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I would like to know how to calculate the volume of the union of multiple possibly intersecting spheres (using Sphere[]). Please see the figure:

enter image description here

I was trying to get a mesh/delaunayMesh out of the spheres and apply "Volume" which didn't work.

Edit: Fixed spelling (Feb 6th).

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  • $\begingroup$ Please include any code you have developed to try to solve this problem. Also, please reformat your code you have displayed according to the guidelines in meta1027. Doing so will encourage more people to consider your question. $\endgroup$ Commented Feb 5, 2015 at 0:50
  • $\begingroup$ Do you want the volume of the intersection of the spheres, or the volume of the union of the spheres (which happen to be intersecting)? $\endgroup$ Commented Feb 5, 2015 at 1:22
  • $\begingroup$ The volume of the union of the spheres. $\endgroup$ Commented Feb 5, 2015 at 1:23
  • $\begingroup$ Most of the solutions posted here work fine for 2 to 4 nodes (spheres). But any number of spheres which is higher brings up the following message:RegionMeasure::nmet: Unable to compute the measure of region RegionUnion[Ball[{100.,100.,100.},30.],Ball[{120.,120.,120.},30.],Ball[{130.,130.,130.},30.],Ball[{140.,140.,140.},30.],<<1>>,<<1>>,Ball[{190.,190.,190.},30.],Ball[{200.,200.,200.},30.],Ball[{210.,210.,210.},30.],Ball[{230.,230.,230.},30.]]. >> Any idea what it could be? $\endgroup$ Commented Feb 6, 2015 at 20:55
  • $\begingroup$ NIntegrate might be better for many spheres. you might want to provide a real example. $\endgroup$ Commented Feb 7, 2015 at 2:45

5 Answers 5

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spheres = { Sphere[{50, 50, 50}, 25], Sphere[{70, 70, 70}, 25]}; rgn = RegionUnion @@ (spheres /. Sphere -> Ball); RegionMeasure[rgn // N]

122585.

Volume[rgn // N]

122585.

EDIT: Example with more elements

rgn2 = RegionUnion[ Ball[{100., 100., 100.}, 30.], Ball[{120., 120., 120.}, 30.], Ball[{130., 130., 130.}, 30.], Ball[{140., 140., 140.}, 30.], Ball[{190., 190., 190.}, 30.], Ball[{200., 200., 200.}, 30.], Ball[{210., 210., 210.}, 30.], Ball[{230., 230., 230.}, 30.]]; RegionMeasure[rgn2]

590768.

Volume[rgn2]

590768.

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  • $\begingroup$ This helped me! The calculation is fast and I can extend the solution to multiple spheres. Many thanks! $\endgroup$ Commented Feb 5, 2015 at 17:51
  • $\begingroup$ Again, thank you, this helped me. It works nicely for up to 4 to 5 randomly distributed spheres. But any number of spheres which is higher brings up the following message:RegionMeasure::nmet: Unable to compute the measure of region RegionUnion[Ball[{100.,100.,100.},30.],Ball[{120.,120.,120.},30.],Ball[{130.,130.,130.},30.],Ball[{140.,140.,140.},30.],<<1>>,<<1>>,Ball[{190.,190.,190.},30.],Ball[{200.,200.,200.},30.],Ball[{210.,210.,210.},30.],Ball[{230.,230.,230.},30.]]. >> Any idea what it could be? I have this problem also for two more solution proposed here. $\endgroup$ Commented Feb 6, 2015 at 20:51
  • $\begingroup$ Example above with eight balls worked without a problem. $\endgroup$ Commented Feb 7, 2015 at 0:50
  • $\begingroup$ Bob, yes, it worked for me too. I included more two balls and it didn't work anymore. Same error. I think it must be my system. Does it work for you with 10? rgn2 = RegionUnion[Ball[{100., 100., 100.}, 30.], Ball[{120., 120., 120.}, 30.], Ball[{130., 130., 130.}, 30.], Ball[{140., 140., 140.}, 30.], Ball[{190., 190., 190.}, 30.], Ball[{200., 200., 200.}, 30.], Ball[{210., 210., 210.}, 30.], Ball[{230., 230., 230.}, 30.], Ball[{135., 130., 130.}, 30.], Ball[{140., 140., 140.}, 30.], Ball[{195., 190., 190.}, 30.], Ball[{200., 200., 200.}, 30.]]; RegionMeasure[rgn2] $\endgroup$ Commented Feb 7, 2015 at 2:38
  • $\begingroup$ Neither RegionMeasure nor Volume worked for nine or more Balls on my system. Perhaps the level of complexity causes the calculation to exceed some time limit. You might try subdividing the region, sum the volume of the subregions, and subtract the intersections of the subregions. $\endgroup$ Commented Feb 7, 2015 at 3:52
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Let the spheres have radius $r1$ and $r2$ and their centers be separated by distance $d$. There are four cases:

  • $r1+r2 < d$ (separate spheres): $V = {4 \pi \over 3} (r1^3 + r2^3)$
  • $r1 > r2 \wedge d + r2 < r1$ (sphere 2 within sphere 1): $V = {4 \pi \over 3} r1^3$
  • $r2 > r1 \wedge d + r1 < r2$ (sphere 1 within sphere 2): $V = {4 \pi \over 3} r2^3$
  • $r1 + r2 <d \wedge (d + r2 > r1 \vee d + r1 > r2)$ (partially intersecting spheres): $V = {4 \pi \over 3}(r1^3 + r2^3) - V_{cap1} - V_{cap2}$ (see below).

This last result comes from realizing that when two spheres partially intersect, we can define a plane through the circle defined by the spheres' intersecting surfaces. Then there are two "caps" that are "overcounted": the "cap" of sphere 1 within sphere 2, and the "cap" of sphere 2 within sphere 1. The cap of sphere 1 within sphere 2 has volume

$V_{cap1} = \int_{h1}^r \pi r^2(x) dx$ ,

where

$r^2(x) = r1^2 - x^2$.

Thus $V_{cap1} = \frac{1}{3} \pi (\text{h1}-\text{r1})^2 (\text{h1}+2 \text{r1})$.

Likewise, we have

$V_{cap2} = \frac{\pi (d-\text{h1}+2 \text{r2}) \left(d^2-2 d \text{r2}+\text{r1}^2-\text{r2}^2\right)^2}{12 d^2}$.

Here $h1$ is the distance from sphere 1's center to the plane, and likewise for sphere 2. Note that here $h1 + h2 = d$. We solve for $h1$ by $\sqrt{r1^2 - h1^2} = \sqrt{r2^2 - (d - h1)^2}$ then the total volume is the sum of the volumes of the individual spheres (${4 \pi \over 3}ri^3$) minus the two overcounted "caps" given by $h1$ and $h2 = d - h1$.

enter image description here

Of course this computes lightning fast.

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  • $\begingroup$ the analytic approach should be preferred ( for only two spheres in any case). You ought to post the mathematica code though.. $\endgroup$ Commented Feb 5, 2015 at 20:20
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Unfortunately as of 10.0.2, RegionIntersection is not implemented for MeshRegion nor BoundaryMeshRegion objects embedded in 3D. But you could use ImplicitRegion[] as follows:

r1 = ImplicitRegion[(x - 50)^2 + (y - 50)^2 + (z - 50)^2 <= 25^2, {x, y, z}]; r2 = ImplicitRegion[(x - 70)^2 + (y - 70)^2 + (z - 70)^2 <= 25^2, {x, y, z}]; Volume[RegionIntersection[r1, r2]] RegionPlot3D[{r1, r2}, PlotPoints -> 50]
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    $\begingroup$ Note that calls to Volume[] can sometimes take a long time! $\endgroup$ Commented Feb 5, 2015 at 0:40
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    $\begingroup$ I run the code. It really took long. :-( At the end, I got the following message: Unable to compute the volume of region. Any idea? $\endgroup$ Commented Feb 5, 2015 at 1:17
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ball[{x0_, y0_, z0_}, r_] := (x - x0)^2 + (y - y0)^2 + (z - z0)^2 <= r^2.; region = ImplicitRegion[ball[{50, 50, 50}, 25] && ball[{75, 75, 75}, 25], {x, y, z}]; Volume[region] (*1683.46*)
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    $\begingroup$ ball[{x0_, y0_, z0_}, r_] := (x - x0)^2 + (y - y0)^2 + (z - z0)^2 <= r^2.; region = ImplicitRegion[ ball[{50, 50, 50}, 25] && ball[{75, 75, 75}, 10], {x, y, z}]; Volume[region] (* 0 *) which is incorrect. You probably want NAND. $\endgroup$ Commented Feb 5, 2015 at 2:41
  • $\begingroup$ There is no intersection in your case. OP asked for the intersection not the union. $\endgroup$ Commented Feb 5, 2015 at 3:49
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 NIntegrate[ Boole[ Norm[ {x, y, z} - {50, 50, 50}] < 25 || Norm[ {x, y, z} - {70, 70, 70}] < 25 ] , {x, 0, 100}, {y, 0, 100}, {z, 0, 100}] // Chop

122585.

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