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:In article < This e-mail address is being protected from spam bots, you need JavaScript enabled to view it : This e-mail address is being protected from spam bots, you need JavaScript enabled to view it (Michael A. Stueben) writes: : : Does anyone know an intuitive derivation for the surface area : of a sphere (4 pi r^2) for high school students. I know a : simple derivation for the volume, but it DEPENDS of the formula : for the surface area. I always end my geometry classes with the : remark If you ever find a simple proof for surface area, come : back and tell me so I can teach it to my future students. : Alas, no student ever returned with any derivation. : :If they know a bit of calculus then it is fairly easy to do the :integral. That is the simplest proof that I know for a sphere. : ::: snip ::: We can build on that, calculating the volume independently: An old rule for finding the volume of barrels is derived from Simpson's Rule (the barrel is standing, not rolling): Volume = (height/6) * (bottom area + 4 * central crossection area + top area) If the crossection area as a function of altitude of the cutting plane is at most cubic then the volume area is exact. (The formula works for pyramids, frusta, ellipsoids, cones, ...) Exercise: The crossection area of a sphere is a quadratic function of the altitude. Hence, the sphere being an extreme case of a barrel with top and bottom areas equal to 0, we calculate: Volume = (2r/6) * (0 + 4 * pi * r^2 + 0) = (4/3) * pi * r^3 , as required. Now the surface area is intuitively obtained by taking two concentric spheres, of radii r and r+h, with h0, calculating the volume of the solid shell  (4/3) * pi * ((r+h)^3 - r^3),  dividing by the thickness of the shell (that is h) and taking the limit as h approaches 0. The more observant students will recognize the derivative.                 - Have fun,      ( )    (an ASCII attempt at a barrel) Slavek (ZVK).   -

Does anyone know an intuitive derivation for the surface area of a sphere (4 pi r^2) for high school students. I know a simple derivation for the volume, but it DEPENDS of the formula for the surface area. I always end my geometry classes with the remark If you ever find a simple proof for surface area, come back and tell me so I can teach it to my future students. Alsa, no student ever returned with any derivation. You could look at how the 1/4 the area of a circle can be divided to cover 1/16 the surface of a sphere. 1/4 the area of a circle is a quadrant. The quadrant of the circle is the section of the 1/8th part of the sphere. you could then compare this to the formula for the area of a triangle which is equal to 1/2 the _base_ times the height. Steve

In article < This e-mail address is being protected from spam bots, you need JavaScript enabled to view it This e-mail address is being protected from spam bots, you need JavaScript enabled to view it (Michael A. Stueben) writes: Does anyone know an intuitive derivation for the surface area of a sphere (4 pi r^2) for high school students. If they know a bit of calculus then it is fairly easy to do the integral. That is the simplest proof that I know for a sphere. The first person to solve this was Archimedes. Until then it was thought to be a great and difficult problem. Archimedes' solution essentially summed thin parallel slices (he understood limits well enough to take the limit of the sum). Enclose the sphere in a the smallest cylinder (tin can) with the ends parallel to the slices. Then it's easy to see that, for a thin slice, the area is the same* as for the projection of that slice on the cylinder. (It's also true for thick slices by integration). Thus the whole cylinder (without the ends) has the same area as the sphere. * to a first order approximation Terry Moore, Statistics Department, Massey University, New Zealand. Imagine a person with a gift of ridicule [He might say] First that a negative quantity has no logarithm; secondly that a negative quantity has no square root; thirdly that the first non-existent is to the second as the circumference of a circle is to the diameter. Augustus de Morgan

Does anyone know an intuitive derivation for the surface area of a sphere (4 pi r^2) for high school students. I know a simple derivation for the volume, but it DEPENDS of the formula for the surface area. I always end my geometry classes with the remark If you ever find a simple proof for surface area, come back and tell me so I can teach it to my future students. Alsa, no student ever returned with any derivation. The sphere of radius 1 is a surface of revolution obtained by rotating the semicircle y = + sqrt( 1 - x^2 ) around the x-axis. The centroid of the semicircle is (0,c) by symmetry, and c is the average y-value on the semicircle, namely     [int[0 to pi]*sin(x)dx] / pi  = 2/pi. Then by Pappus' Theorem, the surface area is obtained from     surf. area =  L * d where L = length of rotated curve = pi, and d = distance through which centroid moves during rotation = (2 pi)*(2/pi) = 4. This gives the surface area as 4pi. Although this method relies on the Pappus Theorem, the latter is interesting enough in itself and could be presented to a highschool class via several examples checkable by other means.

The problem was to find an elementary derivation for the surface area. Benjamin Tilly suggested that the integral was not difficult, and Zdislav Kovarik suggested an application of Simpson's rule. However, I suspect that neither proof would fit into a high school geometry class. One beautiful result known to the ancient Greeks is that the surface area of a cap of a sphere is linear with respect to the height of the cap. In other words, if a cylinder circumscribes a sphere and two planes perpendicular to the axis cut the sphere, then lateral areas of the sections of the sphere and cylinder are the same. The lateral area of the cylinder is 2r*2rpi = 4(r^2)pi. I do not know what rigorous proofs of this do not use calculus, but I do not know of a rigorous definition of surface area which does not use calculus. However, a simple hand-waving argument (easily made rigorous) can be made using no more than the Pythagorean theorem. (The slope cancels out the smaller radius exactly.) Douglas Zare This e-mail address is being protected from spam bots, you need JavaScript enabled to view it

One beautiful result known to the ancient Greeks is that the surface area of a cap of a sphere is linear with respect to the height of the cap. In other words, if a cylinder circumscribes a sphere and two planes perpendicular to the axis cut the sphere, then lateral areas of the sections of the sphere and cylinder are the same. The lateral area of the cylinder is 2r*2rpi = 4(r^2)pi. I do not know what rigorous proofs of this do not use calculus, but I do not know of a rigorous definition of surface area which does not use calculus.  [...] Perhaps it's `easier' to derive the volume of a hemisphere first, using Cavalieri's principle, as the difference in volume between a cylinder and cone, both with radius and height the radius of the hemisphere in question.  Then the surface area of a sphere pops out from an argument rederiving the volume as a limit of infinitely many almost pyramids, all with apex our sphere's center. The students presumably went through parts of this argument in relating the areas and radii of circles, and also in deriving the formulas for the volumes of cylinders and cones, so I think this should go over.