**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 (2

*h*).
Which ring has the larger volume of remaining material?

**Solution**

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,*h*² =*r*² –*a*². 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*r*² –*x*². Hence the area of the material in the cross-section is*p*[(

*r*² –

*x*²) –

*a*²] =

*p*(

*r*² –

*a*² –

*x*²) =

*p*(

*h*² –

*x*²)

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**h*³, 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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