#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#?

Question

Sadie Adams · Accepted Answer

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 after the question went unanswered for over a year. I now provide more information to pique curiosity. Place the joined spheres on a table that is horizontal. At this point, the height of the whole body is 2 units. There is a #sqrt 3# unit long common chord. The great circle planar-section illustrates this. between opposing spheres, via this axis. The interior common-to-all-spheres space's surface appears like graph{((x+1/2)^2+y^2-1) ((x-1/2)^2+y^2-1)(x)=0[-2 2 -1.1 1.1]} is a pumpkin without dimples at the ends of the axis. Naturally, the external surface is similar, but it has dimples. that are #(1-sqrt3 /2)# unit deep. Since my computer has little memory, I would add more to my second response.

Violet Aldrich · Answer

Continuation.

The common-volume surface (CVS) is segmented by #2^N# equal parts, all of which arrive at the common chord (axis of symmetry) via flat sides, resembling the peeling of an orange in sections. In the planar-section, rotate the LHS circle to obtain a typical segment. graph ( in my 1st part answer ) about the chord, through #pi/2^(N-1)# rad. The circle's RHS portion creates a segment for the inner surface, and thus, the bigger LHS portion forms the for the exterior, the opposite segment. Let us study the limit, as #N to oo#. The CVS #to# ( Rugby ball prolate spheroid with semi-axes in shape 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- conical dimples at the poles of axes a = b = 1.5 and c = 1. are #(1 - sqrt 3/2)# unit deep. In the limit, the enclosed volume is almost #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 The expression for spheres is 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 restrictions #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 )# , called the center of the common chord the origin and the z-axis the z-axis. Selecting the origin as the LHS sphere's center makes this become #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))#. Yes, this is a Gordian knot. Consequently, I search for another technique that results in a solution in closed form. The planar section z = 0 for 8 ( N = ) The graph below shows the majority of the aspects, with three spheres visible. 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.

Jayden Jackson · Answer

undefined To find the common volume of $2^N$ unit spheres conjoined such that each passes through the center of the opposite sphere without using integration, you can follow these steps:

1. Consider one of the spheres. It touches $2^{N-1}$ other spheres.
2. The volume of the region occupied by this sphere is $2^{N-1}$ times the volume of a single sphere.
3. Multiply the volume of a single sphere by $2^{N-1}$ to find the total volume occupied by all the spheres.

The volume of a single sphere is $\frac{4}{3}\pi r^3$, where $r$ is the radius (in this case, $r = 1$). So, the volume of one sphere is $\frac{4}{3}\pi$.

The total volume occupied by all the spheres is therefore $2^{N-1} \times \frac{4}{3}\pi$.

For the limit as $N \rightarrow \infty$:
$$
\lim_{N \to \infty} 2^{N-1} \times \frac{4}{3}\pi
$$
As $N \rightarrow \infty$, $2^{N-1}$ grows very large. So, the limit of $2^{N-1}$ as $N \rightarrow \infty$ is infinity.

Thus, the limit of the total volume as $N \rightarrow \infty$ is also infinity.