Thursday, November 3, 2016
[Enrich20161103NRP] The Napkin Ring Problem
Two rings are made by drilling a cylindrical hole through a small sphere and a hole through the large sphere, such that the resulting rings have the same height (2h).
Which ring has the larger volume of remaining material?
The answer is: both rings have the same volume. How can we know?
There is a way to show this using integration. But calculus is not necessary.
Let r be the radius of any chosen sphere and let a be the radius of the cylindrical hole. By Pythagoras’ Theorem, h² = r² – a². Consider a cross-section of the ring sliced a distance x from the centre of the sphere, perpendicular to the axis of the cylindrical hole. The outer radius of this cross section is the square root of r² – x². Hence the area of the material in the cross-section is
p [(r² – x²) – a²] = p (r² – a² – x²) = p (h² – x²)
Note that r does not appear in the formula. That means the cross-section does not depend on r. A bigger (or smaller) sphere would have the same cross-sectional area for each distance x away from the centre. By Cavalieri'sPrinciple, the other sphere will have the same volume!
By the way what is this volume? It is the same as that of a sphere without hole i.e. where a = 0 and r = h. This works out to be 4/3 p h³, where h is half the height of the ring.
H02. Use a diagram / model
H09. Restate the problem in another way
H11. Solve part of the problem
H13* Use Equation / write a Mathematical Sentence
* GCE ‘A’ Levels H2 Mathematics (Number patterns, with algebra)
* revision for IB Mathematics HL & SL
* Advanced Placement (AP)
& BC Calculus
* University / College Calculus
* other syllabuses that involve volumes and Pythagoras’ Theorem
* any learner who is interested