The ideal cannoli

Given an unit circle as shown, let $x$ be the distance from the center to red vertical line.

The length of this red vertical line is $2\sqrt{1-x^2}$ .

When wrapped on a cylinder of radius $r=\dfrac{1}{\pi}$ , this red line subtends an angle

$\theta = 2\pi \sqrt{1-x^2}$

so that the total cross section area is (pay attention to sign!)

$\dfrac{1}{2} \theta r^2-r^2 Sin\left(\dfrac{1}{2}\theta\right)Cos\left(\dfrac{1}{2}\theta\right)$

We integrate this from $x=-1$ to $x=1$ and get the numerical approximation of $0.5338...$

The edge of the cannoli can be given by the parametric equations $\left\{\begin{array}{c}l x=\cos t \\ y=\frac{1}{\pi}\sin(\pi \sin t) \\ z=-\frac{1}{\pi}\cos(\pi \sin t)\end{array}\right.$ over $0 \leq t \leq 2\pi$ .

Since the cannoli is symmetric over both the planes $x=0$ and $y=0$ , we can consider the total volume as simply $4$ times the volume such that $x,y\geq 0$ .

Now just note that for any given value of $z$ , the cross section of the cannoli (over $x,y\geq 0$ ) is a rectangle with area $xy$ . Putting all this into an integral, we find the volume of the cannoli is

$4\int_{\substack{x,y\geq 0 \\ -1/\pi \leq z \leq 1/\pi}} xy\,dz = 4\int_0^{\pi/2} x(t)y(t)z'(t)\,dt = 4\int_0^{\pi/2} \frac{1}{\pi}\cos^2t \, \sin^2(\pi \sin t)\,dt = \frac{4}{\pi}\int_0^1 \sqrt{1-u^2} \sin^2(\pi u)\,du \approx 0.533801729488$

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I don't know if anyone else shares my excitement about that last integral, but it's interesting to note that this integral relates the answer to the first few Fourier coefficients of the circle, which are given in terms of the Bessel function of the first kind. In other words, it is incredibly unlikely (as a couple others have already suggested) that an elementary expression exists for this answer. Rather, the exact answer would be in terms of the Bessel function (at $2\pi$ , I think)

Since the unit circle is wrapped about a cylinder, we can use cylindrical coordinates to find the volume. The volume is given by

$V = \displaystyle \int_{t=0}^{2 \pi} \int_{r=0}^{a} f(t, r) \, r \, {dr} \, {dt}$

where $a = \dfrac{1}{\pi}$ .

Now we have to find the function $f$ .

From the attached figure, we can deduce the following relations:

$f = 2 \sin s$

where,

$\cos s = a t'$

and,

$a \cos t' = r \cos t$

Combining these relations, we obtain

$f(t, r) = 2 \sin( \cos^{-1}(a \cos^{-1} (r \cos(t) / a ) ) )$

And the volume integral becomes

$V = \displaystyle \int_{t=0}^{2 \pi} \int_{r=0}^{a} 2 \sin( \cos^{-1}(a \cos^{-1} (r \cos(t) / a ) ) ) \, r \, {dr}\, {dt}$

Integrating numerically, we obtain, $V \approx 0.5338$

I was trying to find an analytic solution and concluded that using partial circular cross sections was not going to work (see Michael Mendrin for that method). So I integrated using rectangles whose length is 2x and width 2w. Here $w = r sin(\alpha)$ with $r = 1/\pi$ and $r\alpha = y = \sqrt{(1-x^2)}$ . Integration was along $0 \le z \le2r$ . Here $z = r(1 - cos(\alpha))$ so $dz = rsin(\alpha)d\alpha$ . The volume element is $dV = 4r^2\sqrt{(1-\alpha^2/\pi^2)}sin^2(\alpha)d\alpha$ . I changed variable to $t = \alpha/\pi$ and went to Wolfram for help with the integral. I have attached an image of the Wolfram results. Wolfram did find a recognizable function (Bessel).

oh my god i guess it with random banging to the keyboard. can someone calculate the probability of this situation? :D

The area of the full cylinder is 2 pi (1/pi)^2, which is approximately 0.637 units^3. The cannoli is about 85% of that volume, which gives me an approximation of 0.541 units^3.

I did numerical integration too. Is there any analytic solution?