## Sunday, November 6, 2016

### [AM_20161105ITFF] False Friends in Integration (Calculus)

Question

Introduction
False friends are words in two languages that look/sound alike, but differ significantly in meaning.  Do you know that there are also false friends in mathematics?  Can you distinguish and explain the difference between the two integrals?

Solution

The integrand on the left has the variable  x  as the base and the constant  e  as the index.  So we integrate it using the Power Law.
By contrast, for the integrand on the right, the base  e  is a constant whereas the index is the variable  x.  Integrating  e  to the power of  x  is the eeeeeeeeeeeeeeeasiest.  You just get back the same thing, plus the arbitrary constant of course.

Remark
Many students make the mistake of trying to apply the Power Law for the exponential.  As a learner of mathematics, one needs to cultivate the habit of being observant and paying attention to detail.  This is part of developing one’s identity and character which is important in life.

Suitable Levels
GCE ‘O’ Level Additional Mathematics
GCE ‘A’ Levels (revision)
* revision for IB Mathematics HL & SL (revision)
* Advanced Placement (AP) Calculus AB & BC
* University / College Calculus
* other syllabuses that involve integral calculus

* whoever is interested

## Thursday, November 3, 2016

### [Enrich20161103NRP] The Napkin Ring Problem

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?

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,  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