A smooth bounded solid of finite volume and infinite surface is constructed. It is a variant of the classical Gabriel’s horn that is often taught in Calculus classes

Krenicky, Joseph

Rychtář, Jan

English

Furman University

Furman UniversityElectronic Journal of Undergraduate MathematicsVolume 11, 4 – 7, 2006NOTE ON GABRIEL’S HORNJOSEPH KRENICKY AND JAN RYCHTÁŘAbstract. A smooth bounded solid of finite volume and infinite surface isconstructed. It is a variant of the classical Gabriel’s horn that is often taughtin Calculus classes.Gabriel’s horn, [ABD], is a classical example from Calculus. It is a solid withfinite volume and infinite surface obtained by rotating the graph of the functionf(x) = 1/x for x ≥ 1 about the x-axis. As it is taught in Calculus classes, theobject can be filled with paint but its surface cannot be painted.The classical Gabriel’s horn is an unbounded object. In reality, as some studentspoint out ([L]), it cannot be filled with paint. Lynch [L] constructed a boundedvariant of Gabriel’s horn. He constructed such an object by rotating the graph ofa certain piecewise linear function defined on the interval [0, 1]. The graph of thefunction had an infinite arclength; and thus this object could indeed be filled withpaint and yet its surface could not be painted.In this note, Lynch’s example is modified to get a solid with a smooth surface.The core of the construction is the following Theorem.Theorem 1. There exists a bounded function f that is differentiable on a closedinterval [0, 0.5] and whose arclength is infinite. Moreover, f is infinitely many timesdifferentiable on (0, 0.5) and satisfies 1 ≤ f(x) ≤ 3 for all x ∈ [0, 0.5].Proof. Define a functionf(x) =2 +xlnxsin(1x), x ∈ (0, 0.5],2, x = 0.To get a better idea about the function f(x), consider the function sin(1x)whosegraph appears in Figure 1.The function sin(1x)oscillates between −1 and 1, and has an extremely smallperiod in the neighborhood of x = 0. The function f is nothing more than thefunction sin(1x)squeezed in between the function 2 + xln x and 2−xln x on the giveninterval x ∈ [0, 0.5], see Figure 2.Note that 1 ≤ f(x) ≤ 3.First, we will establish the differentiability of the function f . Clearly, the functionis infinitely differentiable on (0, 0.5]. The function f is differentiable at x = 0 fromReceived by the editors August 28, 2006.2000 Mathematics Subject Classification. 28A75, 51M25.Key words and phrases. Gabriel’s horn, smooth surfaces, solids of revolution.Joseph Krenicky is an undergraduate research assistant of Jan Rychtář. The work was sup-ported by the UNCG New Faculty Grant held by the second author. Jan Rychtář will be theauthor for correspondence and proofs.4NOTE ON GABRIEL’S HORN 5–1–0.500.510.05 0.1 0.15 0.2 0.25xFigure 1. Graph of sin(1x).1.71.81.922.12.22.3 0 0.05 0.1 0.15 0.2 0.25 Figure 2. Graph of f(x) = 2 + xln x sin(1x).the right (and thus continuous at x = 0). Indeed,f ′+(0) = limh→0+f(0 + h)− f(0)h= limh→0+sin(1h)lnh= 0,where the last equality follows by the Squeezing Theorem (see [ABD]) because−1lnh≤sin(1h)lnh≤ 1lnh, andlimh→0+±1lnh= 0Hence, the function f is differentiable on [0, 0.5].In order to establish that the arclength L of the graph of f is infinite we will be usethe following easy observation (whose proof is straightforward and it is omitted).Fact 2. The function g(x) = 1/(x ln(x)) is a decreasing function on [2,∞) satis-fying limx→∞g(x) = 0.Now, consider pointsxk =1kπ + π/2, k = 1, 2, . . . .6 JOSEPH KRENICKY AND JAN RYCHTÁŘFor the arclength Lk of the graph between points xk and xk+1 we have the followingestimatesLk ≥ |f(xk)− f(xk+1|=∣∣∣∣ − sin(kπ + π/2)(kπ + π/2) ln(kπ + π/2) − − sin((k + 1)π + π/2)((k + 1)π + π/2) ln((k + 1)π + π/2)∣∣∣∣=∣∣(−1)kg((kπ + π/2)− (−1)k+1g((k + 1)π + π/2)∣∣≥ 2g((k + 1)π + π/2)≥ 2g((k + 2)π),where we used (twice) the monotonicity of the function g. Hence, the arclength Lof the graph of f is at leastL ≥∞∑k=1Lk ≥∞∑k=12g((k + 2)π) =∞∑k=12(k + 2)π ln((k + 2)π).The series diverges by the limit comparison test (see [ABD]) with the series∞∑k=31k ln kwhere the later series diverges by the integral criteria because g is a strictly de-creasing function and∫ ∞21x lnxdx substitution=u = ln xdu = 1xdx∫ ∞ln 21udu = ∞.Thus, the arclength L of the graph of f is infinite.Corollary 3. There exists a bounded smooth solid with finite volume and infinitesurface.Proof. Consider the solid given by rotating the graph of f (given in the aboveTheorem) about the x-axis. The volume of the solid is finite. Indeed,f(x) ≤ 3and thus the solid is contained in a cylinder of radius 3 and height 0.5. It remainsto show that the surface S of the solid is infinite. This is done by the followingestimate, where we used the inequality f ≥ 1.S =∫ 0.502πf(x)√1 + f ′(x)2dx ≥ 2π∫ 0.50√1 + f ′(x)2dx = 2πL = ∞.Remark. Note, as already observed in Lynch’s paper [L], that the solid could notbe constructed by using a function h : [a, b] 7→ R with a continuous derivative.Indeed, we would haveS =∫ ba2πh(x)√1 + h′(x)2dx ≤ (b− a)2πM√1 + D2 < ∞,NOTE ON GABRIEL’S HORN 7whereM = max{h(x), x ∈ [a, b]},D = max{|h′(x)|, x ∈ [a, b]},with both M and D being finite by the continuity of h and h′.References[ABD] H. Anton, I. Bivens and S. Davis, Calculus Eight edition, John Wiley & Sons, Inc., 2005.[L] M. Lynch: A Paradoxical Paint Pail, The College Mathematics Journal, No. 5, 2005,402–403.Department of Mathematics and Statistics, University of North Carolina Greens-boro, Greensboro, NC 27402, USAE-mail address: jnkrenic@uncg.eduDepartment of Mathematics and Statistics, University of North Carolina Greens-boro, Greensboro, NC 27402, USAE-mail address: rychtar@uncg.edu

Notes on Gabriel\u27s Horn

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Notes on Gabriel\u27s Horn

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