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

0:00Pi times the radius squared. You have known the area of a circle since you were small, and you have never once needed to derive it. So deriving it now looks like a waste of a lesson. It is not, and here is why. You have just been handed a new machine for measuring regions: strips standing on the axis, each one running up to the curve, added along the axis as an integral.

0:28A new machine has to be tested against something you already know the answer to, because that is the only kind of test you cannot fudge. The circle is that test. And because everybody knows where it ends, it is very easy to walk past the one step in it that is not routine at all. That step is the whole video. Then a second curve arrives, an ellipse, and repeats every line of the argument with exactly one thing changed.

1:02Begin by trying the obvious thing, so that you can watch it fail. Here is a circle centred at the origin. Feed it straight to the area formula: strips across the whole width, from the leftmost point to the rightmost. Take one strip and ask how long it is. Slide an upright line across the circle and watch where it meets the curve. It meets the curve twice. Not once. Twice, at every single place strictly inside.

1:34Swept across the whole width, at twenty-one interior places the equation hands back two values. At each of the two extreme ends it hands back one, and outside the circle, at four places tried, it hands back nothing at all. The formula needs one length per strip. The circle offers two. There is no integrand to write down, because there is no function here. A circle is not the graph of anything. That is the obstruction, and it is not a technicality to be waved through.

2:12Now the repair, and it is the step everyone skips. Do not take the whole circle. Take only the piece sitting in the corner where both coordinates are positive: the first quadrant, one quarter of the disc. Sweep the same upright line across that quarter and ask the same question. Over the quarter, the equation returns exactly one value at all twenty-three places the curve reaches, and two values at none of them.

2:43One value. A length. A function. Now the strip has a length and the integral exists. And that is the point that gets lost. Taking a quarter is not a shortcut to save arithmetic. It is the thing that makes the arithmetic possible at all. Without it the integrand does not exist. With it, everything after this is routine. It is also not a fluke of the circle we happened to draw. A circle of a different size answers exactly the same way.

3:20So we can measure the quarter. We wanted the whole disc. The bridge is the sentence that the figure repeats itself across both axes. Reflect the quarter across the upright axis and you have half the disc. Reflect that half across the flat axis and you have all of it. Four congruent pieces, so the whole is four times one of them. Notice the justification: it is the repetition, not the picture looking like four equal slices.

3:53Worth checking rather than believing. Take six different radii, and for each one measure all four quarters separately, each over its own stretch, as its own sum of strips. Twenty-four measurements. On every radius the four quarters come to the same number. Not one wrong. And to be sure that is a reading and not a routine that says yes to anything, hold each quarter up against a different radius instead.

4:24Against a quarter of a different size it agrees with none of the twenty-four. So the four really are equal, and the number four in front is earned. Now rearrange the circle equation to get the length of the strip. The sum of the two squared coordinates equals the radius squared. Move one across, take the square root, and the root comes with a plus-or-minus in front of it. Two arcs on the same circle. The positive root traces the upper half. The negative root traces the lower half.

5:02The usual line at this point is that we drop the minus because areas are positive. That is the wrong reason, and it will fail you the first time a region sits somewhere else. The right reason is about the region, not the answer. The piece being measured lies where both coordinates are positive, so the upward coordinate is positive, so the positive root is the one that describes it.

5:31Check it at a hundred and twenty-six places inside the quarter: the kept root is positive at every one of them, and the discarded root is negative at every one of them. The discarded root points out of the region. It would give the strip a length running the wrong way. And it is not harmless. Summed over the same stretch it returns the quarter number with its sign turned over, on all six radii. It is dropped for a reason and the reason is written on the picture.

6:08Everything is now in place, so write the set-up down in one line and read it back. Four, outside, for the four quarters. Then the integral of the positive root, from nought at the origin out to the radius. Those limits come off the figure: the quarter runs from the centre out to where the curve meets the flat axis, and there the strip has shrunk to nothing. Only that first move, the four, is particular to a circle. Everything after it is the strip formula applied exactly as it was given to you.

6:46Which is the good news, because it means the next curve will reuse all of it. The integral itself needs an antiderivative for a square root of a constant minus a square, and this is where a line arrives from somewhere else. It is written down in a single step, as a sum of two terms: the variable times the root, halved, plus an inverse sine scaled by half the squared constant.

7:16No derivation is offered here, and none is needed. That result was obtained earlier, by a substitution, in the work on integration itself. This is a borrowed rule. Say so plainly, rather than letting it look as though the two terms were conjured out of the circle. You should not accept a borrowed rule on faith either. An antiderivative has exactly one job: differentiate it and you must get back the thing you started with.

7:49Tested at five places on each of three different radii, it does. Every time. And tested against a root that has been doubled, it does not, on any of the three. So the check is a real check and not a formality that would have agreed with anything. One more reassurance before we use it. Read that rule at the two ends and subtract, and it returns the same number that four hundred strips returned independently, on all six radii. The shortcut and the slow way agree.

8:26Now evaluate. Two ends, two terms each, so four terms are written down, and most of them are about to disappear. At the upper end, where the variable equals the radius, the root is the root of nothing, so the first term is nothing at all. At the lower end, where the variable is nought, the first term has a nought in front of it, so it goes too. And the inverse sine of nought is nought, so the second term at that end goes as well.

8:58Three terms gone, one survivor. Across six radii that is twenty-four terms written down, eighteen of them nothing, and six standing. The survivor is half the squared radius, times the inverse sine of one. The inverse sine of one. Every instinct says ninety, and ninety is the answer to a different question. In this work every angle is measured in turns of the circle, not in degrees. The sine reaches one at a quarter turn, and it climbs steadily all the way up to it, at every one of twenty-four steps checked along the way, so no earlier angle can be the answer.

9:44A quarter turn. Not ninety. And it matters, because that number is about to be multiplied by a length squared. Substitute ninety instead and the evaluation misses the area on all six radii. It does not miss it by a rounding error either. It hands you a number that is not an area of anything. So the quarter is half the squared radius times a quarter turn. Multiply by the four waiting outside.

10:18Four times a quarter is one. Half a turn times the squared radius. And half a turn is pi. Pi times the radius squared. The formula you already had, arriving from the other direction. Confirmed on all six radii: four times the first quadrant sum lands exactly on the school formula, every time. And the four is doing real work, not decoration. Multiply by three instead and it lands on none of the six. The quarter on its own is a quarter of the answer, on all six, exactly as it should be.

10:59The machine passed the test it could not fudge. One quick sanity check that costs a line and is never run. Double the radius and the area should quadruple, because the radius is squared. Treble it and the area should be nine times as much. Read that straight off the strips themselves, not off the formula: eighteen readings, doubling and trebling across a range of circles. Not one of them wrong.

11:30You have believed that since you were nine years old. Now it has fallen out of an integral, which is a different kind of knowing. Here is the same argument as a question rather than a derivation. A region in the first quadrant, bounded by a circle of radius two and by upright lines. This is the last example with the four removed and a number put where the letter was.

11:59The region is simply the quarter disc, and its area is pi. Measured as strips, exactly pi. And the whole disc of that radius is four times as much, which is the same relationship as before, now with numbers in it. There is a trap hidden in how such a question is worded, and it is worth seeing once. Try four upright lines against that circle. The line at nought meets it at two points. The line at one meets it at two points. The line at two meets it at exactly one point, and the line beyond that meets it nowhere.

12:39A line that meets a circle at a single point touches it. It contributes a point to the boundary, and a point has no length. It bounds nothing. So do not measure a region by reading its list of boundaries off the wording. Draw it, look at it, and describe what you actually see. Now the second curve, and the reason both live in one video. An ellipse centred at the origin: a squared coordinate over one number, plus the other squared coordinate over a second number, adding to one.

13:17The two letters underneath are the semi-axes: half the width across, and half the height up. And they are squared under there, which is the single most expensive thing to forget in this whole topic. Everything we did for the circle happens again. The whole curve is not a graph. The first-quadrant quarter is. The figure repeats across both axes, so the whole is four times that quarter. The rearrangement brings back a plus-or-minus and the same clause settles it, word for word: the piece lies where both coordinates are positive.

13:58Four moves, and three of them you have already watched. So we are not going to walk them again. One thing is new, and it is one line long. Rearrange the ellipse equation for the strip length and look carefully at what comes out. A constant, standing in front of a square root. The constant is the ratio of the two semi-axes: the upright one over the across one. And the root behind it is not a new root. It is the circle root, for the circle whose radius is the across semi-axis.

14:35Checked at a hundred and five places across seven different pairs of semi-axes: at every one of them, the thing behind the constant is exactly the circle root. And it is genuinely that constant, not the other way up. Try the reciprocal instead and it agrees at only fifteen of those places out of a hundred and twenty, and all fifteen belong to one pair. That pair is the one whose two semi-axes are equal, where the ratio and its reciprocal are the same number, so it cannot tell you anything. Everywhere else the reciprocal is simply wrong.

15:19A constant multiplying everything inside an integral can be lifted out in front of the integral sign. You have that rule already. Lift it. And now look at what is left standing inside. What is left is the integral finished a few minutes ago. Not something like it. The same integral, the same limits, the circle quarter at the across semi-axis, whose value we already know. That is true on all seven pairs. Nothing further needs to be computed at all.

15:53So the answer is the constant, times the four outside, times a quarter of pi times the across semi-axis squared. The squares cancel against the ratio and what survives is pi, times one semi-axis, times the other. Confirmed on all seven pairs. Compare that with taking the reciprocal outside instead: on six of the seven, what is left inside is not the integral already finished. The seventh, again, is the pair whose semi-axes are equal.

16:30That is the whole of what is new in the second curve. One constant, lifted out, once. The constant is carried through those lines and never named, so let us name it. It is a stretch. Take the circle whose radius is the across semi-axis, and multiply every upright coordinate by that ratio, leaving the across coordinates alone. Every point of the circle lands on the ellipse. A hundred and five points across seven pairs, every one of them on the curve.

17:07With the reciprocal, ninety of them land off it, and the fifteen that survive are the equal pair once more. And the effect on area is as clean as it possibly could be: stretching in one direction multiplies the area by exactly the stretch. Not more, not less. Verified on all seven. Which is really why the constant appears out front and why nothing else changes. An ellipse is a stretched circle, and the integral is only telling you what stretching does.

17:44Before using a formula, make it prove itself against a case you already trust. This costs one line. Set the two semi-axes equal. The ellipse equation collapses into the circle equation, and pi times one semi-axis times the other collapses into pi times a radius squared. On the pair where the two are genuinely equal, the ellipse answer closes exactly onto the circle answer. On the other six pairs the two answers differ, which is what tells you the check is looking at something.

18:22Now two questions of the kind you will be set. The first has sixteen and nine underneath. The semi-axes are their square roots, four and three, and the area is twelve pi. The second has four and nine. Semi-axes two and three, area six pi. Read those numbers as the semi-axes themselves, forgetting to take the roots, and you get a hundred and forty-four pi and thirty-six pi: twelve times and six times too large.

18:57And look at that second one again. Its upright semi-axis is the larger, so that ellipse is taller than it is wide, and every ellipse you have been shown so far has been drawn wider than tall. The formula does not care. Of seven pairs tried, four are wider than tall, two are taller than wide, and one is neither, and the formula multiplies the two semi-axes together without ever asking which is bigger.

19:30One last thing, which closes both curves at once. Everything so far used upright strips standing on the flat axis. You could instead lay the strips flat against the upright axis and integrate up the other way. Same quarter, same figure, strips turned through a right angle. Rearrange for the other variable, and the borrowed rule comes back with its two letters exchanged. For the ellipse, the constant that appears is the other one: the across semi-axis over the upright one, the reciprocal of the one before.

20:09And the integral it leaves standing is the circle of the upright semi-axis, rather than the circle of the across one. The circle stood on end. The two constants multiply to one on all seven pairs, and are the same number on exactly one of them, the equal pair. That is what reciprocal means, checked rather than asserted. And the flat version, a separate sum over a separate stretch with a separate rule, returns the same seven numbers as the upright one.

20:44Which is a consistency check, not new information. There is one derivation here, and you may run it in either direction. So: a whole circle is not a function, a quarter is, four quarters make a disc, the positive root is chosen by where the region sits and not by the sign of an area, the angle is a quarter turn and never ninety, and an ellipse is that same argument with one stretch carried through it.