[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

AJC 2009/I/14(a)(ii) Application of Integration: Area

figure 0 – problem statement

Introduction

     In this article, we look at question 14 part (a)(ii) taken from the preliminary examination paper of Anderson Junior College H2 Mathematics Paper 1 in year 2009.  In Singapore schools, you either get challenging questions or very challenging questions.  This part of the question is challenging, and yet is worth only worth 3 marks’ credit.  We can apply the same processes of metacognition (self-monitoring, self-awareness, self-questioning) and heuristics (guidelines, rules-of-thumb, tactics) to solve this problem.  These processes are generally applicable in problem-solving, and more important than the mathematical content itself (which you probably will forget anyway after you graduate from school).  One should learn mathematics not just for the sake of clearing examinations, but to get educated.

     “Education is what remains after one has forgotten what one has learned in
     school. ”                                                                              – Albert Einstein

Stage 1 – Understanding the problem

What topic is this under?

     Integration Applications (Area)

What are you trying to find?

     The area of the shaded region.

Stage 2 – Planning

What methods can you use?  Which is easier?

     We can integrate by cutting the area vertically or horizontally.  [Imagine cutting the area into very thin rectangular strips.]  Horizontal slicing looks easier.

What is the correct formula for that?

     Observing that the area is the area between two curves/lines, the formula is

Which means … ?

     Obviously, y_upper = ⁸/₃  and  y_lower = ¹/₆.  These integration limits correspond with the variable of integration ‘y’.  If the integration is ‘dy’ (with respect to y), then the upper and lower values must be  y  values.  If the integration is ‘dx’ (with respect to x), then the definite integral’s limits must be  x  values.

     x_right is the equation of the (straight) oblique line on the right boundary of the region, with  x  expressed in terms of  y.

     x_left is the equation of the elliptical curve on the left boundary of the region, with  x  expressed in terms of  y.

What heuristics can you use?

     1.  I can split this task into smaller sub-tasks.

     2.  Try to use the result in the previous part, part(a)(i).

What do you need to do?

     I need to find the x_right and x_left and then do the calculation.

Stage 3 – Execution

     Let us find the  x-formulas for the right and left lines.

figure 1 – determining the formulas for right and left parts

Remark: Most of this is straightforward for a JC student.  You are expected to be very familiar with secondary school algebra by now.  Regarding the last two lines: there are two choices for the equation of the curve.  Since we are using the left side of the ellipse (x < 1), we choose the one with the ‘–’ square root instead of the one with the ‘+’.  This is a common trick that schools like to catch students with.  Make sure you don’t stumble on this point.

     With these formulas, we can now work out our solution.

     Applying the formula (line #2) we have line #3 and taking out the brackets leads to line #4.  We do a bit of algebra, and then split the integration into two parts (#line 5).  For lines #6 and #7, the left integral is a straight-forward calculation and the right integral is the answer from part (a)(i).  In line #8, we consolidate our answer by pulling out 25 as common factor.

Stage 4 – Evaluation

Is your answer correct?

     The upper and lower limits match the variable of integration and they make sense.

     Although an exact answer (i.e. non-decimal) is required, we can use the Graphing Calculator to check the calculation numerically.  Here is one way (there are other ways too) to do it:-

We are correct.  The slight differences in the 7^th decimal place is because the Graphing Calculator itself uses an approximation to numerically calculate this definite integral.

Stage 5 – Reflection

What lessons did you learn by solving this question?

     ·  decide whether do to the area by slicing “horizontally” or by slicing “vertically”.  Whichever is easier.

     ·  if doing the integral by slicing “horizontally” always take the right curve
         minus the left one.

     ·  split the task into smaller sub-tasks:-

         1.  determine the upper and lower limits of the definite integral

         2.  if you integrate by slicing “horizontally”, determine the equations of the right
              and left curves, with  x  as the subject.

     ·  for the equation of the elliptical curve with  x  as the subject: choose ‘–Ö’ for the left

         half and ‘+Ö’ for the right half.  [But usually Singapore schools like to set ‘–Ö’ to
         catch unwary students off-guard, and that becomes so predictable: If you do not
         know which to choose, just choose the ‘–’ and you’d probably be right!  LOL! J ]

What if we took the left minus the right?

What will happen if we did the integration by slicing vertically (i.e. with respect to x)?

     Type your comments below.