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  rA 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

Suitable Levels
GCE ‘A’ Levels H2 Mathematics (Number patterns, with algebra)
* revision for IB Mathematics HL & SL
* Advanced Placement (AP) Calculus AB & BC
* University / College Calculus
* other syllabuses that involve volumes and Pythagoras’ Theorem
* any learner who is interested

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