Volume of a ConeDate: 02/19/99 at 11:35:04 From: Jodi Anderson Subject: Volume of a Cone I'm teaching a tenth grade geometry class, and I'm not happy with my discussion on finding the volume of a cone. We have actually measured the volume of a cone and compared it to the volume of a cylinder with the same height and radius, and found that it is about 1/3 the volume of the cylinder, but my students still want to know why. I do not remember ever seeing a proof of this, except possibly in Calculus. Do you have any suggestions for a proof of this formula? Thanks. Date: 02/19/99 at 13:16:48 From: Doctor Rob Subject: Re: Volume of a Cone Slice the cone into n pieces with n-1 equally spaced planes parallel to its base. Each of these pieces is a frustum of a cone, and has volume that is greater than the volume of a cylinder whose radius is the radius of the smaller base, and smaller than the volume of a cylinder whose radius is the radius of the larger base. That is because the small cylinder is contained in the frustum of the cone contained in the large cylinder. Here is a diagram of a cross-section of the kth piece, from the side: A B C D E o--o---------------o---------------o--o | /| | |\ | |/ | | | \| o--o---------------o---------------o--o F G H I J The cross-section of the frustum is FJDB; of the larger cylinder it is FJEA; and of the smaller cylinder it is GIDB. A segment of the axis of the cone and the cylinders is CH, whose length is h/n, the height of each frustum and each cylinder. For the kth piece from the top, the radius of the smaller base BC = CD = r*(k-1)/n, and the radius of the larger base FH = HJ = r*k/n. Then, if V(k) is the volume of the kth frustum, we have the inequality Pi*[r*(k-1)/n]^2*(h/n) < V(k) < Pi*[r*k/n]^2*(h/n), (k-1)^2*Pi*r^2*h/n^3 < V(k) < k^2*Pi*r^2*h/n^3, Now, add up these inequalities for k = 1, 2, 3, ..., n to get the volume of the cone in the middle, V = V(1) + ... + V(n). Divide by Pi*r^2*h, and multiply by n^3, and you have: 0^2 + ... + (n-1)^2 < V*n^3/(Pi*r^2*h) < 1^2 + ... + n^2. Now you have to know the sum of the first n squares is equal to n*(n+1)*(2*n+1)/6, so the sum of the first n-1 squares is equal to (n-1)*n*(2*n-1)/6. This can be proved by Mathematical Induction if you don't already know this formula. So, (n-1)*n*(2*n-1)/6 < V*n^3/(Pi*r^2*h) < n*(n+1)*(2*n+1)/6. Now, divide by n^3, and rearrange things a little: (1-1/n)*(1-1/[2*n])/3 < V/(Pi*r^2*h) < (1+1/n)*(1+1/[2*n])/3. Now recall that n could be any positive integer, and the above inequalities must be true. As n gets large, the quantity on the left gets closer and closer to 1/3, and the quantity on the right also gets closer and closer to 1/3. The difference between these two quantities is just 1/n. As n gets large, this difference gets small, approaching zero. In fact, if you want the left side to be larger than 1/3 - a, for some real number a > 0, just pick n > 1/(2*a), and if you want the right side to be smaller than 1/3 + a, just pick n > 1/(3*a). That means that for any positive real number a, we have 1/3 - a < V/(Pi*r^2*h) < 1/3 + a The only real number x that satisfies 1/3 - a < x < 1/3 + a, for every positive real number a, is 1/3 itself. Thus V/(Pi*r^2*h) = 1/3, V = (1/3)*Pi*r^2*h. By the way, a very similar proof, depending on the same formula for the sum of the first n squares, works for pyramids, too. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/