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I'm in Calculus 2, and we were first given the problem to find the intersection of two perpendicular cylinders of equal radius.

Two intersecting cylinders creates this shape

This breaks down into eight times the volume of a quarter circle (with radius r) with perpendicular square cross sections.

$$V=8\int_0^r \sqrt{r^2-x^2}^2dx=8\int(r^2-x^2)dx=8\left[ r^2x - \frac{1}{3}x^3 \right]^{r}_{0}=\frac{16}{3}r^3$$

After this question on the problem set, my teacher has written "Aren't you glad I didn't have you find the intersection of ten cylinders?"

Assuming the ten cylinders intersect in an equal way, like the faces of an icosahedron, I assume this would make some sort of curvy-face icosahedron.

My question is two parts

  1. Can I find the volume using a Calculus II base of knowledge (including a bit of multivar)?
  1. What is the volume of the intersection of ten cylinders of equally radius equally spaced?

Edit: The question should be so that the axis of each cylinder is perpendicular to the face of an icosahedron- because this is 10 pairs of parallel sides, that should be ten cylinders.

Edit 2:
Question 1 is answered: No, but maybe. (That wasn't the important part anyway)
Question 2 is still hanging, as I'd like to see the methodology involved, I'll restate the problem with my current understanding of it.

Ten cylinders, each of radius r intersect along the lines that are perpendicular to the faces of a regular icosahedron at the center of each face. What is the volume of the intersection?

I have created rather crude pictures with my limited Geogebra knowledge: Icosahedron visible for reference. Without Icosahedron visible

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  • $\begingroup$ It's not directly relevant, but you might find this interesting: math.stackexchange.com/questions/50953 $\endgroup$ – joriki Apr 11 '16 at 15:11
  • $\begingroup$ The lines that are central lines of the ten cylinders of radius $r$ are spaced so that neighboring lines would land on adjacent points along a regular 20-gon. This allows you to compute via trigonometry the necessary lengths to find the volume of a piece of the whole volume via similar method as with case you already worked out. There are a total of 40 such pieces. $\endgroup$ – Justin Benfield Apr 14 '16 at 2:07
  • $\begingroup$ To clarify, are all of the cylinders positioned so that their central lines are in the $x$-$y$ plane? $\endgroup$ – Justin Benfield Apr 14 '16 at 2:13
  • $\begingroup$ @JustinBenfield No, the axis of each cylinder is spread "evenly" through $x-y-z$ space so that each would intersect the face of a regular icosahedron twice. As each icosahedron has 20 sides, and the axis intersects twice, this is ten cylinders. $\endgroup$ – Noah Harris Apr 14 '16 at 2:32
  • $\begingroup$ @Noah I noticed that later...the version where they are all on $x$-$y$-plane is quite doable w/ Calc 2 knowledge and some ingenuity. The problem at hand though, is much more complex. $\endgroup$ – Justin Benfield Apr 14 '16 at 2:34
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The picture below illustrates what one will get if one intersect ten infinite long cylinders of unit radius, whose axes are aligned along the ten diagonals of a dodecahedron, against each other.

$\hspace1in$ Intersection of ten cylinders

The resulting figure is very complicated. It consists of $180$ quadrilateral faces and each cylinder contribute $18$ faces. Faces coming from same cylinder has been colored with same color. For example, all the red faces lie on a cylinder whose axis is pointing along the $(-1,1,1)$ direction. The $18$ faces from any cylinder fall into two groups. Up to mirror reflection, $12$ of them are congruent to each other. The remaining $6$ faces are congruent to each other directly.

If one study the figure carefully, one will notice the quadrilaterals arrange themselves into $12$ pentagons. Each pentagon carries $15$ quadrilaterals and these pentagons forming the faces of a dodecahedron. As a "dodecahedron", one vertex $U$ of it is lying along the direction $(-1,1,1)$ and another nearby one $V$ is lying along the direction $(0,\phi, \phi^{-1})$ where $\phi$ is the golden ratio.

To simplify analysis, choose a new coordinate system such that $U$ lies along the $z$-axis and $V$ in the $yz$-plane. i.e.

$$\begin{array}{rcl} (x,y,z)_U^{old} = \sqrt{\frac38} (-1,1,1) &\mapsto& (x,y,z)_U = \frac{3}{\sqrt{8}}(0,0,1)\\ (x,y,z)_V^{old} = \sqrt{\frac38} ( 0,\phi,\phi^{-1}) &\mapsto& (x,y,z)_V = \frac{3}{\sqrt{8}}(0,\frac23,\frac{\sqrt{5}}{3})\\ \end{array} $$

If one "zoom in" the figure from the direction of new +ve $x$-axis and perform an orthographic projection to the new $yz$-plane, one see something like below:

$\hspace1in$ Orthographic projection of intersections

The $18$ red faces now lies along the equator. The cylinder holding them becomes $$\mathcal{C} \stackrel{def}{=} \{ (x,y,z) : x^2 + y^2 = 1 \}.$$ Furthermore, the $18$ red faces can be viewed as the union of $12$ non-simple polygons. Each of them is congruent to either the non-simple polygon $\mathcal{P}$ with vertices $AHDIGDF$ (the one highlighted by a white border) or its mirror image.

To compute the volume of the intersection, we first need to figure out the area of $\mathcal{P}$. As shown in figure above, we can break $\mathcal{P}$ into $6$ right angled triangles:

$$\mathcal{P} = \triangle ABF \cup \triangle BDF \cup \triangle AHC \cup \triangle HDC \cup \triangle DEG \cup \triangle DIE$$

It turns out it is not that hard to compute the area of these sort of right angled triangle on a cylindrical surface. Let me use $\triangle ABF$ on $\mathcal{C}$ as an example.

First, the curve $AF$ lies on the intersection of two cylinders. The axes of these two cylinders are pointing along the direction $OU$ and $OV$ respectively ($O = (0,0,0)$ is the origin, right behind $A$ in above figure). From above figure, it is easy to see $AF$ lies on the plane equal distance between $U$ and $V$. Let $\alpha = \angle BAF$ and $\beta = \angle VOU$. The slope of $AF$ with respect to the equator is then given by

$$\tan\alpha = \cot\frac{\beta}{2} = \frac{1+\cos\beta}{\sin\beta} = \sqrt{\frac{1 + \cos\beta}{1 - \cos\beta}} = \sqrt{\frac{3+\sqrt{5}}{3-\sqrt{5}}} = \frac{3+\sqrt{5}}{2} = \phi^2$$

The point $F$ is one of the vertex of the dodecahedra, it is not hard to see $\;z_F = \frac{3}{\sqrt{8}}\cdot \frac13 = \frac{1}{\sqrt{8}}$.
We can parametrize $AF$ by the map $$ [0,\theta_F] \ni \theta\; \mapsto\; (x,y,z) = (\cos\theta,\sin\theta,\tan\alpha\sin\theta ) \in \mathcal{C} \quad\text{ where }\quad \tan\alpha\sin\theta_F = z_F $$ With this parametrization, the area of the $\triangle ABF$ on $\mathcal{C}$ is given by:

$$\int_0^{\theta_F} \tan\alpha \sin\theta d\theta = \tan\alpha - \tan\alpha \cos\theta_F = \tan\alpha - \sqrt{\tan\alpha^2 - z_F^2} = \phi^2 - \sqrt{\phi^4 - \frac18 } $$

As one can see from this example, given the slope $k$ and height $h$ of such a right angled triangle, its area on the cylinder can be computed using following function: $$A(k,h) = k - \sqrt{k^2 - h^2}$$

Since we are dealing with cylinders with unit radius, the volume of the cone span by $O$ and such a right angled triangle is simply $\frac13 A(k,h)$.

By brute force, one can work out the slopes and heights of remaining $5$ right angled triangles.
To summarize, we have:

$$ \begin{cases} \tan\angle BAF = \phi^2,\\ \tan\angle HAB = \frac{1}{\phi^2},\\ \tan\angle FDB = \tan\angle IDE = \sqrt{2},\\ \tan\angle CDH = \tan\angle EDG = \frac{1}{\sqrt{2}} \end{cases} \quad\text{ and }\quad \begin{cases} |z_F| = \frac{1}{\sqrt{8}},\\ |z_G| = |z_H| = \frac{1}{4\phi^2}\\ |z_I| = \frac{1}{2\phi^2} \end{cases} $$ From this, we find the volume of the intersection is given by

$$\verb/Volume/ = \frac{10 \times 12}{3}\left[ \begin{align} & A\left(\phi^2,\frac{1}{\sqrt{8}}\right) + A\left(\sqrt{2},\frac{1}{\sqrt{8}}\right) + A\left(\frac{1}{\phi^2},\frac{1}{4\phi^2}\right)\\ + & 2 A\left(\frac{1}{\sqrt{2}},\frac{1}{4\phi^2}\right) + A\left(\sqrt{2},\frac{1}{2\phi^2}\right) \end{align} \right] $$ With help of a CAS, one can simplify this to $$\begin{align} \verb/Volume/ &= 5\left(24 + 24 \sqrt{2} + \sqrt{3} - 4\sqrt{6} - 7\sqrt{15} - 4\sqrt{30}\right)\\ &\approx 4.277158048659416687225951566030890254054503016349939576882... \end{align} $$ which is about $2\%$ larger than the volume of unit sphere.

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    $\begingroup$ Approximately $4\pi r^3/3$. (A little more.) $\endgroup$ – Yves Daoust Apr 11 '16 at 21:26
  • $\begingroup$ @YvesDaoust yeah that's basically what I said initially $\endgroup$ – Noah Harris Apr 11 '16 at 21:35
  • $\begingroup$ This is doable analytically if one can compute the volume under a triangular facet on a cylindre (intersection of the cylindre and a pyramid with the apex on the axis). But the complexity is tremendous for the whole computation. A Mathematica guru is required. $\endgroup$ – Yves Daoust Apr 11 '16 at 21:49
  • $\begingroup$ @achillehui (or anybody who knows), With what did you make that image/where did you find it? $\endgroup$ – Noah Harris Apr 12 '16 at 15:21
  • $\begingroup$ @Noah I write a program to compute the intersection of the cylinders, generate a 3-d model in X3D format, use the x3dom library to render it in my browser and dump it to an image. $\endgroup$ – achille hui Apr 12 '16 at 19:07
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Question Two

Consider the example you gave: orienting the poles of the cylinders in a perpendicular fashion must give the smallest possible intersection volume. If the cylinders point in the same direction, you'd have the largest intersection, because the cylinders would fully intersect, and as you rotate away from this extreme, you would decrease the volume of intersection.

You can imagine a skewer going through the middle of a sphere (which you can imagine as the long side of a cylinder), how would you insert another skewer so that the points which the second skewer enter and exit the sphere are as far away from the others as possible? You should see that they would be perpendicular there.

If you had ten cylinders, that would be trying to solve the problem where you have ten skewers (cylinders) to poke into the sphere, now the problem of arranging them has become far more difficult :)

But you can convince yourself (or calculate), the fact that these skewer entry and exit points are as far away from each other as possible, when they are equally spaced by symmetry; since the surface of a sphere is regular.

Question One

So since orienting the cylinders is a problem in itself, I don't believe there will be an easy way to solve this with Calculus II knowledge. But in principle it could be done. You'd just take the integral of the cylinders one at a time.

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  • $\begingroup$ I was trying to express that each cylinder would intersect so that the axis of the cylinder is perpendicular to the faces of a regular icosahedron- is that not as well defined as I though? $\endgroup$ – Noah Harris Apr 11 '16 at 16:02
  • $\begingroup$ "You'd just take the integral of the cylinders one at a time": can you elaborate, I don't think this is right. $\endgroup$ – Yves Daoust Apr 11 '16 at 21:30
  • $\begingroup$ If you integrate the first two cylinders, you'd get a region that intersects the first two integrals. After that, you'd integrate the intersection shape against the next cylinder. How you'd do that is of course non-obvious, and I can not think of a simple analytical way to do it. I'd like to know if anybody has one :) $\endgroup$ – user2662833 Apr 11 '16 at 23:50

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