4
$\begingroup$

Imagine a sphere with a hole that has been drilled clean through its center (i.e. a cylindrical piece of the sphere is now missing). This new shape, with the core missing, has height of 6 when standing on the ground (i.e. flat side on the ground). What is the volume of this shape?

$\endgroup$
  • 1
    $\begingroup$ Are you sure this has a unique solution? My first guess is that it depends on the diameter of the drill. $\endgroup$ – Glorfindel Sep 4 '15 at 5:47
  • $\begingroup$ @Glorfindel It does indeed have a unique solution and no additional information is needed to solve it. The simplicity of this question is why I love it so much $\endgroup$ – Dsel Sep 4 '15 at 5:50
  • $\begingroup$ en.wikipedia.org/wiki/Napkin_ring_problem Here is the answer. $\endgroup$ – CodeNewbie Sep 4 '15 at 5:58
  • 1
    $\begingroup$ @Dsel: Don't be disheartened. I don't think this is a bad question, it reminded me of this problem that I had encountered years ago, so that was a good trip down memory lane. What you could do in the future is rehash the question by adding some creative fluff, so that the underlying problem is not immediately visible. $\endgroup$ – CodeNewbie Sep 4 '15 at 6:39
  • 2
    $\begingroup$ @CodeNewbie I'll definitely incorporate that advice in the future. And don't worry, not discouraged at all. Thanks for being patient! $\endgroup$ – Dsel Sep 4 '15 at 6:44
2
$\begingroup$

This is a popular problem commonly referred to as the Napkin ring problem.

Essentially, irrespective of the diameter of the drilled cylinder, the volume of the resultant object will always be equal to the volume of a sphere whose diameter is the same length as the height of the object. In this case, the volume of the sphere is $4/3\pi (6/2)^3$ or roughly 113.1 cubic units, no matter what the diameter of the original sphere was. As long as it was drilled out perfectly till the height of the new object is 6, the volume remains the same.

Here is the Wikipedia entry outlining the problem and its solution.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.