Volume of a ConeDate: 01/29/2001 at 21:55:37 From: cari Subject: Volume of cones Dr. Math, I know HOW to find the volume of a cone(1/3area of base times height divided by three) but my teacher wants to know WHY. We know that filling the cone with water can prove it, but how does it work with the actual shapes? If you put a cone inside a cylinder, then you obviously have extra room, but how do two other cones fit in there? I understand that they won't keep their regular shape, but please explain. Thank you, Cari Date: 01/29/2001 at 23:29:46 From: Doctor Peterson Subject: Re: Volume of cones Hi, Cari. We have explanations of many such formulas in our archives; but the cone is probably one of the hardest to explain without using calculus. Here's one answer I've given about pyramids, which are closely related: Volume of a Pyramid http://mathforum.org/dr.math/problems/terence9.1.98.html You'll see there how you can fit three pyramids with the same volume into a prism; from there, geometrical knowledge lets us build up to any pyramid, and then to the cone. The best proof I'm familiar with comes very close to the spirit of calculus, without requiring you to know any of it. Let's try doing this directly, rather than starting with a pyramid. You can't actually fit three cones together into a cylinder. Instead, we can dismantle a cone into lots of little near-cylinders. Picture one of those baby toys that look like a cone made up of several rings stacked up; or imagine a cone sliced like a pineapple, and the slices trimmed to make flat cylinders. You can imagine that if you make the slices thin enough, the scrap from the trimming will be as little as you like; so the sum of the volumes of the cylinders will be very close to the volume of the cone itself. How can we find the volume of those slices? Here's a cross-section of the cone, showing the slices: + /|\ / |h\ +--+--+ /| |h |\ / | |r1| \ +--+--+--+--+ /| |h |\ / | | r2 | \ +--+-----+-----+--+ /| |h |\ / | | r3 | \ +--+--------+--------+--+ /| |h |\ / | | r4 | \ +--+-----------+-----------+--+ R If the cone has base radius R and height H, and we've cut it into N slices (including that empty slice at the top, with radius r0 = 0), then each cylinder will have height h = H/N, and radius r[k] = kR/N, where k is the number of the cylinder, starting with 0 at the top and ending with N-1 for the bottom cylinder. The volume of cylinder k will be pi r[k]^2 h = pi (kR/N)^2 (H/N) = pi R^2 H * k^2/N^3 The total volume will be the sum of these, for all k from 0 to N-1; since only k is different from one cylinder to the next, we can factor everything else out from the sum and get V = pi R^2 H / N^3 Sum(k^2) = pi R^2 H / N^3 (0 + 1 + 4 + ... + (N-1)^2) At this point I have to either do some magic and tell you the formula for the sum of squares, and hope you trust me, or try to convince you. The formula is: 0 + 1 + 4 + ... + N^2 = N(N+1)(2N+1)/6 I show a proof by induction in the page I referred to above; another proof can be found here: Formula For the Sum Of the First N Squares http://mathforum.org/dr.math/problems/sandin2.20.98.html If we replace N with N-1, we get 0 + 1 + 4 + ... + (N-1)^2 = (N-1)(N)(2N-1)/6 Put this into our formula and you get V = pi R^2 H (N-1)(N)(2N-1)/(6N^3) = pi R^2 H/6 (N-1)/N N/N (2N-1)/N = pi R^2 H/6 (1-1/N)(1)(2-1/N) Now, if N is very large, 1/N is very small, in fact, as close to zero as you want if N is large enough. So to find the volume of the cone itself, we can just replace it with 0. (Proving this thoroughly is where calculus begins.) We get V = pi R^2 H/6 (1)(1)(2) = 1/3 pi R^2 H Whew! There's the formula. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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