$2^{N}$ $2^N$ 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 $N \to \infty$ $N to oo$ ?

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$2^{N}$ $2^N$ 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 $N \to \infty$ $N to oo$ ?

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Answer 1 · 2018-06-14T05:22:35

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 $2^{N}$ $2^N$ spheres ( on par with the common volume $π (4 \frac{\sqrt{3}}{9} - \frac{1}{4})$ $pi(4sqrt3/9-1/4)$ 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 $\sqrt{3}$ $sqrt 3$ 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 $(1 - \frac{\sqrt{3}}{2})$ $(1-sqrt3 /2)$ unit deep.

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

second answer.

Answer 2 · 2018-06-14T06:38:55

Continuation.

Explanation:

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

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 $\frac{π}{2^{N - 1}}$ $pi/2^(N-1)$

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 $N \to \infty$ $N to oo$ . The CVS $\to$ $to$ ( Rugby ball

shaped ) prolate spheroid of semi-axes

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

$\frac{4}{3} π (\frac{1}{2}) (\frac{1}{2}) (\frac{\sqrt{3}}{2}) = π \frac{\sqrt{3}}{6}$ $4/3 pi (1/2)(1/2)(sqrt3 /2) = pi sqrt 3 / 6$ cu.

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

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

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

The volume enclosed, in the limit, is nearly

$\frac{4}{3} π (\frac{3}{2}) (\frac{3}{2}) (1) - 2 (\frac{1}{3} π (\frac{1}{2}) (\frac{1}{2}) (1 - \frac{\sqrt{3}}{2}))$ $4/3 pi (3/2)(3/2)(1) - 2 (1/3pi(1/2)(1/2)(1-sqrt3/2) )$ cu

$= π (\frac{17}{6} + \frac{\sqrt{3}}{12})$ $= pi (17/6 + sqrt3/12)$ cu.

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

spheres is expressed as a double integral

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

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

with limits

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

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

I had used cylindrical polar coordinates $(ρ, θ, z)$ $( rho, theta, z )$ ,

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

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

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

$θ$ $theta$ from $0 \to α = \frac{π}{2^{N}} - {sin}^{- 1} (\frac{1}{2} sin (\frac{π}{2^{N}}))$ $0 to alpha = pi/2^N- sin^(-1)(1/2 sin (pi /2^N))$ .

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 = $\pm \frac{\sqrt{3}}{2}$ $+- sqrt3/2$ . This height is $1 \pm \frac{\sqrt{3}}{2}$ $1 +- sqrt3/2$ 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.

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$2^{N}$ $2^N$ 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 $N \to \infty$ $N to oo$ ?

2 Answers

Explanation:

Explanation:

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2^N2N2^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 ooN→∞ N to oo?

2 Answers

Explanation:

Explanation:

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$2^{N}$ $2^N$ 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 $N \to \infty$ $N to oo$ ?