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Let's say I have two spheres whose center's coordinates (cartesian) are

0,0,0
d,0,0

and both have radius R.

I want to analytically calculate the total surface area of both spheres. When d>2R it is obviously two times the surface of one sphere, so 8*Pi*R^2. When d is smaller than 2R, spheres overlap, but using surface of revolution integrals, I can easily derive the expression for the surface area.

Before I had a "one-dimensional" array of spheres. But in a "two-dimensional case", let's say I have four spheres, with coordinates

   0,0,0
   0,d,0
   d,0,0
   d,d,0

when d is smaller than 2R (and therefore overlapping spheres) I do not see clearly how can I calculate the total surface. And same problem in the case of 8 spheres in a similar "three-dimensional" arrange.

Anybody could give me any hints?

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1  
Two times the surface area of one sphere is $8\pi R^2$. And the spheres overlap if $d\lt 2R$, not $d\lt R$. –  joriki Sep 27 '11 at 14:57
    
    
yes, you are right joriki, thanks –  flow Sep 27 '11 at 15:13

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