BREADCUT BRIEF ABSTRACT The claim is that by slicing parallel sections of equal altitudes from a sphere we get identical surface areas of these sections. The problem is posed in terms of equity of crust distribution for French bread. GENERAL INFORMATION FileName: BREADCUT Full title: How to cut a hemispherical loaf of French bread so that each of 8 guests gets the same amount of crust. Last Revision Date: 27 May 1996. Developer: Brian J. Winkel, Department of Mathematical Sciences, United States Military Academy, West Point NY 10996 USA. Phone: 914-938-3200. Email: ab3646@usma2.usma.edu. FAX: 914-938-2409. Contact: Brian J. Winkel, Department of Mathematical Sciences, United States Military Academy, West Point NY 10996 USA. Phone: 914-938-3200. Email: ab3646@usma2.usma.edu. FAX: 914-938-2409. Aaron D. Klebanoff, Department of Mathematics, Rose-Hulman Institute of Technology, Terre Haute IN 47803 USA. Phone: 812-877-8151. Email: Aaron.Klebanoff@Rose-Hulman.Edu. FAX: 812-877-3198. Support: The production of this material is supported by the National Science Foundation under Division of Undergraduate Education grant DUE-9352849: Development Site for Complex, Technology-Based Problems in Calculus with Applications in Science and Engineering and the Arvin Foundation of Columbus IN. STATEMENT OF PROBLEM Consider a hemispherical loaf of crusty French bread of radius 8". Alice and Bill just love the crust of the bread and are having 6 friends over for dinner. They wish to slice the bread so that each of the 8 dinner party members will get an equal amount of crust. Alice suggests that they just cut the bread vertically (parallel to a plane going through the center) into pieces each one-eighth of the diameter wide. She claims that in this way each person will have the same amount of crust. Bill says, "Just because the width of each piece is the same does not mean the amount of crust on each piece is the same." Formulate this problem mathematically and settle the argument. KEYWORDS Surface area of rotation. TEACHER NOTES ISSUES RELATED TO THE PROBLEM Prerequisites Knowledge of arc length formula for function of one variable and surface area of rotation. Time allotment - time management 15 minutes in one class, more if you permit discussion to wander or if you introduce physical objects and permit a great deal of conjecturing. Students will debate the reasonableness of Alice's conjecture and be pressed to conjecture another approach, e.g., cutting wider sections at end or in the middle. Even more time can be used if you pose the problem as open ended without positing the cutting scheme offered above. Expectations We expect the students to argue about this in class and to really find the mathematical analysis as a proof of one side or the other. Students will see Students will see how analysis confirms conjecture and settles debate. Future payoffs Students will improve their communication skills as they are forced to argue their point of view based on a mathematical model. Extensions One could request other subdivisions, e.g., in proportion 1:2:3:4:5:6:7:8. Or one could examine other shapes, e.g. ellipsoid to see how complicated the problem might really be if the geometry is different. References and Sources POSSIBLE SOLUTION(S) We consider the problem of a hemisphere of radius 8" as a half of the shape of a rotated circle, x^2 + y^2 = 8^2, of radius 8" about the x axis. To obtain this we shall take one half of the volume obtained from rotating the top half of x^2 + y^2 = 8^2 about the x-axis. f[x_] = Sqrt[64 - x^2]; We obtain the surface area of the slice of width a going from x value h to x value h + a. We do this by rotating an element of arc length Sqrt[1 + f'[x]^2] through a radius f[x] and taking one-half of this amount. va[h_] = 1/2 Integrate[2 Pi f[x] Sqrt[1 + f'[x]^2], {x,h,h+a}] 1 2 (-16 h Sqrt[-------] Sqrt[64 - h ] Pi + 2 64 - h 2 2 (a + h) Sqrt[64 - (a + h) ] -64 Sqrt[--------------] Pi) / 2 2 -64 + (a + h) In our case we are suggesting that we cut the total width of the 16" diameter hemisphere of bread into 8 equal width pieces of length a = 2". Thus we compute the surface area from h to h + 2. Here h is in the domain h = -8" to h = 6". party[h_] = va[h]/.{a->2} 1 2 (-16 h Sqrt[-------] Sqrt[64 - h ] Pi + 2 64 - h 2 2 (2 + h) Sqrt[64 - (2 + h) ] -64 Sqrt[--------------] Pi) / 2 2 -64 + (2 + h) We simplify this by noting the square root terms cancel appropriately. p[h_] = Expand[(- 16 h Pi + 2 (2 + h) 8 Pi)/2] 16 Pi And we see that no matter where we start cutting (h in the domain h = -8 to h = 6) when we cut a width of bread 2" long the surface area is always 16 Pi ~ 50.2655 in^2. The plot confirms this. 16 Pi//N 50.2655 Plot[p[h],{h,-8,6},PlotRange->{0,60}] -Graphics- ISSUES IN SOLUTION This is an immediate application of the principle of rotating an element of arc length about an axis (half way, not a full rotation) to determine the surface area. The shape rotated is the top half of a circle of radius 8" and we are taking horizontal slices of this surface area, by integrating over regular intervals. When students confront the intuitive responses they have concerning their opinions they will readily go to analysis to "sovle the question once and for all."