Recovering the areas of a circle and an ellipse by integration, using their own symmetry

What to assume they know

• Both area formulas of §8.2, and the strip construction behind them
• The standard equations of a circle and of an ellipse centred at the origin, from Class XI, and reading the ellipse's two semi-axes off its denominators
• The antiderivative of the square root of a constant minus a square, from the previous chapter of this book
• Taking a constant factor outside a definite integral, also from the previous chapter
• The inverse sine, its principal values, and its value at one, from Chapter 2 in Part I
• The four quadrants, and which signs the two coordinates take in each
• The area of a circle from school geometry — first as the thing to be recovered, and then as the case the ellipse's answer has to reduce to

What they should be able to do

• Say why the whole circle cannot be handed to the area formula directly
• Identify the property that lets one quarter stand for all four, and state it as the chapter states it
• Rearrange the circle's equation for the strip's length and explain why two values appear
• Choose which of the two to keep, and quote the clause in the chapter that settles the choice
• Write the vertical-strip integral for the quarter, with its limits read off the figure
• Name the antiderivative the chapter uses without proving it, and say which chapter proved it
• Evaluate at both limits, account for every term that vanishes, multiply back by four and check that the recovered formula is the one school geometry gives
• Set up the area of a quarter disc described by a circle and two straight lines
• Read the two semi-axes off a printed ellipse equation whose denominators are squares
• Say what the ellipse example reuses from the circle and what is genuinely new in it
• Take the constant outside the integral and recognise what is left as the circle's own integral
• Arrive at the product of the two semi-axes with a factor of pi, and check it by setting the two semi-axes equal
• Run either derivation with horizontal strips, say which step changes, and say where the reciprocal constant comes from
• Compute the area of an ellipse from a printed equation, for both orientations, and say why the chapter's figures draw only one of them
• Answer the three bracketed questions the chapter leaves unanswered across these pages

Where it usually goes wrong

• "The integral gives the area of the circle." The integral gives the area of one quarter. The factor of four is a separate step, taken before the integral and justified by the figure, and a student who forgets it reports a quarter of the true area with no internal sign that anything is wrong.
• "You could just integrate the whole circle from one side to the other." There is no single function to integrate. Across the full width the circle supplies two values at every place, and the area formula needs one. This is the obstruction the quarter argument exists to remove.
• "Symmetry is a shortcut to save work." Here it is not optional. Without it the strip has no well-defined length. Presenting it as labour-saving teaches students to skip it when they are not in a hurry, which is exactly when it matters.
• "The quarter is a quarter because the picture looks like four equal pieces." It is a quarter because the curve repeats across both axes, which the chapter states as its justification in both examples. The picture illustrates the claim; it does not establish it.
• "The plus-or-minus can be dropped because areas are positive." It is dropped because the region being measured sits where the coordinate is positive. Those are different reasons, and only the second survives contact with a region in another quadrant.
• "The antiderivative of that root is something you should be able to see." It is a result from the previous chapter, obtained by a substitution. Nothing in this chapter derives it and nothing in this chapter needs to.
• "The inverse sine of one is ninety." In this chapter every angle is in radians and the value is a quarter turn. A student who works in degrees gets a number that is not an area at all.
• "A circle's area comes from integration." Historically it does not — it comes from exhausting the disc with polygons, which the chapter's own historical passage describes. What integration supplies is a second route that also handles shapes exhaustion cannot reach.
• "An ellipse needs its own derivation." It needs one new line — a constant taken outside the integral. Everything else is the circle's derivation reused. Presenting it as a fresh argument spends half the running time on the part a student has just watched and no time on the part that is new. This is the whole reason the two curves share one video.
• "The numbers under the squares are the semi-axes." They are the squares of the semi-axes. With sixteen and nine printed, the area is twelve pi and not one hundred and forty-four pi. This single omission is the most common way to get Exercise 8.1 question 1 wrong.
• "The first letter is always the bigger one." It is not. In Exercise 8.1 question 2 the upright semi-axis is the larger, and the chapter's two ellipse figures both draw the other case, so the page teaches the wrong expectation by picture while the algebra stays neutral.
• "The constant can be left inside and dealt with at the end." It can, and then the integral is no longer the one already computed, and the student loses the only structural insight the example carries. Take it out first, deliberately, and say why.
• "Setting the two semi-axes equal is a special case not worth checking." It is the check that the whole derivation is right, and it costs one line. A formula that fails to reduce to the case you trust is a formula with an error in it.
• "The area formula for an ellipse means the perimeter formula is similar." It is not. The area is a clean product; the perimeter has no elementary expression at all. The chapter says nothing about perimeter and an explanation should not invite the question without closing it.
• "Doing it with horizontal strips is a different derivation, and both must be memorised." Each flat version is its own upright version with the two variables exchanged, and the fact that it lands on the same number is a consistency check, not new information. Memorise the structure and derive whichever you need. The chapter presents both alternatives and lets the reader decide what they are for.

Questions to check understanding

• State the symmetry claim a circle centred at the origin permits, and use it to reduce an area calculation to one quadrant
• Rearrange a circle's equation for one variable and justify the sign chosen for a named region
• Write the vertical-strip integral for a quarter disc, with limits, without evaluating it
• Evaluate that integral and recover the school formula for the area of a circle
• Choose the area of a first-quadrant region bounded by a circle and two lines from four options — the form of Exercise 8.1 question 3
• Say which result from the previous chapter both derivations depend on, and where it enters
• Read the two semi-axes off a printed ellipse equation and state its area — the form of Exercise 8.1 questions 1 and 2
• Rearrange an ellipse equation for one variable and identify the constant factor
• Show that setting the two semi-axes equal recovers the circle's area
• Set up either area with horizontal strips instead, and confirm the two directions agree
• Given an ellipse whose upright width is the larger, say which of its two numbers goes where in the formula and why the answer is unchanged

Examples worth working on the board

Values marked verified are worked out here from the chapter's own printed data; neither answers file was opened, and this chapter prints no answers to its exercises.

• Example 1 and its solution (Part II p. 294). The task is the area shut in by a circle centred at the origin. The solution has four moves: declare the whole area to be four times a first-quadrant piece, justified by the figure repeating across both axes; write that piece as a vertical-strip integral; substitute the rearranged equation for the strip's length; and evaluate. Only the first move is particular to this example. The other three are §8.2 applied — and, as Example 2 shows on the next page, the first move is not particular either.
• Fig 8.5 (Part II p. 294). Read off a 300 dots per inch the printed page: the full circle is drawn, centred on the origin, with the topmost point lettered and given its coordinates and the rightmost point lettered and given its coordinates. Only the first-quadrant quarter is filled. One vertical strip is drawn inside the filled quarter, running from the horizontal axis up to the arc, with its length labelled inside the quarter to the left of the strip and its width labelled below the axis at its foot. The three unfilled quarters are drawn but empty. That contrast is the figure's whole argument and the redraw must keep it.
• The chapter's naming of the two quarters (Part II pp. 294–295). Both examples name their first-quadrant piece by walking the same three lettered points and returning to the first, inside a bracket that also names the two boundary ordinates; the ellipse's bracket names the quadrant and the circle's does not, though both examples name it again in the prose lower down the page, in the clause that picks the root. The two quarters carry the identical four-letter name. Read closely: on the circle's figure only two points beyond the origin are lettered, the topmost and the rightmost, while on the ellipse's all four extreme points are, and Example 2 additionally names the whole ellipse by walking all four of them and returning to the first — so the quarter's name is built from the origin and two lettered points either way. Use the letters; they recur in both horizontal-strip alternatives, so a student who learns them once is set for the whole video.
• The two roots, and the clause that picks one (Part II p. 294). The chapter rearranges the equation, writes the result with a plus-or-minus in front of the root, and then disposes of the ambiguity in a single clause: the piece being measured lies where both coordinates are positive, so the positive root is the one taken. Verified: the negative root traces the lower half of the circle, which is not part of the quarter and would give the strip a length pointing the wrong way. This one clause is the most skipped line on the page. It is printed a second time, almost word for word, inside Example 2 on Part II p. 295, and it answers two of the chapter's three unanswered bracketed questions — the two that mark a rearrangement. Say it once, carefully, and then point back to it twice.
• The antiderivative (Part II p. 294). The chapter writes down the antiderivative of the root expression in one step, as a sum of two terms — a product of the variable with the root, halved, and an inverse sine scaled by half the squared constant. It offers no derivation. That result belongs to the previous chapter of this book, where it is obtained by substitution; this chapter simply uses it, and uses the very same two terms again in Example 2 and in both alternatives, with only the letters exchanged. Say so.
• The evaluation (Part II p. 294). Verified: at the upper end the root vanishes, so the first term is nothing at all, and the inverse sine is taken at one, which is a quarter turn; at the lower end both terms are nothing at all. What survives is half the squared constant times a quarter turn, and the factor of four outside turns that into the school formula. Every term that disappears should be watched disappearing — three of the four do, and the same three disappear again in Example 2.
• The circle's alternative, with horizontal strips (Part II p. 295). The same quarter, the strips turned flat, the equation rearranged for the other variable, and the identical antiderivative with the two variables exchanged. Verified: the value is the same. Read off a 300 dots per inch the printed page of Fig 8.6: the same circle, the same single filled quarter, one horizontal strip running from the vertical axis across to the arc, its length labelled above it inside the quarter and its thickness labelled to the left of the vertical axis, outside the figure.
• Exercise 8.1 question 3 (Part II p. 296). A first-quadrant region described by a circle of radius two and two vertical lines, with four options. Verified: the region is the quarter disc and its area is pi, which is the first option. It is Example 1 with the factor of four removed and a number put in place of the letter. Note the printed oddity about the second of the two named lines, recorded under Notes.
• The scaling check worth running once (not in the book). Verified: setting the constant to one gives a unit circle of area pi, and setting it to two gives four pi, which is four times as much for twice the radius — the scaling every student already believes, now falling out of an integral. The chapter offers no such sanity check anywhere.
• Example 2 and its solution (Part II p. 295). Structurally identical to Example 1: a first-quadrant quarter declared to be a quarter of the whole, justified by the figure repeating across both axes; a vertical-strip integral; a rearrangement of the printed equation for the strip's length; and an evaluation. The only new step is the constant. Rearranging leaves the ratio of the two semi-axes multiplying a root that is exactly the circle's root, and that constant is then carried through four lines unchanged. The explanation has already worked all four moves through in full on the circle, so this is the one place it should refuse to slow down.
• Fig 8.7 (Part II p. 295). Read off a 300 dots per inch the printed page: the same construction as Fig 8.5 — full curve drawn, only the first-quadrant quarter filled, one vertical strip inside it with its length labelled to its left and its width labelled below the axis at its foot. What differs is the curve and the lettering: the ellipse is drawn wider than it is tall, and all four extreme points are lettered with their coordinates, the two on the horizontal axis and the two on the vertical, with the origin lettered as well.
• The constant, and where it goes (Part II p. 295). Verified: the printed rearrangement gives the strip's length as the ratio of the second semi-axis to the first, multiplying the square root of the difference between the first semi-axis squared and the variable squared. The chapter pulls that ratio outside the integral without comment. What is then inside the integral sign is character for character the integral finished on the page before, whose value the explanation has already computed, and section 10 exists to make a student notice that rather than watch four more lines of algebra. The reading of that ratio as a stretch factor is added here and is flagged again in Key terms and in Notes.
• The ellipse's evaluation and answer (Part II pp. 295–296). Verified: the surviving integral is a quarter of pi times the square of the first semi-axis; multiplying by the ratio and by four gives pi times the product of the two semi-axes. Three of the four terms in the evaluated bracket vanish at the two ends, exactly as in Example 1.
• The reduction check the chapter never runs (not in the book). Verified: setting the two semi-axes equal turns the ellipse's equation into the circle's and turns the product into the square, recovering Example 1's answer. This is the cheapest possible confirmation that nothing was dropped, and it takes one line. Give it attention; a formula that reduces correctly to the case you already trust is a formula you can use.
• The ellipse's alternative, with horizontal strips (Part II p. 296). The strips turned flat, the equation rearranged for the other variable, and the constant arriving as the reciprocal ratio — the first semi-axis over the second — multiplying the root of the second semi-axis squared minus the variable squared. Verified: the value is the same product. Read off a 300 dots per inch the printed page of Fig 8.8: the same ellipse, wider than tall, all four extreme points lettered with coordinates, only the first-quadrant quarter filled, one horizontal strip running from the vertical axis across to the curve with its length labelled above it and its thickness labelled outside the vertical axis. Note the typesetting slip on that page, recorded under Notes, before the display is shown.
• Why the two constants are reciprocals and the answers agree (not in the book). Verified: the upright version scales the circle whose radius is the horizontal semi-axis by the ratio of vertical to horizontal, and the flat version scales the circle whose radius is the vertical semi-axis by the reciprocal ratio, and both land on pi times the product. Which of the two semi-axes is the larger does not enter, and must not be said to — see Exercise 8.1 question 2 on Part II p. 296, where the vertical one is larger and both routes still work. Saying this out loud closes the alternative properly instead of letting it be four unexplained lines that happen to finish in the same place.
• Exercise 8.1 questions 1 and 2 (Part II p. 296). Two ellipses, given as printed equations, differing only in the number under the first square. Verified: question 1 has sixteen and nine under the two squares, so the semi-axes are four and three and the area is twelve pi; question 2 has four and nine, so the semi-axes are two and three and the area is six pi. Both are Example 2 with numbers, and the only skill they add is taking the square root of each denominator before using the formula — which is precisely the step students omit. In question 2 the longer width is the upright one, so that ellipse is taller than it is wide — and neither of the chapter's two figures draws that shape. The formula is unaffected, because it multiplies the two semi-axes and does not care which is larger, but a student reasoning from the printed picture will hesitate. Part of section 11 exists for that item alone.
• The three bracketed questions (Part II pp. 295 and 296). The chapter marks three steps with a question mark and answers none of them: the rearrangement opening the circle's flat derivation and the rearrangement opening the ellipse's, both asking why the positive root is taken, and the evaluated bracket in the ellipse's upright derivation, asking for the antiderivative. Verified: two of the three are answered by the clause already printed in both worked examples — the quarter lies where both coordinates are positive — and the third by the previous chapter of this book. Answer all three; leaving them hanging is the chapter's habit and it should not be the explanation's.

Figures to have open

• A redraw of Fig 8.5 (Part II p. 294) with the full circle drawn, only the first-quadrant quarter filled, both lettered points with their coordinates, one vertical strip with both labels. The chapter's own diagram; it carries sections 3 and 5, and the three empty quarters must stay drawn and empty.
• A redraw of Fig 8.6 (Part II p. 295) with the same circle, the same single filled quarter and one horizontal strip with both labels. The chapter's own; it carries section 12 and should sit beside the Fig 8.5 redraw at the same scale.
• A redraw of Fig 8.7 (Part II p. 295) with the full ellipse drawn wider than tall, all four extreme points lettered with their coordinates, only the first-quadrant quarter filled and one vertical strip with both labels. The chapter's own diagram; it carries sections 9 and 10, and should be drawn to the same construction as the Fig 8.5 redraw so the two read as one method.
• A redraw of Fig 8.8 (Part II p. 296) with the same ellipse and one horizontal strip, drawn to the same scale as the Fig 8.7 redraw so all four figures can sit in one grid in section 12. The chapter's own.
• A sweeping vertical line across the whole circle for section 2, with both meeting points marked and following the sweep. An added device; the chapter states the two-valued problem in algebra and never draws it.
• The two arcs of the two roots drawn on one circle for section 4, upper and lower in different weights. Not in the book.
• The exercise region for section 8: a circle of radius two, the vertical axis, the tangent line, and the single contact point marked. The chapter prints no figure for any exercise item, so this is entirely added here.
• A two-column diff for section 9, the circle's printed lines against the ellipse's, with the identical lines visibly identical. Not in the book.
• A morph for section 11: the ellipse's upright semi-axis growing until it equals the horizontal one, the outline closing into a circle while the product closes into the square. Not in the book; the chapter never relates its two pairs of figures to one another.
• A tall ellipse for section 11, drawn to the numbers of the second exercise item, beside the chapter's wide one. This drawing does not exist anywhere in the chapter and it is the single most useful new figure in this brief.

Where this sits in the book

• NCERT Class 12 Mathematics, Chapter 8 "Application of Integrals", Example 1 with Fig 8.5, §8.2, Part II p. 294
• The circle's alternative derivation with Fig 8.6, Part II p. 295
• Example 2 with Fig 8.7, §8.2, Part II p. 295
• The ellipse's alternative derivation with Fig 8.8, Part II p. 296
• The two area formulas both examples apply, §8.2, Part II pp. 292–293
• Exercise 8.1, questions 1, 2 and 3, Part II p. 296
• Summary, first bullet, Part II p. 298