Volume of a PyramidDate: 05/16/99 at 14:32:53 From: Jad Aoun Subject: Volume of a Pyramid (Clarify) I am a student in Lebanon and I am in grade 11. I have a huge project about the surface area and volume of solids. I was searching on your site for the derivation of the formula for the surface area and volume of a pyramid when I stumbled on a similar question someone else had asked at Volume of a Cone or Pyramid http://mathforum.org/dr.math/problems/swan3.30.98.html It is about the proof of the volume of a pyramid. You responded by saying that V = a^2*b*[1^2 + 2^2 + 3^2 + ... + n^2] = a^2*b*n*(n+1)*(2n+1)/6 (which can be proved by induction) = a^2*b*n^3*(1/3 + 1/[2*n] + 1/[6*n^2]). How exactly was this done? Date: 05/17/99 at 13:15:26 From: Doctor Peterson Subject: Re: Volume of a Pyramid (Clarify) Hi Jad, I helped another student understand this same explanation, here: Volume of a Pyramid http://mathforum.org/dr.math/problems/terence9.1.98.html There I explained the induction proof this way: There is a formula for the sum of squares, which is: 1^2 + 2^2 + ...+ N^2 = N*(N + 1)*(2N + 1)/6 This can be proved by induction; that is, by showing that it is true for N = 1, and that if it is true for N, it is also true for N+1. Once that is proved, it must be true for all N. Here is a quick proof: For N = 1, 1^2 = 1*2*3/6 is true. If true for N, then 1^2 + ... + N^2 + (N+1)^2 = N*(N + 1)*(2N + 1)/6 + (N+1)^2 = (N+1) * [N*(2N+1)/6 + (N+1)] = (N+1) * [(2N^2 + N) + (6N + 6)] / 6 = (N+1) * [2N^2 + 7N + 6] / 6 = (N+1) * (N+2) * (2N+3) / 6 = (N+1) * ((N+1) + 1) * (2*(N+1) + 1) / 6 so the formula is true for N+1. The last step in the part you want help on is to divide each of the three factors in this formula by N: N*(N + 1)*(2N + 1)/6 = N * N(1 + 1/N) * N(2 + 1/N) / 6 = N^3 * (1 + 1/N)(2 + 1/N) / 6 = N^3 * (2 + 3/N + 1/N^2) / 6 = N^3 * (2/6 + 1/(2N) + 1/(6N^2)) I hope that helps fill in the gaps for you. I can send you need the derivation of other formulas, such as the area and volume of a sphere, if you'd like. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 05/18/99 at 09:00:34 From: (anonymous) Subject: Re: Volume of a Pyramid (Clarify) About the other derivations, that would be such a great help if you can send them to me. Thank you for all the help. You guys are great. Date: 05/18/99 at 12:08:16 From: Doctor Peterson Subject: Re: Volume of a Pyramid (Clarify) I'll assume that the formulas you are interested in are for the sphere; if there are others, let me know. Here's what I wrote recently about spheres: Usually these formulas are derived using calculus, but they were first worked out by the Greeks, notably Archimedes, using geometrical methods. The complete proof of the formulas can be a little involved, but I can give you the basic idea. First let's look at the surface area of a sphere. Picture cutting the sphere into many horizontal slices, making a set of sloped rings like this: **************** ***** ***** * + *------- / ***** | ***** \ l | / **************** \ |h /. | .\ | / ..... | ..... \ | * .................. *----- * | C *| ***** | ***** | ****************** | |<------x------>| Each is almost the frustum of a cone. Let's ignore the curvature and pretend it is. The area of this piece is the product of its height l (along the slant) and its length, that is, its circumference C around the middle. If we cut all the pieces so their actual heights h are the same, then the slant height l of each piece will get larger as it gets closer to the top or bottom and becomes more slanted. In fact, the ratio of the slant height to the actual height is the same as the ratio of the radius of the sphere, r, to the radius of the ring, x: *********** ***** | ***** ****----------+----------****--+ ** | **|h *----------------+----------------* ** | / |** * | r / | * * | / | * * | / | * * | / | * * +----------------+ * * | x * * | * * | * * | * ** | ** * | * ** | ** **** | **** ***** | ***** *********** l r --- = --- h x So the area of the ring is r * h A_ring = C * l = 2 pi x * ----- = 2 pi r h x This means that each ring has the same area as a cylinder with radius r and height h. If you put all these cylinders together, you get the cylinder that circumscribes the sphere: --------------------------- ------- ------- +- *********** -+ | ------- ***** | ***** ------- | | *---------------------------* | | ** | ** | | * | * | | ** | ** | | * | * | |* | *| |* | *| * | * * + * 2r * | * |* | *| |* | *| | * | * | | ** | ** | | * | * | | ** | ** | | **** | **** | | ***** | ***** | +- *****+*****----------------+ ------- r ------- --------------------------- The height of this cylinder is 2r, so its lateral area is A = 2 pi r * 2r = 4 pi r^2 This is the area of the sphere. Now, for the volume, just picture cutting the sphere into lots of little polygons, and connect each of these to the center of the sphere to make a pyramid. *********** ***** ***** **** **** ** ** * * ** ** * * * * * * * * * + * * \ * * \r * * \\ * * \\\ * ** \ \_\ ** * \|_| * ** ** **** **** ***** ***** *********** The area of each pyramid is 1/3 the product of the base area and the height. But the heights are all the same (if they're small enough): the radius of the sphere. And the sum of their base areas is the surface area of the sphere: 4 pi r^2. So the volume is the same as that of a pyramid whose base is the surface area of the sphere and whose height is its radius: V = 1/3 4 pi r^2 * r = 4/3 pi r^3 What I've described makes a lot of assumptions; to turn it into a careful proof I would have to deal with all those curves I had to ignore, and talk more carefully about pieces being "small enough." But this should give you enough reason to believe the formulas are right. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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