[H220151129IAAV] Integration for an Eclipsed Ellipse

Problem

Introduction

     Here we have a challenging H2 Mathematics question that tests students on the technique of substitution, finding area and volume of solid of revolution.  Note that the diagram is not drawn to scale and the line actually intersects the ellipse (oval shape) at the point (-2, 2).  Also the region  R  is the shaded area, but the label “R” is put outside of it.  Students should do their due diligence in ascertaining the intersection point themselves.

Solution

Remarks

     Since the word “exact” is not used in the instructions to part (iii), one can also use the Graphing Calculator to obtain the approximate answer  3.14  to 3 significant figures.  Part (iii) can also be done by another method of integration called the Shell Method, but this is not in the H2 syllabus.  Nevertheless, the student who uses it would not be penalised, unless explicitly forbidden in the rubric.

     The word “ellipse” means an oval shape, and is not to be confused with “eclipse” which means to occlude or hide (e.g. eclipse of the sun, eclipse of the moon).  By the way, the moon orbits around the earth in an ellipse and planets revolve around the sun in ellipses (ovals).
     Although not part of the syllabus, it may be advantageous to know that the area of a right ellipse is  pab,  where  a  and  b  are the semi-axes.  Imagine a circle of radius  a  being stretched by  ^b/_a.  Then its area  pa²  will be multiplied by the same factor  ^b/_a.  So this is not too difficult actually.  With the formula, one can verify one’s answer obtained by integration.

Heuristics Used

H02. Use a diagram / model

H04. Look for pattern(s)

H05. Work backwards        [ e.g. calculating new limits for substitution ]

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

Suitable Levels

* GCE ‘A’ Level H2 Mathematics

* IB Mathematics HL

* AP Calculus BC

* University / College calculus

* other syllabuses that involve applications of integration

* whoever is interested

[IBHL_SOTA201304_1B10c] Quadratic Discriminants

Question

Important Reminders

Solution

Suitable Levels

* GCE ‘O’ Level Additional Mathematics

*  IB Mathematics HL / SL

* other syllabuses that involve quadratics

[IB-HL H&H_Rev6C Q11] Factors of a Complex Polynomial

Question

Introduction

     This question is taken from the Haese & Harris textbook for IB HL Mathematics and is rather challenging.  Here I present two solutions.  In the first solution, I use substitution to make a variable “disappear”.  [ heuristics: H10. Simplify the problem, H11. Solve part of the problem]  And that allowed me to crack the rest of them problem.  In the second solution, I rephrased the problem in terms of tangents to curves. [ heuristic H09. Restate the problem in another way ] This gives another angle from which to tackle the problem.

Solution 1

Solution 2

Thinking Back

     Both solutions are related in the sense that they hinge on some form of the relation maked as [*].  That led to an equation in  a.  Once a  is found,  k  can be found, and the solutions proceed similarly.  The equation [*] is a manifestation of the fact that for a repeated root, both P(x) and P’(x) share a common factor.