Advanced view of the napkin ring problem? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T20:32:40Z http://mathoverflow.net/feeds/question/37295 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/37295/advanced-view-of-the-napkin-ring-problem Advanced view of the napkin ring problem? Michael Hardy 2010-08-31T19:01:02Z 2011-04-03T17:00:50Z <p>The <a href="http://en.wikipedia.org/wiki/Napkin_ring_problem" rel="nofollow">"napkin-ring problem"</a> sometimes shows up in 2nd-year calculus courses, but it can fit quite neatly into a high-school geometry course via <a href="http://en.wikipedia.org/wiki/Cavalieri%27s_principle" rel="nofollow">Cavalieri's principle</a>.</p> <p>However, the conclusion remains astonishing. Is there some advanced viewpoint from which it becomes obvious from some sort of symmetry that's not visible in the naive formulation? </p> http://mathoverflow.net/questions/37295/advanced-view-of-the-napkin-ring-problem/37315#37315 Answer by David Eppstein for Advanced view of the napkin ring problem? David Eppstein 2010-08-31T21:04:17Z 2010-09-01T21:43:41Z <p>I think it's a bit more elegant to use a different version of Cavalieri's principle: instead of taking cross-sections, sweep the napkin ring by half-disks, bounded by semicircles on the outer sphere whose diameter is the same as the napkin ring's height. The shape of the half-disk is independent of the outer radius. The instantaneous volume swept by the half-disk, per unit angle, just looks (in the limit as the angle goes to zero) as a wedge of a sphere, so it's also independent from the outer radius.</p> <p>Edited to add: I think this is the argument in the following reference from the Wikipedia article: Levi, Mark (2009), "6.3 How Much Gold Is in a Wedding Ring?", The Mathematical Mechanic: Using Physical Reasoning to Solve Problems, Princeton University Press, pp. 102–104, ISBN 978-0-691-14020-9.</p> http://mathoverflow.net/questions/37295/advanced-view-of-the-napkin-ring-problem/37316#37316 Answer by Steve Huntsman for Advanced view of the napkin ring problem? Steve Huntsman 2010-08-31T21:09:34Z 2010-09-01T00:23:39Z <p>Let $V(r,z)$ denote the volume of a napkin ring of outer radius $r$ and height $2z$. We have that $V(r,rz) = r^3 V(1,z)$ and $V(1,az) = a^3 V(1,z)$. The former equality is trivial. To see the latter one, note that $V(1,z) = 4\pi z^3/3$.* </p> <p>It follows that $V(r,z) = r^3 V(1,r^{-1}z) = r^3 r^{-3} V(1,z) = V(1,z)$.</p> <hr> <p>*This might want some fleshing out: e.g. decomposing the complementary part of the sphere into two spherical cones, a cylinder, and two "negative" cones. The volumes of the cylinder and cones can be trivially computed; the volume of a spherical cone can be computed without calculus using its <a href="http://en.wikipedia.org/wiki/Solid_angle#Cone.2C_spherical_cap.2C_hemisphere" rel="nofollow">solid angle</a> $\Omega$ and taking the proportion $\Omega/4\pi$.</p> http://mathoverflow.net/questions/37295/advanced-view-of-the-napkin-ring-problem/37338#37338 Answer by Joseph O'Rourke for Advanced view of the napkin ring problem? Joseph O'Rourke 2010-09-01T01:53:21Z 2010-09-01T10:22:56Z <p>This is my attempt to understand David Eppstein's construction. The green segment is the "stick connecting the center of the sphere to a point halfway from top to bottom on the inner surface of the hole." The half-disk that gets angularly swept around the center of the sphere is purple. The napkin ring ("hole") is red.</p> <p><b>Edit.</b> Altered as per David's comment.</p> <p><img src="http://cs.smith.edu/~orourke/MathOverflow/EppsteinNapkin.jpg" alt="alt text"></p> http://mathoverflow.net/questions/37295/advanced-view-of-the-napkin-ring-problem/37355#37355 Answer by Aaron Meyerowitz for Advanced view of the napkin ring problem? Aaron Meyerowitz 2010-09-01T07:44:59Z 2010-09-01T14:56:57Z <p>EDIT: More briefly than before . Here is a naive physical argument which might meet the request of <em>a point of view from which the result would immediately appear to be just what one would expect before going through the argument.</em></p> <ul> <li>A blob of (incompressible) fluid volume V will form a spherical ball of radius (what it needs to be) if uncontrained</li> <li>A blob of fluid volume V constrained between two parallel plates at z=r and z=-r will form the height 2r central slice of a sphere of radius R where R is just right so that the slice has volume V (provided r is not too large relative to V in which case we get a sphere)</li> <li>A blob of fluid of volume V constrained between two parallel plates at z=r and z=-r and with a cylinder of height 2r and radius q imposed in the middle will form (along with the cylinder) the height 2r central slice of a sphere of radius R where R is just right so that the slice has volume $V+\pi q^2 h$. This shape might not have the curved boundry reach the cylinder</li> <li>Imagine that the volume is just right to get that napkin ring. Now start shrinking q. We will still have a napkin ring. Keep going until q=0 and we see that the volume was that of a sphere of radius r.</li> </ul> <p>(read the previous version if you wish, it might not be worth it)</p> http://mathoverflow.net/questions/37295/advanced-view-of-the-napkin-ring-problem/37365#37365 Answer by sleepless in beantown for Advanced view of the napkin ring problem? sleepless in beantown 2010-09-01T11:07:05Z 2010-09-01T11:24:21Z <p>The key in seeing this lies in breaking it down into a series of stacked washers or discs. </p> <ul> <li><p>Define the napkin ring of height $h$ as being hollowed out of a sphere of radius centered at the origin with radius $R=\alpha h$, with $\alpha>=1$</p></li> <li><p>Define the hole as being drilled along the $z$-axis, creating the napkin ring with an outer diameter $r_{out}$ and inner diameter $r_{in}$ defined as functions of $z$.</p></li> <li><p>$r_{out}=\sqrt(R^2-z^2)$ and inner diameter $r_{in}=\sqrt(R^2-(h/2)^2)$</p></li> <li><p>Define the volume as the integral over $z$ ranging over ($-h/2, +h/2$)</p></li> </ul> <p>$$\pi \int_{-h/2}^{h/2} (r_{out}^2-r_{in}^2) dz$$</p> <p>as the volume being height ($dz$) multiplied by the area ($\pi r_{out}^2 - \pi r_{in}^2)$ of the outer disk minus the inner disk. Note that the $R^2$ cancels out, showing that the radius of the sphere of material does not affect the <strong>AREA</strong> of the annulus at height $z$.</p> <p>This is the intuitive step that may be hard to grasp. The <strong>thickness</strong> of the napkin ring ($r_{out}-r_{in}$) certainly <em>does</em> change with the radius $R$ of the carving sphere. As $R$ increases, the thickness of the napkin ring decreases inversely proportional to the square of $R$. This inverse relationship is what keeps the areas of the annuli at height $z$ at a value that is independent of $R$.</p> <p>The intuitive mis-step is in thinking that even as $R$ increases, the thickness of the napkin ring stays the same. This is incorrect. A similar intuitive mis-step also occurs with this example.</p> <hr> <p>You are wearing a belt of circumference 1 meter. You let out the belt ~31.416 centimeters (~ ten belt notches). How far over your body will the belt float? (also, how much larger can you now become?) The answer is 10 centimeter increase in radius.</p> <p>The earth is wearing a belt at its equator of circumference $C$. How much do you have to let the belt out to create a an increase in radius of 10 centimeters for the earth? The answer is the same ~31.416 centimeters, since circumference is linearly proportional to radius. $C=2\pi r$</p> <p>In the napkin ring problem, the key is that the <strong>area</strong> of the annuslus as a function of distance along the longitudinal-axis remains constant despite any change in $R$ because of the inverse-square proportionality. Intuition fools us into thinking that the thickness must remain constant for such a thing.</p> http://mathoverflow.net/questions/37295/advanced-view-of-the-napkin-ring-problem/60452#60452 Answer by Ashu Prakash for Advanced view of the napkin ring problem? Ashu Prakash 2011-04-03T17:00:50Z 2011-04-03T17:00:50Z <p>Volume of a napkin ring = (4/3)h^3</p>