# Volume of cannoli

Problem statement (see pic below):
1. Wrap unit circle over cylinder with radius $r=1/\pi$, edges of unit circle will just touch each other.
2. Remove cylinder, fill cannoli with cream)).
3. Scrape off the excess of cream with straight edge held perpendicular to the cannoli axis and touch symmetic points of the edges.

Find volume of cannoli.
This lead me to a difficult integral which I can't solve so far.

Here are my steps:

1. since figure is simmetrical we can integrate from 0 to R=1:
$V=2\int_0^R S(h)dh$, where $S(h)$ is area at heigh $h$
2. $S(h)$ is a segment of circle with radius $r=1/\pi$
3. Length of this segment is: $l(h)=2\sqrt{R^2-h^2}$ (top pic)
4. The angle of corresponding segment is $\phi={l(h)/ r}$ and it's area is:
$S(h)={1\over 2}r^2(\phi-\sin\phi)=r\sqrt{R^2-h^2}-{1 \over 2}r^2 \sin{2\sqrt{R^2-h^2} \over r}$ (right picture)

5. $V=2r\int_0^R \sqrt{R^2-h^2}dh-r^2\int_0^R\sin{2\sqrt{R^2-h^2} \over r}dh$ Am I being wrong somewhere? If 'no' then how to take 2nd integral?

Your set up seems completely correct but it seems also that the second integral can't be simplified in order to be solved in an easy way.

We can evaluate that by numerical analisys setting $x=h/r$, since $R/r=\pi$

$$r^2\int_0^R\sin{2\sqrt{R^2-h^2} \over r}dh=r^3\int_0^\pi \sin({2\sqrt{\pi^2-x^2}})\, dx$$

and we obtain

$$I=-\int_0^\pi \sin({2\sqrt{\pi^2-x^2}})\, dx\approx 1.048066\dots$$

integral evaluation

Therefore

$$V=\frac{\pi}2rR^2+I\cdot r^3\implies \frac{V}{R^3}=\frac12+\frac{I}{\pi^3}$$

• I found an analytical way to calculate the integral. It requires two variable substitutions and repeated integration by parts, but no advanced techniques. See my answer below. – Connor Harris May 20 '18 at 5:20
• @ConnorHarris That seems very nice! – user May 20 '18 at 6:09
• @gimusi: There is an error: there shouldn't be x under integral when you replaced variables from dh to dx. wolfram evaluation – IgorStack May 20 '18 at 19:31
• @IgorStack Opssss...of course! I fix, Thanks. – user May 20 '18 at 19:42