Ungula

{{Short description|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.<ref>[https://www.webster-dictionary.org/definition/ungula Ungula] at Webster Dictionary.org</ref> 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.<ref>Gregory of St. Vincent (1647) ''Opus Geometricum quadraturae circuli et sectionum coni''</ref> Two cylinders with equal radii and perpendicular axes intersect in four double ungulae.<ref>Blaise Pascal [http://babel.hathitrust.org/cgi/pt?id=ucm.5325890510;view=1up;seq=567 Lettre de Dettonville a Carcavi] describes the onglet and double onglet, link from HathiTrust</ref> 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.<ref name =Baron>Margaret E. Baron (1969) ''The Origins of the Infinitesimal Calculus'', Pergamon Press, republished 2014 by Elsevier, [https://books.google.com/books?id=OcTSBQAAQBAJ Google Books preview]</ref>{{rp|146}}

==Cylindrical ungula== thumb|300px|Ungula of a right circular cylinder. A cylindrical ungula of base radius ''r'' and height ''h'' has volume :<math> V = {2\over 3} r^2 h</math>,.<ref name=cyl>[https://www.engineeringtoolbox.com/surface-volume-solids-d_322.html Solids - Volumes and Surfaces] at The Engineering Toolbox</ref> Its total surface area is :<math> A = {1\over 2} \pi r^2 + {1\over 2} \pi r \sqrt{r^2 + h^2} + 2 r h</math>, the surface area of its curved sidewall is :<math> A_s = 2 r h </math>, and the surface area of its top (slanted roof) is :<math> A_t = {1\over 2} \pi r \sqrt{r^2 + h^2} </math>.

===Proof=== Consider a cylinder <math>x^2 + y^2 = r^2</math> bounded below by plane <math>z = 0</math> and above by plane <math>z = k y</math> where ''k'' is the slope of the slanted roof: :<math> k = {h \over r}</math>. Cutting up the volume into slices parallel to the ''y''-axis, then a differential slice, shaped like a triangular prism, has volume :<math> A(x) \, dx</math> where :<math> 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)</math> is the area of a right triangle whose vertices are, <math> (x, 0, 0)</math>, <math> (x, \sqrt{r^2 - x^2}, 0) </math>, and <math> (x, \sqrt{r^2 - x^2}, k \sqrt{r^2 - x^2})</math>, and whose base and height are thereby <math>\sqrt{r^2 - x^2}</math> and <math>k \sqrt{r^2 - x^2}</math>, respectively. Then the volume of the whole cylindrical ungula is :<math> V = \int_{-r}^r A(x) \, dx = \int_{-r}^r {1\over 2} k (r^2 - x^2) \, dx </math> :<math> \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 (2 r^3 - {2\over 3} r^3) = {2 \over 3} k r^3</math> which equals :<math> V = {2\over 3} r^2 h </math> after substituting <math> r k = h</math>.

A differential surface area of the curved side wall is :<math> dA_s = k r (\sin \theta) \cdot r \, d\theta = k r^2 (\sin \theta) \, d\theta </math>, which area belongs to a nearly flat rectangle bounded by vertices <math>(r \cos \theta, r \sin \theta, 0)</math>, <math>(r \cos \theta, r \sin \theta, k r \sin \theta)</math>, <math>(r \cos (\theta + d\theta), r \sin (\theta + d\theta), 0)</math>, and <math>(r \cos (\theta + d\theta), r \sin (\theta + d\theta), k r \sin (\theta + d\theta))</math>, and whose width and height are thereby <math>r \, d\theta</math> and (close enough to) <math> k r \sin \theta</math>, respectively. Then the surface area of the wall is :<math> A_s = \int_0^\pi dA_s = \int_0^\pi k r^2 (\sin \theta) \, d\theta = k r^2 \int_0^\pi \sin \theta \, d\theta </math> where the integral yields <math> -[\cos \theta]_0^\pi = -[-1 - 1] = 2 </math>, so that the area of the wall is :<math> A_s = 2 k r^2 </math>, and substituting <math> r k = h </math> yields :<math> A_s = 2 r h</math>.

The base of the cylindrical ungula has the surface area of half a circle of radius ''r'': <math> {1\over 2} \pi r^2 </math>, 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 <math> r \sqrt{1 + k^2}</math>, so that its area is :<math> A_t = {1\over 2} \pi r \cdot r \sqrt{1 + k^2} = {1\over 2} \pi r \sqrt{r^2 + (k r)^2}</math> and substituting <math> k r = h</math> yields :<math> A_t = {1\over 2} \pi r \sqrt{r^2 + h^2}</math>. ∎

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

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

==Conical ungula== thumb|300px|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 :<math> V = {r^3 k H I \over 6} </math> where :<math> H = {1 \over {1\over h} - {1\over r k}} </math> is the height of the cone from which the ungula has been cut out, and :<math> I = \int_0^\pi {2 H + k r \sin \theta \over (H + k r \sin \theta)^2} \sin \theta \, d\theta </math>.

The surface area of the curved sidewall is :<math> A_s = {k r^2 \sqrt{r^2 + H^2} \over 2} I </math>.

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: :<math>\lim_{H\rightarrow \infty} \Big(I - {4\over H}\Big) = \lim_{H\rightarrow \infty} \Big({2 H \over H^2} \int_0^\pi \sin \theta \, d\theta - {4\over H} \Big) = 0</math> so that :<math>\lim_{H\rightarrow \infty} V = {r^3 k H \over 6} \cdot {4\over H} = {2\over 3} k r^3 </math>,

:<math>\lim_{H\rightarrow \infty} A_s = {k r^2 H \over 2} \cdot {4\over H} = 2 k r^2 </math>, and

:<math> \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 + (r k)^2}</math>, which results agree with the cylindrical case.

===Proof=== Let a cone be described by :<math> 1 - {\rho \over r} = {z \over H} </math> where ''r'' and ''H'' are constants and ''z'' and ''&rho;'' are variables, with :<math> \rho = \sqrt{x^2 + y^2}, \qquad 0 \le \rho \le r </math> and :<math> x = \rho \cos \theta, \qquad y = \rho \sin \theta </math>.

Let the cone be cut by a plane :<math> z = k y = k \rho \sin \theta</math>. Substituting this ''z'' into the cone's equation, and solving for ''&rho;'' yields :<math> \rho_0 = {1 \over {1\over r} + {k \sin \theta \over H}}</math> which for a given value of ''&theta;'' 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 ''&theta;'' from the ''x''-axis. The cylindrical height coordinate of this point is :<math> z_0 = H \Big(1 - {\rho_0 \over r}\Big) </math>. So along the direction of angle ''&theta;'', a cross-section of the conical ungula looks like the triangle :<math> (0,0,0) - (\rho_0 \cos \theta, \rho_0 \sin \theta, z_0) - (r \cos \theta, r \sin \theta, 0) </math>. Rotating this triangle by an angle <math>d\theta</math> about the ''z''-axis yields another triangle with <math>\theta + d\theta</math>, <math>\rho_1</math>, <math>z_1</math> substituted for <math>\theta</math>, <math>\rho_0</math>, and <math>z_0</math> respectively, where <math>\rho_1</math> and <math>z_1</math> are functions of <math>\theta + d\theta</math> instead of <math>\theta</math>. Since <math>d\theta</math> is infinitesimal then <math>\rho_1</math> and <math>z_1</math> also vary infinitesimally from <math>\rho_0</math> and <math>z_0</math>, 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 <math>r d\theta</math>, a length at the top of <math>\Big({H - z_0 \over H}\Big) r d\theta</math>, and altitude <math>{z_0 \over H} \sqrt{r^2 + H^2}</math>, so the trapezoid has area :<math>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 {(2 H - z_0) z_0 \over 2 H^2} \sqrt{r^2 + H^2}</math>. An altitude from the trapezoidal base to the point <math>(0,0,0)</math> has length differentially close to :<math>{r H \over \sqrt{r^2 + H^2}}</math>. (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: :<math> V = \int_0^\pi {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} r\,d\theta = \int_0^\pi {1\over 3} r^2 {(2 H - z_0) z_0 \over 2 H} d\theta = {r^2 k \over 6 H} \int_0^\pi (2 H - k y_0) y_0 \,d\theta</math> where :<math> y_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}}</math> Substituting the right hand side into the integral and doing some algebraic manipulation yields the formula for volume to be proven.

For the sidewall: :<math>A_s = \int_0^\pi A_T = \int_0^\pi {(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_0^\pi (2 H - z_0) y_0 \,d\theta </math> and the integral on the rightmost-hand-side simplifies to <math>H^2 r I</math>. ∎

As a consistency check, consider what happens when ''k'' goes to infinity; then the conical ungula should become a semi-cone. :<math> \lim_{k\rightarrow \infty} \Big(I - {\pi \over k r}\Big) = 0 </math>

:<math> \lim_{k\rightarrow \infty} V = {r^3 k H \over 6} \cdot {\pi \over k r} = {1\over 2} \Big({1\over 3} \pi r^2 H\Big) </math> which is half of the volume of a cone.

:<math> \lim_{k\rightarrow \infty} A_s = {k r^2 \sqrt{r^2 + H^2} \over 2} \cdot {\pi \over k r} = {1\over 2} \pi r \sqrt{r^2 + H^2} </math> which is half of the surface area of the curved wall of a cone.

===Surface area of top part=== When <math>k = H / r </math>, the "top part" (i.e., the flat face that is not semicircular like the base) has a parabolic shape and its surface area is :<math> A_t = {2\over 3} r \sqrt{r^2 + H^2} </math>.

When <math> k < H / r </math> then the top part has an elliptic shape (i.e., it is less than one-half of an ellipse) and its surface area is :<math> A_t = {1\over 2} \pi x_{max} (y_1 - y_m) \sqrt{1 + k^2} \Lambda </math> where :<math> x_{max} = \sqrt{{k^2 r^4 H^2 - k^4 r^6 \over (k^2 r^2 - H^2)^2} + r^2} </math>,

:<math> y_1 = {1 \over {1 \over r} + {k \over H}} </math>,

:<math> y_m = {k r^2 H \over k^2 r^2 - H^2} </math>,

:<math> \Lambda = {\pi \over 4} - {1\over 2} \arcsin (1 - \lambda) - {1\over 4} \sin (2 \arcsin (1 - \lambda)) </math>, and

:<math> \lambda = {y_1 \over y_1 - y_m} </math>.

When <math> k > H / r </math> then the top part is a section of a hyperbola and its surface area is :<math> A_t = \sqrt{1 + k^2} (2 C r - a J) </math> where :<math> C = {y_1 + y_2 \over 2} = y_m </math>,

:<math> y_1 </math> is as above,

:<math> y_2 = {1 \over {k\over H} - {1\over r}} </math>,

:<math> a = {r \over \sqrt{C^2 - \Delta^2}} </math>,

:<math> \Delta = {y_2 - y_1 \over 2} </math>,

:<math> J = {r\over a} B + {\Delta^2 \over 2} \log \Biggr|{{r\over a} + B\over {-r \over a} + B}\Biggr| </math>,

where the logarithm is natural, and

:<math> B = \sqrt{\Delta^2 + {r^2 \over a^2}} </math>.

==See also== * Spherical wedge * Steinmetz solid

==References== {{Reflist}}

==External links== * William Vogdes (1861) [https://books.google.com/books?id=TihRAAAAYAAJ&pg=PA139 An Elementary Treatise on Measuration and Practical Geometry] via Google Books

Category:Geometric shapes Category:Euclidean solid geometry