Overlap volume of three spheres 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:
ANALYTICAL TREATMENT OF THE VOLUME AND SURFACE-AREA OF MOLECULES FORMED BY AN ARBITRARY COLLECTION OF UNEQUAL SPHERES INTERSECTED BY PLANES
Molecular physics [0026-8976] DODD yr:1991 vol:72 iss:6 pg:1313-1345
Lawrence R. Dodda & Doros N. Theodoroua
 A: 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!
Derivation
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.
Discussion
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.

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

