Surface Area of a Sphere
Date: 10/03/97 at 09:21:54 From: Stephanie Kaszuba Subject: Surface Area of a Sphere I would like help in assisting my daughter in answering the question, "How is the surface area of a sphere calculated, and why?". Bob Kaszuba
Date: 10/03/97 at 11:29:10 From: Doctor Rob Subject: Re: Surface Area of a Sphere Lengths, areas, and volumes of curved objects are hard to compute with the tools of ordinary geometry. That is one reason that the calculus was invented. Without resorting to the calculus, however, one could establish some of the formulas by the following sort of reasoning. Construct two polyhedral objects, one contained in the curved object, and one containing it. The surface area of the curved object must exceed that of the contained object, and be exceeded by that of the containing object. Measure the surface area of the polyhedral objects. Construct two more complicated polyhedral objects, whose surface area lies between those ofthe previously constructed ones and the curved object. Measure their surface areas. Repeat this until the approximation is sufficiently good for your purposes. If you can do this symbolically, you may be able to get formulas for the surface areas of the sequences of contained and containing polygonal or polyhedral objects. Then you may be able to prove that the difference between the surface areas of the corresponding containing and contained objects gets arbitrarily small as you step through better and better approximating objects. In fact, you may be able to determine the limit that both sequences approach, which is the surface area of the curved object. In the case of a sphere, you could start with an octahedron, composed of eight equilateral triangles, formed by putting two square pyramids base-to-base. If the radius of the sphere is r, then the length of each side of the triangles of the inscribed octahedron will be r*Sqrt, and their areas will be r^2*Sqrt/2, for a total surface area of 4*r^2*Sqrt. For the circumscribed octahedron, the length of each side of the triangles will be r*Sqrt, and their areas will be 3*r^2*Sqrt/2, for a total surface area of 12*r^2*Sqrt. Now on each face of the inscribed octahedron build a triangular pyramid, the height of which is just enough to make the vertex be on the surface of the sphere, and whose lateral sides are isosceles triangles. The surface area of this new polyhedron with 24 congruent isosceles triangle faces will be larger than that of the octahedron, but still smaller than that of the sphere. Using solid geometry, one can figure out the length of the new edges to be 2*r*Sqrt/3. This allows one to determine the areas of each of the 24 isosceles triangles, and the total surface area of this polyhedron. A construction on the circumscribing octahedron which chops off the six corners with planes tangent to the sphere will create a polyhedron with 14 faces, eight squares, six equilateral triangles, called a "cuboctahedron," which circumscribes the sphere and whose surface area is less that that of the octahedron but more that that of the sphere. Its surface area can be computed in a similar way. Repeating this process of approximating the sphere by inscribing and circumscribing polyhedra with more and more faces, by adding pyramids inside and chopping off pyramids outside, will give better and better approximations to the surface area of the sphere. This is the kind of approach used by Archimedes. Among other things, he used it to show that the area of a circle is the square of the radius times the constant we call Pi. His approach leads to the following facts. Let x(0) = 0, and for every n > 1, let x(n) = 2 + Sqrt[x(n-1)]. Then as n grows without bound, the expression Sqrt[4-x(n)]*2^n approaches Pi as a limit. This is derived from using inscribed regular polygons with 2^n sides. Archimedes used this kind of argument (but not exactly this) to show that 3+10/71 < Pi < 3+1/7. Using the calculus to find this surface area involves using spherical coordinates centered at the center of the sphere. The surface area is given by 2*Pi Pi S = Integral Integral r^2 sin(phi) d(phi) d(theta), 0 0 = 2*Pi*r^2*[cos(0) - cos(Pi)], S = 4*Pi*r^2. This is much quicker, and it gives an exact answer, not an approximate one, but you have to know enough of the calculus to understand why this double integral equals the surface area, and to evaluate it. I hope this helps. If this doesn't answer the question, write again. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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