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A $10 \times 10 \times 10$ mm cube has $1000000$ 2-micron spheres randomly distributed thru out it. If a random cross-section was done. What would be the area of the cross-sectioned spheres and how many portions of spheres could be counted?

David G. Stork
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James S
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    Is 2-micron the radius or the diameter of the sphere? Is a cross-section found intersecting a random plane and the cube? – Giulio Scattolin Aug 03 '18 at 01:37
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    The odds that a random chord in a unit circle has length less than sqrt(3) is either 1/2, 1/3, or 1/4, as shown in the random chord paradox. The random cross section is worse. – Ed Pegg Aug 03 '18 at 02:06
  • @EdPegg For that paradox it is a matter of choosing the method for drawing the chords. By the way I think the "random cross section" calculus could be simplified observing that any area obtained by the intersecting a plane and a sphere (provided this is the method intended) is a circle whose radius can be desumed from the distance between the section and the center of the sphere. – Giulio Scattolin Aug 03 '18 at 09:11
  • Lets say it's the radius – James S Aug 04 '18 at 11:20
  • @JamesS It would be helpful if you defined how a "random cross-section" was done. As Ed Pegg noted, different ways in cutting the cube could yield different values for the probability of what you're asking. – Giulio Scattolin Aug 04 '18 at 12:33
  • If the spheres are randomly dispersed, wouldn't the count at any one cross section be more or less the same? – James S Aug 06 '18 at 00:41
  • Let me put out the real life problem: I work in the metals business and inclusions(oxides) are not a good thing normally. We took cross sections of ingots, polished and had them scanned for inclusions. The gave us the total count of inclusions, the area scanned and the %area covered by inclusions. I would like to determine the volume of the inclusions so that I can back calculate the weight and knowing what type of oxide it is I can then call the wt% of oxygen. The inclusions are globular and randomly dispersed and relatively the same size. – James S Aug 08 '18 at 23:12

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