All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 The Holy Cannoli (Posted on 2006-11-01)
Take a cookie dough rolled flat into a perfect circle of radius R, and wrap it around a cylinder of radius R/Pi , such that opposite points of the original circle now meet at the top. After the cookie is baked and hard, remove the cylinder and fill with cream cheese.

Scrape off the excess filling using a straight edge held perpendicular to the long axis and connecting symmetric points of the edges as you scrape.

What is the volume of one of these theoretical cannoli?

 No Solution Yet Submitted by Larry Rating: 4.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 a couple of extra significant digits (using Bessel J1) Comment 7 of 7 |
(In reply to re: solution--but there's plenty of room for error by Joel)

Following up on Joel's Bessel J1 solution, I looked into Abramowitz and Stegun's Handbook of Mathematical Functions, and found on page 370 a polynomial approximation, with error limits for the J1 function, which is coded below:

DEFDBL A-Z
pi = ATN(1) * 4
x = 2 * pi
f1 = .79788456# + .00000156# * 3 / x + .01659667# * (3 / x) ^ 2 + .00017105# * (3 / x) ^ 3
f1 = f1 - .00249511# * (3 / x) ^ 4 + .00113653# * (3 / x) ^ 5 - .00020033# * (3 / x) ^ 6
th1 = x - 2.35619449# + .12499612# * 3 / x + .0000565# * (3 / x) ^ 2 - .00637879# * (3 / x) ^ 3
th1 = th1 + .00074348# * (3 / x) ^ 4 + .00079824# * (3 / x) ^ 5 - .00029166# * (3 / x) ^ 6
j1 = (f1 + 4E-08) * COS(th1 + 9E-08) / SQR(x)
PRINT j1; (pi - j1) / (2 * pi)
j1 = (f1 + 4E-08) * COS(th1 - 9E-08) / SQR(x)
PRINT j1; (pi - j1) / (2 * pi)
j1 = (f1 - 4E-08) * COS(th1 + 9E-08) / SQR(x)
PRINT j1; (pi - j1) / (2 * pi)
j1 = (f1 - 4E-08) * COS(th1 - 9E-08) / SQR(x)
PRINT j1; (pi - j1) / (2 * pi)

`output:-.2123825198284913  .5338017278570168-.2123825628618599  .5338017347059902-.2123824986322003  .5338017244835224-.2123825416655646  .533801731332495`

The four lines of output contain among them the largest and smallest values possible for J1(2pi) (first column) and for (Pi-J1(2Pi))/(2Pi) (second column).

So the coefficient of r^3 is most likely .53380173, but possibly closer to .53380172.  The full output of the numerical integration program in my first post results in .5338017294051739, when the step value of the integration is stp = .000001, in agreement with the .53380173 value.

BTW, the notation of (Pi-J1(2Pi))/2Pi for (Pi-J1(2Pi))/(2Pi), reminds me that the non-standard practice of assuming multiplication takes precedence over division is somewhat common, and I think useful, but the "official" stance, also incorporated in most programming languages, is that neither takes precedence and the evaluation is left-to-right.

Edited on November 4, 2006, 11:05 am
 Posted by Charlie on 2006-11-04 10:59:39

 Search: Search body:
Forums (0)