituting r k = h .

A differential surface area of the curved side wall is

dA _{s} = k r (\sin \theta) ⋅ r d\theta = k r ^{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, k r \sin \theta) , (r \cos (\theta + d\theta), r \sin (\theta + d\theta), 0 ) , and (r \cos (\theta + d\theta), r \sin (\theta + d\theta), k r \sin (\theta + d\theta)) , and whose width and height are thereby r d\theta and (close enough to) k r \sin \theta , respectively.Then the surface area of the wall isA _{s} = \int ^{\pi} _{0} dA _{s} = \int ^{\pi} _{0} k r ^{2} (\sin \theta) d\theta = k r ^{2} \int ^{\pi} _{0} \sin \theta d\theta where the integral yields - [\cos \theta] ^{\pi} _{0} = - [ -1 - 1 ] = 2 , so that the area of the wall isA _{s} = 2 k r ^{2} ,and substituting r k = h yieldsA _{s} = 2 r h .

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 isA _{t} = { 1 \over 2 } \pi r ⋅ r \sqrt{ 1 + k ^{2}} = { 1 \over 2 } \pi r \sqrt{r ^{2} + (k r) ^{2}} and substituting k r = h yieldsA _{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

2 kr ^{2} , multiplying it by dr gives the volume of a differential half-shell, whose integral is { 2 \over 3 } k r ^{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} k H I \over 6 } whereH = { 1 \over { 1 \over h} - { 1 \over r k}} is the height of the cone from which the ungula has been cut out, andI = \int ^{\pi} _{0} { 2 H + k r \sin \theta \over (H + k r \sin \theta) ^{2}} \sin \theta d\theta .

The surface area of the curved sidewall is

A _{s} = {k r ^{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 →} ( I - { 4 \over H} ) = \lim _{H →} ( { 2 H \over H ^{2}} \int ^{\pi} _{0} \sin \theta d\theta - { 4 \over H} ) = 0 so that\lim _{H →} V = {r ^{3} k H \over 6 } ⋅ { 4 \over H} = { 2 \over 3 } k r ^{3} ,\lim _{H →} A _{s} = {k r ^{2} H \over 2 } ⋅ { 4 \over H} = 2 k r ^{2} , and\lim _{H →} 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} + (r k) ^{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},} 0 \le \rho \le r andx = \rho \cos \theta, y = \rho \sin \theta .

Let the cone be cut by a plane

z = k y = 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 isz _{0} = H ( 1 - {\rho _{0} \over r} ) .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

r d\theta , a length at the top of ( {H - z _{0} \over H} ) r d\theta , and altitude {z _{0} \over H} \sqrt{r ^{2} + H ^{2}} , so the trapezoid has area A _{T} = {rd\theta + ( {H - z _{0} \over H} ) rd\theta \over 2 } {z _{0} \over H} \sqrt{r ^{2} + H ^{2}} = rd\theta {( 2 H - z _{0)} z _{0} \over 2 H ^{2}} \sqrt{r ^{2} + H ^{2}} .An altitude from the trapezoidal base to the point ( 0 , 0 , 0 ) has length differentially close to{r H \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:V = \int ^{\pi} _{0} { 1 \over 3 } {r H\over \sqrt{r ^{2} + H ^{2}}} {( 2 H - z _{0)} z _{0} \over 2 H ^{2}} \sqrt{r ^{2} + H ^{2}} rd\theta = \int ^{\pi} _{0} { 1 \over 3 } r ^{2} {( 2 H - z _{0)} z _{0} \over 2 H} d\theta = {r ^{2} k \over 6 H} \int ^{\pi} _{0} ( 2 H - k y _{0)} y _{0} d\theta wherey _{0} = \rho _{0} \sin \theta = {\sin \theta \over { 1 \over r} + {k \sin \theta \over H}} = { 1 \over { 1 \over r \sin \theta} + {k\over H}} Substituting the right hand side into the integral and doing some algebraic manipulation yields the formula for volume to be proven.

For the sidewall:

A _{s} = \int ^{\pi} _{0} A _{T} = \int ^{\pi} _{0} {( 2 H - z _{0)} z _{0} \over 2 H ^{2}} r \sqrt{r ^{2} + H ^{2}d\theta} = {k r \sqrt{r ^{2} + H ^{2}} \over 2 H ^{2}} \int ^{\pi} _{0} ( 2 H - z _{0)} y _{0} d\theta and the integral on the rightmost-hand-side simplifies to H ^{2} r I . ∎

As a consistency check, consider what happens when k goes to infinity; then the conical ungula should become a semi-cone.

\lim _{k →} ( I - {\pi \over k r} ) = 0 \lim _{k →} V = {r ^{3} k H \over 6 } ⋅ {\pi \over k r} = { 1 \over 2 } ( { 1 \over 3 } \pi r ^{2} H ) which is half of the volume of a cone.\lim _{k →} A _{s} = {k r ^{2} \sqrt{r ^{2} + H ^{2}} \over 2 } ⋅ {\pi \over k r} = { 1 \over 2 } \pi r \sqrt{r ^{2} + H ^{2}} which is half of the surface area of the curved wall of a cone.

Surface area of top part

When

k = H / r , the "top part" (i.e., the flat face that is not semicircular like the base) has a parabolic shape and its surface area isA _{t} = { 2 \over 3 } r \sqrt{r ^{2} + H ^{2}} .

When

k < H / r then the top part has an elliptic shape (i.e., it is less than one-half of an ellipse) and its surface area isA _{t} = { 1 \over 2 } \pi x _{max} (y _{1} - y _{m)} \sqrt{ 1 + k ^{2}} Λ wherex _{max} = \sqrt{{k ^{2} r ^{4} H ^{2} - k ^{4} r ^{6} \over (k ^{2} r ^{2} - H ^{2)} ^{2}} + r ^{2}} ,y _{1} = { 1 \over { 1 \over r} + {k \over H}} ,y _{m} = {k r ^{2} H \over k ^{2} r ^{2} - H ^{2}} ,Λ = {\pi \over 4 } - { 1 \over 2 } \arcsin ( 1 - λ ) - { 1 \over 4 } \sin ( 2 \arcsin ( 1 - λ )) , andλ = {y _{1} \over y _{1} - y _{m}} .

When

k > H / r then the top part is a section of a hyperbola and its surface area isA _{t} = \sqrt{ 1 + k ^{2}} ( 2 C r - a J) whereC = {y _{1} + y _{2} \over 2 } = y _{m} ,y _{1} is as above,y _{2} = { 1 \over {k\over H} - { 1 \over r}} ,a = {r \over \sqrt{C ^{2} - \Delta ^{2}}} ,\Delta = {y _{2} - y _{1} \over 2 } ,J = {r\over a} B + {\Delta ^{2} \over 2 } log r |{{r\over a} + B\over { - r \over a} + B} r | ,

where the logarithm is natural, and

B = \sqrt{\Delta ^{2} + {r ^{2} \over a ^{2}}} .

See also

• Spherical wedge
• Steinmetz solid

External links

• William Vogdes (1861) An Elementary Treatise on Measuration and Practical Geometry via Google Books

Notes and References

1. https://www.webster-dictionary.org/definition/ungula Ungula
2. [Gregory of St. Vincent]
3. [Blaise Pascal]
4. [Margaret Baron|Margaret E. Baron]
5. https://www.engineeringtoolbox.com/surface-volume-solids-d_322.html Solids - Volumes and Surfaces