#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#?

2 Answers
Jun 14, 2018

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


The closed form answer for this general case of #2^N# spheres ( on par with the common volume #pi(4sqrt3/9-1/4)# 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 #(1-sqrt3 /2)# unit deep.

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

second answer.

Jun 14, 2018



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 #pi/2^(N-1)#

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 oo#. The CVS #to# ( Rugby ball

shaped ) prolate spheroid of semi-axes

a = b = 1/2 and c = #sqrt 3 /2#. Its volume is

#4/3 pi (1/2)(1/2)(sqrt3 /2) = pi 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 #(1 - sqrt 3/2)# unit deep.

The volume enclosed, in the limit, is nearly

#4/3 pi (3/2)(3/2)(1) - 2 (1/3pi(1/2)(1/2)(1-sqrt3/2) )# cu

#= pi (17/6 + sqrt3/12)# 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( 3/4 - rho cos theta - rho^2) rho d theta d rho#,

with limits

#rho# from 0 to #1/2 (sqrt( cos^2theta + 3 ) - cos theta )# and

#theta# from 0 to #pi/2^N#

I had used cylindrical polar coordinates #( rho, theta, z )# ,

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

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

#4/3 2^N int (1- cos^2 alpha sec^2theta)^(3/2) d theta#, with

#theta# from #0 to alpha = pi/2^N- sin^(-1)(1/2 sin (pi /2^N))#.

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 = #+- sqrt3/2#. This height is #1 +- sqrt3/2# 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.