# 2^N unit spheres are conjoined such that each passes through the center of the opposite sphere. Without using integration, how do you find the common volume?and its limit, as  N to oo?

Jun 14, 2018

See explanation. Please do not attempt to edit my answer. I would review and edit my answer, if necessary.

#### Explanation:

The closed form answer for this general case of ${2}^{N}$ spheres ( on par with the common volume $\pi \left(4 \frac{\sqrt{3}}{9} - \frac{1}{4}\right)$ for ( N = 1 ) two spheres ) is elusive. I think that this is the reason for the

question being unanswered, for more than a year. I now give more

details to invoke interest.

Let the conjoined spheres rest on a horizontal table. Now, the

height of the whole body is 2 units. There is a $\sqrt{3}$ unit long

common chord. You can see this in the great circle planar-section

of opposite spheres, through this axis.

The surface of the interior common-to-all-spheres space looks like

a pumpkin, with no dimples at the ends of the axis.
graph{((x+1/2)^2+y^2-1) ((x-1/2)^2+y^2-1)(x)=0[-2 2 -1.1 1.1]}

Of course, the exterior outer surface is similar but with dimples

that are $\left(1 - \frac{\sqrt{3}}{2}\right)$ unit deep.

As I have a low-memory computer, I would add more, in my

Jun 14, 2018

Continuation.

#### Explanation:

The common-volume surface (CVS) is segmented by ${2}^{N}$

equal segments, each reaching the common chord ( axis of

symmetry ) through planar sides, like segments of a peeled orange.

To get a typical segment, rotate the LHS circle in the planar-section

graph ( in my 1st part answer ) about the chord, through $\frac{\pi}{2} ^ \left(N - 1\right)$

rad. The RHS part of the circle generates a segment for the inner

surface, and correspondingly, the larger LHS part forms the

opposite segment, for the outer surface.

Let us study the limit, as $N \to \infty$. The CVS $\to$ ( Rugby ball

shaped ) prolate spheroid of semi-axes

a = b = 1/2 and c = $\frac{\sqrt{3}}{2}$. Its volume is

$\frac{4}{3} \pi \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) \left(\frac{\sqrt{3}}{2}\right) = \pi \frac{\sqrt{3}}{6}$ cu.

The outer surface $\to$ an Earth-like oblate spheroid of semi-

axes a = b = 1.5 and c = 1, with conical dimples at the poles that

are $\left(1 - \frac{\sqrt{3}}{2}\right)$ unit deep.

The volume enclosed, in the limit, is nearly

$\frac{4}{3} \pi \left(\frac{3}{2}\right) \left(\frac{3}{2}\right) \left(1\right) - 2 \left(\frac{1}{3} \pi \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) \left(1 - \frac{\sqrt{3}}{2}\right)\right)$ cu

$= \pi \left(\frac{17}{6} + \frac{\sqrt{3}}{12}\right)$ cu.

Now, the elusive volume of the common-to ${2}^{N}$ conjoined

spheres is expressed as a double integral

V = ${2}^{N}$ ( volume of a typical segment)

$= {2}^{N + 2} \int \int \sqrt{\frac{3}{4} - \rho \cos \theta - {\rho}^{2}} \rho d \theta d \rho$,

with limits

$\rho$ from 0 to $\frac{1}{2} \left(\sqrt{{\cos}^{2} \theta + 3} - \cos \theta\right)$ and

$\theta$ from 0 to $\frac{\pi}{2} ^ N$

I had used cylindrical polar coordinates $\left(\rho , \theta , z\right)$ ,

referred to the common chord as z-axis and its center as origin.

Choosing the center of the LHS sphere as origin, this becomes

$\frac{4}{3} {2}^{N} \int {\left(1 - {\cos}^{2} \alpha {\sec}^{2} \theta\right)}^{\frac{3}{2}} d \theta$, with

$\theta$ from $0 \to \alpha = \frac{\pi}{2} ^ N - {\sin}^{- 1} \left(\frac{1}{2} \sin \left(\frac{\pi}{2} ^ N\right)\right)$.

This is indeed a Gordian knot. So, I look for another method that

leads to a closed form solution. The planar section z = 0 for 8 ( N =

3 ) spheres appears below. The graph reveals most of the aspects

in the description.
graph{((x-0.5)^2+y^2-1)((x+0.5)^2+y^2-1)((y-0.5)^2+x^2-1)((y+0.5)^2+x^2-1)((x-0.3536)^2+(y-0.3536)^2-1)((x+0.3536)^2+(y-0.3536)^2-1)((y+0.3536)^2+(x+0.3536)^2-1)((y+0.3536)^2+(x-0.3536)^2-1)=0[-3 3 -1.5 1.5]}
Now, the common space has just disappeared at z = $\pm \frac{\sqrt{3}}{2}$. This height is $1 \pm \frac{\sqrt{3}}{2}$ above the Table.
graph{((x-0.5)^2+y^2-.25)((x+0.5)^2+y^2-.25)((y-0.5)^2+x^2-.25)((y+0.5)^2+x^2-.25)((x-0.3536)^2+(y-0.3536)^2-.25)((x+0.3536)^2+(y-0.3536)^2-.25)((y+0.3536)^2+(x+0.3536)^2-.25)((y+0.3536)^2+(x-0.3536)^2-.25)=0[-3 3 -1.5 1.5]}
The graphs are on uniform scale.