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?
We want to find the easiest way to integrate the volume. Let the slices be rectangles perpendicular to a line that goes through the meeting point of the oppisite sides of the cookie and through the axis of the cannoli cylinder. Measure x as the distance of this plane from the axis of the cannoli.
The distance horizontally along the surface of the cookie/cannoli to the plane of a given such rectangular element is c = arccos(pi x / R) * R/pi. That is an x coordinate for the cookie laid flat, so the height of the cookie parallel to the axis of the cannoli is 2*sqrt(R^2  c^2). That is also, then, the height of the rectangular volume element.
The width of the rectangular volume element is 2*sqrt(R^2/PI^2  x^2) * h.
The integral is to be evaluated from R/pi to R/pi
I was unable to get a definite integral for this when substituting the above derivation of h, via c to x. However it can be integrated numerically:
DECLARE FUNCTION acos# (x#)
DECLARE FUNCTION asin# (x#)
' here R is assumed = pi
DEFDBL AZ
DIM SHARED pi
pi = 4 * ATN(1)
stp = .000001
FOR x = 1 TO 1 STEP stp
c = acos(x)
h = 2 * SQR(pi * pi  c * c)
t = t + 2 * SQR(1  x * x) * h * stp
NEXT
PRINT t
FUNCTION acos (x)
acos = pi / 2  asin(x)
END FUNCTION
FUNCTION asin (x)
IF ABS(x) < 1 THEN
asin = ATN(x / SQR(1  x * x))
ELSE
asin = pi / 2 * SGN(x)
END IF
END FUNCTION
The above assumes R = pi. It comes out to a volume of 16.5512. As it should be proportional to R^3, it would give a formula of 16.5512 R^3 / pi^3, or about 0.5338 * R^3.

Posted by Charlie
on 20061101 16:05:49 