#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.