What is the volume of spheres in higher dimensions?
[Math] Volume of spheres in higher dimensions?
differential-geometrygeometry
Related Solutions
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.
Yes, you get ellipses and spheres. There are also 3D equivalents to a parabola and hyperbola. There are actually two types of 4D conic surfaces: one is a single connected surface, and the other has two separate surfaces. Here are some animations I just made on the topic:
3D View of Conic Sections from $x^2+y^2=z^2$
2D View
4D Conic Sections of $x^2+y^2=z^2+w^2$
4D Conic Sections of $x^2+y^2+z^2=w^2$
Best Answer
Did you try the obvious?