# #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

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

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

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

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

second answer.

Continuation.

#### Explanation:

The common-volume surface (CVS) is segmented by

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

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

shaped ) prolate spheroid of semi-axes

a = b = 1/2 and c =

The outer surface

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

are

The volume enclosed, in the limit, is nearly

Now, the elusive volume of the common-to

spheres is expressed as a double integral

V =

with limits

I had used cylindrical polar coordinates

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

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

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 =

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.