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Volume of a napkin ring using cylindrical shells

  1. Sep 18, 2007 #1

    tony873004

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    Suppose you make napkin rings by drilling holes with different diameters through two wooden bals (which also have different dimeters). You discover that both napkin rings have the same height h

    a) Guess which ring has more wood in it. I guess equal.

    b) Check your guess using cylindrical shells to compute the volume of a napkin ring created by drilling a hole with radius r through the center of a sphere of radius R and express the answer in terms of h.

    [​IMG]

    I found the answer on this web site: http://www.jimloy.com/puzz/napkin.htm
    It says it is pi*h^3/6 . This is very similar to my answer. I only did the top half since it's symmetrical. So that web site's answer, when adjusted for the top half only should be pi*h^3/12.

    I get 2*pi*h^3/3. I must have done something right for my answer to take the same form as his, with the exception of a constant. But I can't find my mistake >:( . Is it possible that website has the wrong answer, and I'm right?

    Thanks :)

    [​IMG]
     
    Last edited: Sep 18, 2007
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  3. Sep 18, 2007 #2

    dynamicsolo

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    I'm not sure why you decided to replace R^2 so early in the evaluation of the limit, particularly when the upper limit is R, which would have immediately given you a term of zero.

    I did the integral using the transformed limits for u [lower: 0 , upper: (R^2)-(r^2)] for the full sphere and confirm your result for the hemisphere. As a sanity check, you can imagine choosing h = R (which would leave an infinitesimal drill hole along a diameter), which would give you the expected total volume (4pi/3)(R^3).

    I find it interesting that the site you referred to shows no work at all. I suspect there's an error there.

    I've always liked this problem because the answer is independent of the size of the sphere, which seems counterintuitive...
     
  4. Sep 18, 2007 #3

    dynamicsolo

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    OK, a couple problems here. First, I went to *this* site to get the volume of a spherical cap: http://mathworld.wolfram.com/SphericalCap.html . To convert the notation, Wolfram's "a" is Loy's "R", the radius of the hole. That taken into account, Loy has omitted a factor of (1/6) in his volume expression. Therefore, he is taking too much volume off with his spherical caps.

    Second, in order to apply Loy's approach (with the correction indicated), you have some further adjustments to make. His "R" is our "r" and vice versa, and the height of the spherical cap "h" becomes for us R-h . Once you get the notation straight (I made two false starts because of the change in interpretation of "h"), you must watch your algebra *very* closely. You must also replace (r^2) with (R^2)-(h^2) at some point. [Who said calculus is harder?] Let's say, after Loy, that "it takes some algebra" (there's an understatement!).

    Cogitating upon his remarks a moment longer, I suddenly realized the other difference. You (and I) are calling the half-height of the hole "h"; Loy is referring to the full height, which he calls "L". If you replace L=2h in his result, you get ours! Your integration is fine...

    Because of that omitted factor, I am suspicious now. Either it's simply a typo on his page or he didn't really do that algebra...
     
    Last edited: Sep 18, 2007
  5. Sep 18, 2007 #4

    tony873004

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    I figured it out. In the first 4 lines, everywhere I had h, I should have had h/2, since if you look at the bottom of the 2 illustrations, b doesn't represent h, it represents half of h, since this illustration is the top half only. So when I plug in later for r, this comes into play.

    You're right, I made it more difficult by replacing R since it zeros out if I leave it in. Re-doing it, the answer does turn out to be pi*h^3/6. Thanks for your help :)

    [​IMG]
     
  6. Sep 18, 2007 #5

    dynamicsolo

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    Good, I'm glad you've sorted it out. As I say, the result you had earlier is equivalent: if you use that formula and replace h=(H/2), H being the full height of the hole, you get (4pi/3)[(H/2)^3] = (pi/6)(H^3), as you have now.

    I think this is a problem where the calculus approach is definitely less trouble than the geometric-algebraic method.
     
  7. Oct 3, 2007 #6
    Quick question though...by substituting 1/2H for b, are you by default solving for the entire volume of the napkin ring, without multiplying the answer by 2?
     
  8. Oct 3, 2007 #7

    tony873004

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    You'll notice that my answer after doing all the heavy math is pi*h^3/12. But you're right, I only integrated the top half the ring. But it's symmetrical. That's why I multiplied it by 2 to get the entire volume. (I almost forgot to do this step!)

    Hey, I just noticed you had the same name as the author of the Calc book this question came from :) Shouldn't that make you the expert??
     
  9. Oct 3, 2007 #8
    Ahhh my mistake...I wasn't thinking that our y doesn't give us the entire solid, it only gives us the top portion. Sorry about that I wasn't thinking
     
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