Volume of a Pyramid
Date: 08/30/98 at 07:45:10 From: Terence Tham Subject: Volume of pyramids How come the volume of a pyramid is equal to 1/3 the volume of a prism of the same base area and height? Can you explain it to me? Thanks.
Date: 08/31/98 at 12:03:31 From: Doctor Peterson Subject: Re: Volume of pyramids Hi, Terence. A lot of students wonder about this, and we seldom bother to give a proof, because it generally is beyond the abilities of students at the level we teach the formula. There is an explanation using infiinite series in our Dr. Math archives at Volume of a Cone or Pyramid http://mathforum.org/dr.math/problems/swan3.30.98.html This is rather brief, so if you need help understanding it, write back! You may also be interested in seeing how Euclid proved it (complete with a picture, if your browser supports Java): Book XII, Proposition 7, Euclid's Elements - David Joyce http://aleph0.clarku.edu/~djoyce/java/elements/bookXII/propXII7.html This shows a way to divide a triangular prism into three pyramids of equal volume; a little more work is needed to complete the proof for all pyramids and for cones, which Euclid does next. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Date: 09/01/98 at 06:56:11 From: Terence Subject: Volume of pyramids I do not understand the information on the Web site you sent me. Can you further explain it to me? Thanks.
Date: 09/01/98 at 12:04:49 From: Doctor Peterson Subject: Re: Volume of pyramids Hi, Terence. I was expecting to hear back from you. This proof is a little complicated, which is why we don't generally bother to teach it. Most students are satisfied either to trust mathematicians, or to try it out by measuring the volume of a pyramid and seeing that the formula works. It's good that you're not satisfied with that, and want to know the reasons behind what you learn. The first page I referred you to, at http://mathforum.org/dr.math/problems/swan3.30.98.html gives this calculation for the volume of a square pyramid: V = a^2*b*[1^2 + 2^2 + 3^2 + ... + n^2] = a^2*b*n*(n + 1)*(2*n + 1)/6 (which can be proved by induction) = a^2*b*n^3*(1/3 + 1/[2*n] + 1/[6*n^2]) Here's what it means: Suppose we try to build a square pyramid out of flat square boxes (whose volume we can calculate as L x W x H). It won't be perfect, but will look like a "step pyramid": *-* ---------------------------------- **-*|----* ^ */ | |* /|----* | *// *-* / * /|----* | *//*--------* / / * /|----* | /// | |/ / / / * /|b H /// *--------* / / / / / * | //*---------------* / / / / / | // | |/ / / a/ / | // *---------------* / / / / v /*----------------------* / / /B ----------------- / | |/ / / / *----------------------* / / *-----------------------------* / | a |/ *-----------------------------* B Let's build it so that the base of each horizontal slab is a cross- section of the pyramid we are trying to measure, so that this step pyramid is larger than the real pyramid. Say the base is B by B, so its area is B^2, and the height is H. If we have divided the height into N slabs, then the height of each slab is H/N, and the width of the K'th slab from the top is B*K/N. (Do you see why? The top slab is B/N, the next is 2B/N, ..., the bottom one is B*N/N = B.) (Note that his a is my B, and his b is my H/N.) Then the volume of the whole thing is the sum of the volumes of the slabs: (B*1/N)^2 * H/N + (B*2/N)^2 * H/N + ... + (B*N/N)^2 * H/N and we can factor out (B/N)^2 * H/N from all the terms to get: (B/N)^2 * H/N * (1^2 + 2^2 + ...+ N^2) 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. If we use this formula, we get the volume of the step pyramid as: V = (B/N)^2 * H/N * N*(N + 1)*(2N + 1)/6 B^2 * H * N * (N+1) * (2N+1) = ---------------------------- N^3 * 6 B^2 * H 2N^3 + 3N^2 + N = ------- * --------------- 6 N^3 B^2 * H 3 1 = ------- * (2 + --- + ---) 6 N N^2 Now, if we use smaller and smaller steps (bigger and bigger N), our step pyramid will get closer and closer to the actual pyramid. But if N is very large, 3/N and 1/N^2 are much smaller than 2, so we can ignore them. (They represent the extra part of the steps that we have to cut off to get the pyramid.) Then the formula becomes: V = B^2 * H / 3 That's what we were looking for! Now, Euclid's proof in: http://aleph0.clarku.edu/~djoyce/java/elements/bookXII/propXII7.html uses no algebra, but more geometric reasoning. The main part of the proof is to show that a triangular prism can be divided into three equal pyramids, so that the volume of one of the pyramids is one third of the volume of the prism that contains it. Given that, you can easily extend the formula to prisms with any shape base, since the base of any pyramid can be divided into triangles. A few propositions later, he extends the formula to cones. Here's how he divides the prism: F + / |\ E / | \D +--------------+ | | | | | | | | | | | | | C| | | + | | / \ | | / \| +--------------+ B A is equal to the sum of these three pyramids: F + / |\ D E D E / | \D * *--------------* *--------------* //| | \ / \ | / / /| | \ / / \ | / / / | | \ / / \ |/ / / | | / / \ |/ / C/ | | / \ / \ | / * | | / * * / / \ | | / / C C / / \| |/ / *--------------+ * B A B Each can be shown to be equal to another because it has the same base area and the same altitude, which Euclid had shown earlier to imply the same volume, even without knowing the actual formula yet. So that gives you two ways to prove the formula. There may still be some hard parts here for you, but work at it and it should become clear. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Date: 09/03/98 at 04:16:42 From: Tham Terence Subject: Re: Volume of pyramids Hi. Can you send me simpler proofs which I can understand? Can the hypoteneuse be shorter than the adjacent side or opposite sides? I am learning the Pythagorean Theorem in school.
Date: 09/03/98 at 08:52:55 From: Doctor Peterson Subject: Re: Volume of pyramids Hi again. As I explained, we usually don't try to prove the volume formulas for cones, spheres, and so on precisely because they are beyond most students of your age. Hold on to that first proof until you have had enough algebra, and the second until you have had enough geometry, then study them and you should be convinced. I think each of them is about as simple as I can make it. I consider the second proof (Euclid's) to be the best, because it is something you can see. If you spend enough time with it (maybe even making models of a prism split into pyramids) you should be able to follow it enough to believe it. If you think about the Pythagorean theorem: a^2 + b^2 = c^2 you can see that the hypotenuse has to be bigger than either side. For example, c must be bigger than a because you are adding a positive number (b^2) to a^2 to get c^2, so c^2 must be bigger than a^2 and c must be bigger than a. I'm glad you want to ask questions beyond your own level. Keep thinking and working at math, and you will find yourself understanding more and more. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
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