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?
Let us assume r=1 and then multiply our final solution by r^3 as the volume will necessarily be proportional.
I propose we stand our cylinder upright and take horizontal cross
sections which will look like circles of radius 1/pi with slices cut
out of them. We can just look at the top half and multiply that result
by two. At y=0 the cross section has area 1/pi and at y=1 it has area
0. In between the cross section is the sum of the arc components
radiating from the central axis and the triangular components. The
expression for the area of the arc components is [sqrt(1y^2)]/pi. The
expression for the area of the triangular components is
(1/pi^2)sin(pi*sqrt(1y^2))cos(pi*sqrt(1y^2) or
(1/(2pi^2))sin(2*pi*sqrt(1y^2)).
Therefore the Volume is
2r^3 times the integral from y=0 to 1 of [[sqrt(1y^2)]/pi]dy plus the
integral of the triangle components which I will look at later.
The arc component expression becomes (y/2pi)sqrt(1y^2) + (1/2pi)ArcSin(y) at y=1 minus the same expression at y=0
or (1/2pi)(pi/2) = 1/4
The Volume of the Arc components evaluates to (r^3)/2.
I am going to need help integrating (1/(2(pi^2)))sin(2*pi*sqrt(1y^2))
from y=0 to 1 explicitly but numerically I get roughly 0.0169. I
see this is also expressed by (1/2(pi^2))*sin[2pi*cos(Arcsin y)] but
can't be sure if that is helpful or not.
This gives a total Volume of 0.5339(r^3)
Edited on November 1, 2006, 6:58 pm
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Edited on November 2, 2006, 12:02 pm

Posted by Eric
on 20061101 18:42:00 