Riddle: Hole in the sphere

March 6, 2018

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A cylindrical hole six centimetres long has been drilled straight through the centre of a solid sphere. What is the volume remaining in the sphere?

This problem is interesting, as it seems to lack information required to be solved. You may wonder how you can ever obtain an answer without knowing the radius of the sphere or the hole. However, as you will find out, obtaining the solution is not that difficult.
 

By calculation

First of all, if we let the radius of the sphere be r, we will find that the radius of the cylindrical hole only depends on this r. By applying Pythagoras’ theorem we easily find this radius:
 
 
 
 
 
 
 

This tells us that the total volume of the hole equals 6\pi(r^2-9)
(height * circle area).

Notice that besides the cylinder itself, also the two caps are missing from the sphere. Acquiring the volume of those caps is a bit tricky, but the formula for the volume of such a spherical cap, in which A is the altitude of the cap and R its radius, is as follows:

    \[\frac{\piA(3R^2+A^2)}{6}=\frac{\pi (r-3)(3(r^2-9)+(r-3)^2)}{6}\]

Subtract the volume of the caps and the volume of the cylindrical hole from the total volume of the sphere, \frac{4\pi r^3}{3}, and we find that everything cancels out except 36π, the remaining volume in cm³!
 

By intuition

When we think about this problem in a proper way, the result is actually not that astonishing. The way that the problem is formulated tells us that the radius of the sphere does not affect the result at all. This can be explained by the fact that this remaining volume depends only on the length of the cylindrical hole. As the original sphere grows, so does the radius of the hole:

Now imagine a sphere that has a radius r such that r=\lim_{h\downarrow 6} h
(r approaches 6 from above). Then the radius of the hole approaches 0 and we can conclude that for a hole of length 6 the remaining volume of the sphere must be equal to that of an intact sphere with a diameter of 6cm : \frac{4\pi 3^3}{3} = 36π.


This article is written by Pieter Dilg

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