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Given three spheres and their coordinates with equal radii that are known to have a triple overlap (a volume contained within all three spheres), is there a known closed form for the calculation of this volume?

Relevant: The two sphere case on Mathworld.

Update: There seems to be a paper that discuss this exact question in general for $n$ three-dimensional spheres. Unfortunately it is behind a paywall that I don't have access to. I'm putting the link here for future reference, please correct me in the comments if the article does not answer the question:


Molecular physics [0026-8976] DODD yr:1991 vol:72 iss:6 pg:1313-1345

Lawrence R. Dodda & Doros N. Theodoroua

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Is there any information on the triangle formed by the centers of the three spheres? – J. M. Sep 29 '11 at 16:57
@J.M. The problem has a few symmetries, you can place the first sphere at the origin, the second along the x-axis and the third in the x-y plane, thus the triangle you mention is fully specified by the sphere coordinates. – Hooked Sep 29 '11 at 18:03
The last sentence of your Mathworld link says: "The surface area of the sphere $R$ that lies inside the sphere $r$ is equal to the great circle of the sphere $r$ , provided that $r \le 2R$ (Kern and Blank 1948, p. 97)." What on earth does this mean? And $-$ whatever it means $-$ how can it possibly be true? – TonyK Sep 29 '11 at 18:05
@TonyK I have no idea - it seems like they are comparing lengths to areas. Maybe someone with the book can provide more insight. – Hooked Sep 29 '11 at 18:10
Yes, sure... so you're saying that triangle could be scalene for instance? – J. M. Sep 29 '11 at 18:14
up vote 3 down vote accepted

The following holds for a specific case where the volume enclosed by 3 spheres is contained within the triangle formed by the 3 sphere centers.

I am hesitant to put this answer up even though I feel it is correct, because the closed form I have found is actually quite ugly. The derivation is quite pretty but the resulting integral is ugly. I will discuss it a bit at the end.

I will post my thoughts anyways in the hope that someone can come along and simplify these formulas into a nicer form, or perhaps find a much simpler formula altogether!


Let $T$ be the solid triangle defined by sides of length $a,b,c$. Define 3 disks of raius $r$: $D,E,F$ and let each vertex of $T$ be the center of a disk in the natural way. See figure included below (at the very bottom) for a graphic description.

Let the wedges be

$$W_1 := D \cap T$$ $$W_2 := E \cap T$$ $$W_3 := F \cap T$$

Let the lenses be

$$L_1 := D \cap E$$ $$L_2 := E \cap F$$ $$L_3 := F \cap D$$

and let the rounded triangle be

$$ R := D \cap E \cap F .$$

It should be noted that each $L_i$ is bisected by a side of $T$. That is

$$ \left| L_i \cap T \right| = \left| L_i \cap \bar{T} \right| $$ where $\bar{T}$ denotes the compliment of $T$ and $| \cdot |$ denotes area. So naturally, define the half-lenses

$$H_1 := T \cap L_1 $$ $$H_2 := T \cap L_2 $$ $$H_3 := T \cap L_3 $$

Let $$ f(r) = \left| R \right| $$, it follows that the volume $V$ contained in the overlap of 3 spheres with equal radii $r_s$ is equal to

$$ V = 2 \int _0 ^{m} f\left(\sqrt{r_s^2-h^2}\right) dh$$

where $m = \sqrt{r_s^2 - \left( \frac {\max\{a,b,c\}} {2} \right)^2}.$

It remains to find a closed form of $f(r)$. We can find one applying the inclusion exclusion principle as

$$ f(r) = |T| - \sum_i |W_i| + \sum_i |H_i| $$

We have $$\sum |W_i| = \frac{\pi}{2} r^2$$

since the sum of the internal angles of $T$ is $\pi.$

By Heron's elegant formula we have

$$ |T| = \sqrt{s(s-a)(s-b)(s-c)}$$ where $s = \frac{1}{2} (a+b+c).$

Lastly, we can compute the area of each half-lense with

$$|H_i| = r^2 \cos^{-1} \left( \frac{d_i}{2r} \right) - \frac{1}{4} d_i \sqrt{4r^2-d_i}$$ where $d_i = c,b,a$ for $i=1,2,3$ using the formulas for lenses but halved.


This seems to be a closed form, as the most difficult integrals to compute, namely those corresponding to the $|H_i|$ terms are found to have a closed form as per a quick query here and here on wolfram's alpha.

But because these indefinite integrals are so ugly, I fear this closed form has little practical use unless a simplification or other formula can be found!

Though one may still be able to find a tight upper and lower bound by simplifiying/approximating these ugly integrals.

Figure depicting 3 disks intersecting

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I should note that this argument can be extended to the case where the region of integration is not contained in the triangle, I'm working on the details at the moment when I have them I'll post an addendum to this answer. – ldog Nov 24 '11 at 18:49

This is an addendum to the previous answer, it handles the general case when $R$ is not contained within the triangle $T$. I use the same approach and notation as in the previous answer.

Instead of using the triangle $T$, we can find a closed form for $f(r)$ using the union of the circles:

$$f(r) = |U| - |D| -|E| -|F| + \sum_i |L_i|$$

where $ |U| = \left| D \cup E \cup F \right| $ .

It remains to find a closed form of $|U|$. In the figure below (at the very bottom) we have that $|U|$ is equal to the area of the 3 triangles plus the area of the 3 circle sectors remaining after taking care of the triangles.

Each new triangle's area can be computed using $$\frac{1}{2}d_i\sqrt{r ^2 - d_i^2/4} $$ for $d_i = a,b,c$.

The area of each circle sector is given by applying the law of cosines. For example, the sector of $F$ is given by

$$\frac{1}{2} r^2 (2 \pi - \theta)$$ where $$\theta = \cos^{-1}\left(\frac{a^2+b^2-c^2}{2ab}\right) + \frac{\pi-\alpha_r}{2} + \frac{\pi-\alpha_s}{2} $$


$$\alpha_r = \cos^{-1} \left( 1 - \frac{b^2}{2r^2} \right)$$ $$\alpha_s = \cos^{-1} \left( 1 - \frac{a^2}{2r^2} \right)$$

Where $\alpha_ \phi$ refers to the angle $\phi$ in the figure below.

Clearly the remaining circle sectors can be solved for in the same fashion.

The general case

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