{"version":1,"lectureId":"01M14TZB450QBG80V6614C1YYK","attempt":0,"publication":{"slug":"everything-is-stokes-theorem","title":"Everything Is Stokes' Theorem","subject":"mathematics","summary":"What Stokes' theorem says, explained from the beginning. It is one sentence: add up how fast something is changing everywhere inside a region, and the answer only depends on what happens on the edge of that region. The line itself goes up in the first minute, and then turns into the three theorems it is usually taught as. We meet the three regions and their boundaries, work the calculus theorem on a concrete interval, stack the line integral version underneath it, bend a route to watch the answer refuse to notice, and follow the tiling argument as it deletes every interior edge until only the rim survives. Then the same line is read one dimension higher, over a surface in space whose boundary is a closed curve, which is where the name Stokes actually belongs.","metaDescription":"What Stokes' theorem says: the fundamental theorem of calculus, line integrals and Green's theorem are one line, read at three dimensions.","transcript":"Let me tell you what Stokes' theorem actually says. It is one sentence. Not a page of conditions, not a machine you turn a handle on. One sentence, which then gets written down three times, in three different numbers of dimensions, and picks up a different name each time. Before the sentence, put its three regions side by side. A chunk of the number line, a path through the plane, and a two dimensional patch. These regions have different dimensions, but each has an edge one dimension smaller. Start with the chunk of line. Its edge is two dots, one at each end. Those two endpoints are the complete boundary of the interval from a to b. The path has the same kind of boundary. Its edge is the point P where the journey starts, together with the point Q where it stops. The patch is different. Its edge is not a pair of points. It is the closed curve running all the way around the outside, walked in one direction. Now look at the pattern. The edge of a piece of line is a couple of points. The edge of a path is its two endpoints. The edge of a blob is a curve. In every case the edge is one dimension smaller, and here is the sentence all three theorems share: add up how fast something changes inside a region, and the answer depends only on its edge. And here is that sentence written in symbols. This is Stokes' theorem. The whole of it, on one line. You are not supposed to be able to read that yet. I only want you to have seen it once. Omega is the region, whatever it happens to be. Del Omega is its edge. And the line says: what you get by integrating inside is decided on the edge. And you have almost certainly met that line three times already, under three different names. The fundamental theorem of calculus. The fundamental theorem for line integrals. And Green's theorem. So let's take them one at a time. One dimension, then a curve through the plane, then a flat patch, and watch the same line come back every time. Start in one dimension, where you have known this theorem the longest. Here is the question it actually answers. You know how fast something is changing at every moment: you're driving, and you can watch the speedometer the whole way. How far did you go? The fundamental theorem of calculus says this. If you integrate the derivative of f from a to b, you don't get anything complicated. You get f at b, minus f at a. Read that out loud in words. Add up all the little changes over the whole interval, and you get the net change from one end to the other. The speedometer tells you the odometer. Let's make that concrete, because a number you can check beats a formula you can only nod at. Take f of x equal to x squared over four. Its derivative is x over two, which is this blue line. Fix the left end at one, and let the right endpoint b start at two. The left hand side is the shaded area from one to b. Now slide b out to four. The region grows with it, and the running total up in the corner counts all the way to three point seven five. Now check the other side. f of four is four. f of one is a quarter. Four minus a quarter is three point seven five. Same number, obviously. But here is the thing I really want you to notice. The right hand side never looks inside the interval at all. It only asks two questions. What is f at the right end, and what is f at the left end? Everything in the middle cancels. So here is the piece of language you need. The interval from a to b is the region. Its boundary is just the two endpoints. That word, boundary, is the whole lecture. The boundary is not the number b minus a. It is a formal sum of two points: plus b at the right hand end, together with negative a at the left hand end. Those signs are exactly why the answer is f of b minus f of a. That sign is orientation, the direction in which you travel. Watch the traveller move from the negative endpoint to the positive endpoint. Keep that direction in mind. It comes straight back in Green's theorem. Now go up a dimension. Same sentence, new region. This time the region is not a chunk of the number line. It is a route, drawn through the plane, from a point P to a point Q. And instead of a function of one variable, there are now two. Picture it as a hill: f of x, y is the height of the ground above the point x, y. These grey rings are its contour lines. A contour is the set of places at one particular height. So put a walker on the ring where f is nine, and send it the whole way round. It climbs nothing at all, because every point of that ring is at the same height. Here is theorem one again, so you can see what is coming. Integrate the derivative across the region, and read the answer off the two ends. And here is theorem two. Integrate the gradient of f along the curve C, and what you get is f at Q minus f at P. Look at those two lines together. It is the same sentence twice. Read the left side as, add up every little bit of climbing as you walk. Read the right side as, the height where you finished minus the height where you started. It is the odometer story again, on a hill. And that hands you something genuinely useful. The route does not matter. Take hold of it here, in the middle, and bend it wherever you like, keeping P and Q where they are. Up over the shoulder of the hill. Or down round the bottom instead. The climb you total up is the same number every single time, because the right hand side never mentions the route at all. Because the formula only ever looks at the endpoints. There is that word again. The boundary of a curve is its two ends, and the answer depends on nothing else. Which hands you a corollary that falls straight out. Suppose the route is a closed loop, like this circle Gamma. You finish exactly where you started, so the start point and the end point are the same point. So the endpoint term is f of P minus f of P. The two copies cancel, and the result is zero. Around any closed loop, a gradient field integrates to zero. You can see it coming without computing anything: closed curve, gradient field, nothing to do. Same theorem. Region, boundary, done. Now the one that looks hardest and really isn't. So, the flat case. Green's theorem. This is the one that looks frightening the first time you meet it, and it is the same sentence again. Our region is now genuinely two dimensional: a flat patch R in the plane, carrying a field of arrows. The rectangle keeps the boundary easy to read, but the theorem does not depend on that shape. Its boundary is the closed curve that runs all the way around the outside. We write it as del R. Orientation matters, exactly as the plus and minus mattered on the interval. Travel counterclockwise, so the region stays on your left. Starting here, one complete trip comes back to the point it left. That choice of direction is the same choice as the plus sign on the interval's right hand end. Here is the statement. The double integral over every point of the region lives on the left. One line integral around its boundary lives on the right. Inside on the left, edge on the right. Read the left side as the total swirl at all points inside R. Read the right side as the total push along one counterclockwise trip around del R. Every arrow inside contributes to the first count; the rim supplies the second. Green's theorem says those two numbers are equal. Total swirl inside equals total push around the rim. That is the whole theorem. Now for the reason. Chop R into little tiles. Each tile contributes a walk around its own boundary. Choose this tile. Its contribution includes this little edge. Adding the tiles means reading that edge once from each tile that touches it. One tile walks the shared edge upward. Its neighbour walks the very same edge downward. The geometric edge is one object, but its two induced directions are opposites. Those two contributions cancel. The same happens at every shared edge, so the entire interior grid deletes itself. Only edges with no neighbour remain: the outside of R. Their induced walk is the surviving boundary, with the region on its left. That is Green's theorem. The equality now reads from the region to its surviving boundary. The derivative is integrated over R on the left; the original field is integrated over del R on the right. If a line integral around a closed curve is horrible, read the equality backward. Swap the loop integral to the left and the double integral to the right. Often that reversal is the whole difference between an integral you can do and one you cannot. Put the three theorems side by side with the three regions we started from. For each theorem, name its region and then its complete boundary. For the fundamental theorem of calculus, the region is the interval from a to b. Its boundary is the two endpoints a and b. For line integrals, the region is the curve C. Its boundary is the point P where the curve starts and the point Q where it ends. For Green's theorem, the region is the patch R. Its boundary is the closed loop del R, walked with the region on the left. Now read the two right hand columns. This one names the region being integrated over. This one names its boundary, always one dimension smaller. Every row is saying the same thing. And that shared structure is the line I showed you at the start. Here it is again, and this time it should read as ordinary English. Omega names the region, whatever dimension it has. Del Omega names its boundary, always one dimension smaller. The symbol d means take the appropriate derivative: an ordinary derivative on an interval, a gradient along a curve, or curl over a patch. It is the operation that turns the original quantity into the inside quantity. The left side integrates that derivative over the region. The right side integrates the original quantity over the boundary. Read at three different dimensions, this one line is all three theorems. And notice what that line does not say. It never says how many dimensions the region has. So take one more: a surface in space, a sheet of it hanging in the air. Let me turn it round, because a curved sheet and a flat one look exactly alike until something moves. Now you can see it is bowed upward, like a cloth held up by its four corners. Its boundary is the closed curve running round the rim, walked so that the surface stays on your left. A region, and an edge one dimension smaller. The same two things as every other picture today. Read that same line over a surface like this one, and what you get is the theorem that actually carries Stokes' name. Read it one dimension higher again and it is the divergence theorem. Nobody had to invent a new idea for either of them. So there it is. That line is Stokes' theorem. Omega is whatever region you have, del Omega is its edge, and adding up the change inside always leaves you something that was settled on that edge.","watch":{"version":1,"scenes":[{"title":"One Line, Three Theorems","start":0,"end":128.0810625,"objects":{"big_idea":"a Panel that says \"Add up how fast something is changing everywhere inside a region, and the answer only depends on what happens on the edge of that region.\"","blob":"a Polygon [blue] labelled \"R\" drawn in trio (vertices=((7.4, 1.1), (9.9, 1.2), (10.2, 2.6), (8.7, 3.4), (7.2, 2.4)), fill_opacity=0.25)","blob_boundary":"a ParametricCurve [red] drawn in trio (function=<function>, breakpoints=(0.2, 0.4, 0.6, 0.8))","blob_edge":"an Orientation [red] drawn in trio (path=((7.4, 1.1, 0.0), (7.5953125, 1.1078125, 0.0), (7.790625, 1.115…, closed=True, arrows=4)","end_a":"a Point [red] labelled \"a\" drawn in trio (location=(0.6, 2.1))","end_b":"a Point [red] labelled \"b\" drawn in trio (location=(2.6, 2.1))","heading":"a Heading that says \"The One Sentence\"","heading_line":"a Heading that says \"The Same Line, Three Times\"","interval_name":"a Math [text] that says \"$[a, b]$\" drawn in trio","master":"a Math [text] that says \"$integral_Omega dif omega = integral_(partial Omega) omega$\"","path":"a ParametricCurve [blue] labelled \"C\" drawn in trio (function=<function>)","path_end":"a Point [red] labelled \"Q\" drawn in trio (location=(6.199999999999999, 3.0))","path_start":"a Point [red] labelled \"P\" drawn in trio (location=(3.9, 1.3))","roster":"a Block [text] that says \"Fundamental Theorem of Calculus Fundamental Theorem for Line Integrals Green's Theorem\"","segment":"a Line [blue] drawn in trio (start=(0.6, 2.1), end=(2.6, 2.1))","title_card":"a Title that says \"Vector Calculus — Everything Is Stokes' Theorem\"","trio":"a Figure (x_range=(0.0, 10.8), y_range=(0.0, 4.2), aspect=(10.8, 4.2))"},"beats":[{"start":0,"say":"Let me tell you what Stokes' theorem actually says. It is one sentence. Not a page of conditions, not a machine you turn a handle on. One sentence, which then gets written down three times, in three different numbers of dimensions, and picks up a different name each time.","live":[],"does":[[0,"title_card is shown on the screen, written out."],[1.5,"title_card: enter:write-left-to-right."],[16.0095,"title_card is hidden from the screen — left the board."]]},{"start":17.2095,"say":"Before the sentence, put its three regions side by side. A chunk of the number line, a path through the plane, and a two dimensional patch. These regions have different dimensions, but each has an edge one dimension smaller.","live":null,"does":[[17.2095,"heading is shown on the screen, written out."],[17.2095,"trio is shown on the screen, written out."],[21.796,"segment is shown on the screen, written out."],[23.038,"path is shown on the screen, drawn."],[25.197,"blob is shown on the screen, written out."]]},{"start":31.787999999999997,"say":"Start with the chunk of line. Its edge is two dots, one at each end. Those two endpoints are the complete boundary of the interval from a to b.","live":["trio","heading","segment","path","blob"],"does":[[34.76,"end_a is shown on the screen, written out."],[35.059999999999995,"end_b is shown on the screen, written out."],[37.395999999999994,"interval_name is shown on the screen, written out."]]},{"start":41.53699999999999,"say":"The path has the same kind of boundary. Its edge is the point P where the journey starts, together with the point Q where it stops.","live":["trio","heading","segment","path","blob","end_a","end_b","interval_name"],"does":[[45.367999999999995,"path_start is shown on the screen, written out."],[48.42199999999999,"path_end is shown on the screen, written out."]]},{"start":50.02049999999999,"say":"The patch is different. Its edge is not a pair of points. It is the closed curve running all the way around the outside, walked in one direction.","live":["trio","heading","segment","path","blob","end_a","end_b","interval_name","path_start","path_end"],"does":[[54.489999999999995,"blob_boundary is shown on the screen, drawn."],[58.007999999999996,"blob_edge is shown on the screen, written out."]]},{"start":59.54849999999999,"say":"Now look at the pattern. The edge of a piece of line is a couple of points. The edge of a path is its two endpoints. The edge of a blob is a curve. In every case the edge is one dimension smaller, and here is the sentence all three theorems share: add up how fast something changes inside a region, and the answer depends only on its edge.","live":["trio","heading","segment","path","blob","end_a","end_b","interval_name","path_start","path_end","blob_boundary","blob_edge"],"does":[[63.042999999999985,"end_a is emphasized."],[63.042999999999985,"end_b is emphasized."],[64.541,"end_a is no longer emphasized."],[64.541,"end_b is no longer emphasized."],[65.65499999999999,"path_start is emphasized."],[65.65499999999999,"path_end is emphasized."],[67.50099999999999,"path_start is no longer emphasized."],[67.50099999999999,"path_end is no longer emphasized."],[68.10499999999999,"blob_boundary is emphasized."],[69.452,"blob_boundary is no longer emphasized."],[72.71399999999997,"trio moves to a new place on the board."],[72.71399999999997,"big_idea is shown on the screen, written out."],[79.95849999999999,"trio moves to a new place on the board."],[79.95849999999999,"big_idea is hidden from the screen — left the board."],[79.95849999999999,"heading is hidden from the screen — left the board."]]},{"start":81.15849999999999,"say":"And here is that sentence written in symbols. This is Stokes' theorem. The whole of it, on one line.","live":["trio","segment","path","blob","end_a","end_b","interval_name","path_start","path_end","blob_boundary","blob_edge"],"does":[[81.15849999999999,"heading_line is shown on the screen, written out."],[82.94699999999997,"master is shown on the screen, written out."]]},{"start":88.27149999999999,"say":"You are not supposed to be able to read that yet. I only want you to have seen it once. Omega is the region, whatever it happens to be. Del Omega is its edge. And the line says: what you get by integrating inside is decided on the edge.","live":["trio","segment","path","blob","end_a","end_b","interval_name","path_start","path_end","blob_boundary","blob_edge","master","heading_line"],"does":[[93.55399999999999,"master (the \"Omega\" part) is emphasized."],[98.49999999999999,"master (the \"Omega\" part) is no longer emphasized."],[98.49999999999999,"master (the \"partial Omega\" part) is emphasized."],[101.98299999999999,"master (the \"integral_Omega dif omega\" part) is emphasized."],[101.98299999999999,"master (the \"partial Omega\" part) is no longer emphasized."],[102.73799999999999,"master (the \"integral_(partial Omega) omega\" part) is emphasized."],[102.73799999999999,"master (the \"integral_Omega dif omega\" part) is no longer emphasized."],[104.37499999999999,"master (the \"integral_(partial Omega) omega\" part) is no longer emphasized."]]},{"start":104.975,"say":"And you have almost certainly met that line three times already, under three different names. The fundamental theorem of calculus. The fundamental theorem for line integrals. And Green's theorem.","live":null,"does":[[109.073,"roster is shown on the screen, written out."],[111.639,"roster (the \"Fundamental Theorem of Calculus\" part) is emphasized."],[114.309,"roster (the \"Fundamental Theorem for Line Integrals\" part) is emphasized."],[114.309,"roster (the \"Fundamental Theorem of Calculus\" part) is no longer emphasized."],[115.96999999999998,"roster (the \"Fundamental Theorem for Line Integrals\" part) is no longer emphasized."],[115.96999999999998,"roster (the \"Green's Theorem\" part) is emphasized."],[117.2115,"roster (the \"Green's Theorem\" part) is no longer emphasized."]]},{"start":117.8115,"say":"So let's take them one at a time. One dimension, then a curve through the plane, then a flat patch, and watch the same line come back every time.","live":["trio","segment","path","blob","end_a","end_b","interval_name","path_start","path_end","blob_boundary","blob_edge","master","roster","heading_line"],"does":[[120.59799999999998,"segment is indicated — a transient flash."],[122.21199999999999,"path is indicated — a transient flash."],[123.52399999999999,"blob is indicated — a transient flash."],[127.03939583333333,"heading_line is hidden from the screen — left the board."],[127.03939583333333,"master is hidden from the screen — left the board."],[127.03939583333333,"roster is hidden from the screen — left the board."],[127.03939583333333,"trio is hidden from the screen — left the board."],[127.03939583333333,"segment is hidden from the screen — trio left the board."],[127.03939583333333,"path is hidden from the screen — trio left the board."],[127.03939583333333,"blob is hidden from the screen — trio left the board."],[127.03939583333333,"end_a is hidden from the screen — trio left the board."],[127.03939583333333,"end_b is hidden from the screen — trio left the board."],[127.03939583333333,"interval_name is hidden from the screen — trio left the board."],[127.03939583333333,"path_start is hidden from the screen — trio left the board."],[127.03939583333333,"path_end is hidden from the screen — trio left the board."],[127.03939583333333,"blob_boundary is hidden from the screen — trio left the board."],[127.03939583333333,"blob_edge is hidden from the screen — trio left the board."]]}]},{"title":"The Fundamental Theorem of Calculus","start":128.0810625,"end":270.32129166666664,"objects":{"area":"an AreaUnder [blue] drawn in axes (x_range=(1.0, <VariableNumber b = 4.0>), target='speed')","axes":"an Axes (x_range=(0.0, 5.0), y_range=(0.0, 3.0), aspect=(5.0, 3.0))","b":"a VariableNumber (initial_value=2.0, format_spec='.1f')","boundary_def":"a Panel that says \"The boundary of a region is what you get when you take its edge. It is always one dimension smaller, and it always comes with signs.\"","boundary_law":"a Math [text] that says \"$partial [a, b] = (+ b) + (- a)$\"","change":"a VariableNumber (initial_value=0.75, format_spec='.2f')","curve_b":"a PlotPoint [yellow] labelled \"2.0\" drawn in axes (target='speed', x=<VariableNumber b = 4.0>)","dot_a":"a Point [red] drawn in axes (location=(1.0, 0.0))","dot_b":"a Point [red] drawn in axes (location=(<VariableNumber b = 4.0>, 0.0))","equations":"a Derivation [text] that says \"$integral_a^b f'(x) thin dif x &= f(b) - f(a) \\ integral_1^4 frac(x, 2) thin dif x &= f(4) - f(1)$\"","heading_boundary":"a Heading that says \"The Boundary Of An Interval\"","line":"a Line [yellow] drawn in axes (start=(1.0, 0.0), end=(0.0, 0.0), dashed=True)","line_2":"a Line [yellow] drawn in axes (start=(2.0, 0.0), end=(0.0, 0.0), dashed=True)","moving_edge":"a Line [red] drawn in axes (start=(<VariableNumber b = 4.0>, 0.0), end=(<VariableNumber b = 4.0>, (b / 2.0)), dashed=True)","nl_a":"a Point [red] drawn in number_line (location=(1.0, 0.0))","nl_b":"a Point [red] drawn in number_line (location=(4.0, 0.0))","nl_brace":"a Brace [text] labelled \"[a, b]\" drawn in number_line (x_start=1.0, x_end=4.0)","number_line":"a NumberLine (x_range=(0.0, 5.0), include_numbers=True)","point":"a Point [yellow] drawn in axes (location=(1.0, 0.0))","point_2":"a Point [yellow] drawn in axes (location=(2.0, 0.0))","point_3":"a Point [yellow] drawn in axes (location=(4.0, 0.0))","point_4":"a Point [yellow] drawn in axes (location=(1.0, 0.0))","speed":"a FunctionPlot [blue] labelled \"f'(x)\" drawn in axes (function=<function>, x_range=(0.0, 5.0))","the_question":"a Panel that says \"You know how fast something is changing at every single moment. How much did it change in total?\"","total_readout":"a Point [red] labelled \"f(b) - f(a) = 0.75\" drawn in axes (location=(1.15, 2.5), show_marker=False)","traveller":"a Point [yellow] labelled \"1.0\" drawn in number_line (location=(<VariableNumber traveller_b = 4.0>, 0.0))","traveller_b":"a VariableNumber (initial_value=1.0, format_spec='.1f')","value_difference":"an Arithmetic [text] that says \"$4 0.25 3.75$\" (operator='-', operands=('4', '0.25'), result='3.75')"},"beats":[{"start":128.0810625,"say":"Start in one dimension, where you have known this theorem the longest. Here is the question it actually answers. You know how fast something is changing at every moment: you're driving, and you can watch the speedometer the whole way. How far did you go?","live":[],"does":[[128.0810625,"the_question is shown on the screen, written out."],[144.4215625,"the_question moves to a new place on the board."]]},{"start":145.02156250000002,"say":"The fundamental theorem of calculus says this. If you integrate the derivative of f from a to b, you don't get anything complicated. You get f at b, minus f at a.","live":["the_question"],"does":[[145.02156250000002,"axes is shown on the screen, written out."],[146.9780625,"axes moves to a new place on the board."],[146.9780625,"equations is shown on the screen, written out."],[148.6030625,"speed is shown on the screen, written out."],[149.4740625,"equations (the \"integral_a^b f'(x) thin dif x\" part) is emphasized."],[152.8530625,"equations (the \"f(b) - f(a)\" part) is emphasized."],[152.8530625,"equations (the \"integral_a^b f'(x) thin dif x\" part) is no longer emphasized."],[155.8480625,"equations (the \"f(b) - f(a)\" part) is no longer emphasized."]]},{"start":156.4480625,"say":"Read that out loud in words. Add up all the little changes over the whole interval, and you get the net change from one end to the other. The speedometer tells you the odometer.","live":["the_question","axes","speed"],"does":[[158.9910625,"equations (the \"integral_a^b f'(x) thin dif x\" part) is emphasized."],[162.4850625,"equations (the \"f(b) - f(a)\" part) is emphasized."],[162.4850625,"equations (the \"integral_a^b f'(x) thin dif x\" part) is no longer emphasized."],[164.7030625,"equations (the \"f(b) - f(a)\" part) is no longer emphasized."]]},{"start":167.48556250000001,"say":"Let's make that concrete, because a number you can check beats a formula you can only nod at. Take f of x equal to x squared over four. Its derivative is x over two, which is this blue line. Fix the left end at one, and let the right endpoint b start at two.","live":null,"does":[[179.60706249999998,"speed is indicated — a transient flash."],[181.9290625,"dot_a is shown on the screen, written out."],[181.9290625,"point is shown on the screen, grown."],[181.9290625,"line is shown on the screen, drawn."],[183.0200625,"dot_b is shown on the screen, written out."],[183.4960625,"curve_b is shown on the screen, written out."],[183.9290625,"point is hidden from the screen."],[183.9290625,"line is hidden from the screen."],[184.1340625,"moving_edge is shown on the screen, written out."],[184.1340625,"total_readout is shown on the screen, written out."],[184.1340625,"point_2 is shown on the screen, grown."],[184.1340625,"line_2 is shown on the screen, drawn."]]},{"start":185.5125625,"say":"The left hand side is the shaded area from one to b. Now slide b out to four. The region grows with it, and the running total up in the corner counts all the way to three point seven five.","live":["the_question","axes","speed","dot_a","dot_b","curve_b","moving_edge","total_readout","point_2","line_2"],"does":[[186.1340625,"point_2 is hidden from the screen."],[186.1340625,"line_2 is hidden from the screen."],[187.2770625,"area is shown on the screen, written out."],[189.9710625,"dot_b is redrawn as the numbers it depends on change."],[189.9710625,"curve_b is redrawn as the numbers it depends on change."],[189.9710625,"moving_edge is redrawn as the numbers it depends on change."],[189.9710625,"total_readout is redrawn as the numbers it depends on change."],[189.9710625,"area is redrawn as the numbers it depends on change."],[189.9710625,"b ticks to 4.0."],[189.9710625,"change ticks to 3.75."],[190.9570625,"equations is shown on the screen, written out."]]},{"start":198.5930625,"say":"Now check the other side. f of four is four. f of one is a quarter. Four minus a quarter is three point seven five. Same number, obviously.","live":["the_question","axes","speed","dot_a","dot_b","curve_b","moving_edge","total_readout","area"],"does":[[200.85706249999998,"value_difference is shown on the screen, written out."],[201.2050625,"equations (the \"f(4)\" part) is emphasized."],[203.3530625,"equations (the \"f(1)\" part) is emphasized."],[203.3530625,"equations (the \"f(4)\" part) is no longer emphasized."],[203.9450625,"value_difference is shown on the screen, written out."],[205.28006249999999,"equations (the \"f(1)\" part) is no longer emphasized."],[205.5940625,"value_difference is shown on the screen, drawn."],[206.39406250000002,"value_difference is shown on the screen, drawn."],[206.6390625,"value_difference is shown on the screen, written out."],[209.4370625,"value_difference is indicated — a transient flash."]]},{"start":211.0705625,"say":"But here is the thing I really want you to notice. The right hand side never looks inside the interval at all.","live":null,"does":[[215.23806249999998,"area is hidden from the screen."]]},{"start":217.7485625,"say":"It only asks two questions. What is f at the right end, and what is f at the left end? Everything in the middle cancels.","live":["the_question","axes","speed","dot_a","dot_b","curve_b","moving_edge","total_readout"],"does":[[221.1670625,"point_3 is shown on the screen, grown."],[222.9550625,"point_4 is shown on the screen, grown."],[223.1670625,"point_3 is hidden from the screen."],[224.9550625,"point_4 is hidden from the screen."],[226.0490625,"axes is hidden from the screen — left the board."],[226.0490625,"speed is hidden from the screen — axes left the board."],[226.0490625,"dot_a is hidden from the screen — axes left the board."],[226.0490625,"dot_b is hidden from the screen — axes left the board."],[226.0490625,"curve_b is hidden from the screen — axes left the board."],[226.0490625,"moving_edge is hidden from the screen — axes left the board."],[226.0490625,"total_readout is hidden from the screen — axes left the board."],[226.0490625,"equations is hidden from the screen — left the board."],[226.0490625,"the_question is hidden from the screen — left the board."],[226.0490625,"value_difference is hidden from the screen — left the board."]]},{"start":226.64906249999999,"say":"So here is the piece of language you need. The interval from a to b is the region. Its boundary is just the two endpoints. That word, boundary, is the whole lecture.","live":[],"does":[[226.64906249999999,"heading_boundary is shown on the screen, written out."],[229.8590625,"number_line is shown on the screen, written out."],[231.3100625,"nl_brace is shown on the screen, written out."],[232.4830625,"boundary_def is shown on the screen, written out."],[233.65606249999996,"nl_a is shown on the screen, written out."],[233.65606249999996,"traveller is shown on the screen, written out."],[233.95606249999997,"nl_b is shown on the screen, written out."]]},{"start":238.71356249999997,"say":"The boundary is not the number b minus a. It is a formal sum of two points: plus b at the right hand end, together with negative a at the left hand end. Those signs are exactly why the answer is f of b minus f of a.","live":["boundary_def","number_line","heading_boundary","nl_brace","nl_a","nl_b","traveller"],"does":[[242.46406249999998,"boundary_law is shown on the screen, written out."],[244.30906249999998,"boundary_law (the \"+ b\" part) is emphasized."],[246.74806249999997,"boundary_law (the \"+ b\" part) is no longer emphasized."],[246.74806249999997,"boundary_law (the \"- a\" part) is emphasized."],[249.55706249999997,"boundary_law (the \"- a\" part) is no longer emphasized."]]},{"start":255.01006249999998,"say":"That sign is orientation, the direction in which you travel. Watch the traveller move from the negative endpoint to the positive endpoint. Keep that direction in mind. It comes straight back in Green's theorem.","live":["boundary_def","boundary_law","number_line","heading_boundary","nl_brace","nl_a","nl_b","traveller"],"does":[[261.1980625,"traveller is redrawn as the numbers it depends on change."],[261.1980625,"traveller_b ticks to 4.0."],[269.279625,"boundary_def is hidden from the screen — left the board."],[269.279625,"boundary_law is hidden from the screen — left the board."],[269.279625,"heading_boundary is hidden from the screen — left the board."],[269.279625,"number_line is hidden from the screen — left the board."],[269.279625,"nl_brace is hidden from the screen — number_line left the board."],[269.279625,"nl_a is hidden from the screen — number_line left the board."],[269.279625,"nl_b is hidden from the screen — number_line left the board."],[269.279625,"traveller is hidden from the screen — number_line left the board."]]}]},{"title":"The Same Theorem, On A Curve","start":270.32129166666664,"end":414.7803124999999,"objects":{"axes":"an Axes (x_range=(0.0, 6.0), y_range=(0.0, 4.0), aspect=(6.0, 4.0))","closed_endpoints":"a Math [text] that says \"$f(P) - f(P)$\"","closed_law":"a Math [text] that says \"$integral.cont_Gamma nabla f dot dif arrow(r) = 0$\"","contours":"a LevelCurves [gray] drawn in axes (function=<function>, values=(9.5, 9.0, 8.5, 8.0))","heading":"a Heading that says \"The Same Theorem, On A Curve\"","heading_closed":"a Heading that says \"A Closed Loop\"","kink_x":"a VariableNumber (initial_value=2.6)","kink_y":"a VariableNumber (initial_value=2.7)","leg_in":"a Line [blue] labelled \"C\" drawn in axes (start=(0.6, 0.6), end=(<VariableNumber kink_x = 4.2>, <VariableNumber kink_y = 0.9>))","leg_out":"a Line [blue] drawn in axes (start=(<VariableNumber kink_x = 4.2>, <VariableNumber kink_y = 0.9>), end=(5.4, 3.5))","loop":"a Circle [magenta] labelled \"Gamma\" drawn in axes (center=(3.0, 2.0))","loop_dir":"an Orientation [magenta] drawn in axes (path=((4.0, 2.0), (3.995184726672197, 2.0980171403295604), (3.980785…, closed=True, arrows=4)","p_end":"a Point [red] labelled \"Q\" drawn in axes (location=(5.4, 3.5))","p_start":"a Point [red] labelled \"P\" drawn in axes (location=(0.6, 0.6))","point":"a Point [yellow] drawn in axes (location=(0.6, 0.6))","point_2":"a Point [yellow] drawn in axes (location=(5.4, 3.5))","walk_angle":"a VariableNumber","walker":"a Point [yellow] labelled \"f = 9\" drawn in axes (location=((3.0 + (1.4142135623730951 * cos(walk_angle))), (2.0 + (1.4142…)","waypoint":"a Point [green] drawn in axes (location=(<VariableNumber kink_x = 4.2>, <VariableNumber kink_y = 0.9>))","work":"a Derivation [text] that says \"$integral_a^b f'(x) thin dif x &= f(b) - f(a) \\ integral_C nabla f dot dif arrow(r) &= f(Q) - f(P)$\""},"beats":[{"start":270.32129166666664,"say":"Now go up a dimension. Same sentence, new region. This time the region is not a chunk of the number line. It is a route, drawn through the plane, from a point P to a point Q.","live":[],"does":[[270.32129166666664,"heading is shown on the screen, written out."],[271.02929166666667,"axes is shown on the screen, written out."],[277.7402916666666,"leg_in is shown on the screen, written out."],[277.7402916666666,"leg_out is shown on the screen, written out."],[280.34029166666664,"p_start is shown on the screen, written out."],[281.16529166666663,"p_end is shown on the screen, written out."]]},{"start":282.38029166666666,"say":"And instead of a function of one variable, there are now two. Picture it as a hill: f of x, y is the height of the ground above the point x, y. These grey rings are its contour lines.","live":["axes","heading","leg_in","leg_out","p_start","p_end"],"does":[[293.02729166666666,"contours is shown on the screen, written out."]]},{"start":295.69379166666664,"say":"A contour is the set of places at one particular height. So put a walker on the ring where f is nine, and send it the whole way round. It climbs nothing at all, because every point of that ring is at the same height.","live":["axes","heading","leg_in","leg_out","p_start","p_end","contours"],"does":[[300.2792916666666,"walker is shown on the screen, written out."],[303.6582916666666,"walker is redrawn as the numbers it depends on change."],[303.6582916666666,"walk_angle ticks to 6.283185307179586."],[308.56929166666663,"walker is hidden from the screen."]]},{"start":309.16929166666665,"say":"Here is theorem one again, so you can see what is coming. Integrate the derivative across the region, and read the answer off the two ends.","live":null,"does":[[310.1442916666666,"axes moves to a new place on the board."],[310.1442916666666,"work is shown on the screen, written out."],[313.32529166666666,"work (the \"integral_a^b f'(x) thin dif x\" part) is emphasized."],[315.48429166666665,"work (the \"f(b) - f(a)\" part) is emphasized."],[315.48429166666665,"work (the \"integral_a^b f'(x) thin dif x\" part) is no longer emphasized."],[317.19129166666664,"work (the \"f(b) - f(a)\" part) is no longer emphasized."]]},{"start":317.79129166666667,"say":"And here is theorem two. Integrate the gradient of f along the curve C, and what you get is f at Q minus f at P. Look at those two lines together. It is the same sentence twice.","live":null,"does":[[318.98729166666664,"work is shown on the screen, written out."],[321.51829166666664,"work (the \"integral_C nabla f dot dif arrow(r)\" part) is emphasized."],[322.7022916666666,"work (the \"f(Q) - f(P)\" part) is emphasized."],[322.7022916666666,"work (the \"integral_C nabla f dot dif arrow(r)\" part) is no longer emphasized."],[326.6732916666666,"work (the \"f(Q) - f(P)\" part) is no longer emphasized."]]},{"start":330.40729166666665,"say":"Read the left side as, add up every little bit of climbing as you walk. Read the right side as, the height where you finished minus the height where you started. It is the odometer story again, on a hill.","live":null,"does":[[331.06929166666663,"work (the \"integral_C nabla f dot dif arrow(r)\" part) is emphasized."],[335.4232916666666,"work (the \"f(Q) - f(P)\" part) is emphasized."],[335.4232916666666,"work (the \"integral_C nabla f dot dif arrow(r)\" part) is no longer emphasized."],[342.24929166666664,"work (the \"f(Q) - f(P)\" part) is no longer emphasized."]]},{"start":343.46479166666666,"say":"And that hands you something genuinely useful. The route does not matter. Take hold of it here, in the middle, and bend it wherever you like, keeping P and Q where they are.","live":null,"does":[[348.7592916666666,"waypoint is shown on the screen, written out."]]},{"start":354.61829166666666,"say":"Up over the shoulder of the hill. Or down round the bottom instead. The climb you total up is the same number every single time, because the right hand side never mentions the route at all.","live":["axes","heading","leg_in","leg_out","p_start","p_end","contours","waypoint"],"does":[[354.87329166666666,"leg_in is redrawn as the numbers it depends on change."],[354.87329166666666,"leg_out is redrawn as the numbers it depends on change."],[354.87329166666666,"waypoint is redrawn as the numbers it depends on change."],[354.87329166666666,"kink_x ticks to 2.0."],[354.87329166666666,"kink_y ticks to 3.4."],[357.0212916666666,"leg_in is redrawn as the numbers it depends on change."],[357.0212916666666,"leg_out is redrawn as the numbers it depends on change."],[357.0212916666666,"waypoint is redrawn as the numbers it depends on change."],[357.0212916666666,"kink_x ticks to 4.2."],[357.0212916666666,"kink_y ticks to 0.9."],[360.66729166666664,"work (the \"f(Q) - f(P)\" part) is emphasized."],[365.4037916666666,"work (the \"f(Q) - f(P)\" part) is no longer emphasized."]]},{"start":366.00379166666664,"say":"Because the formula only ever looks at the endpoints. There is that word again. The boundary of a curve is its two ends, and the answer depends on nothing else.","live":null,"does":[[372.96929166666666,"point is shown on the screen, grown."],[374.96929166666666,"point is hidden from the screen."],[375.2692916666666,"point_2 is shown on the screen, grown."],[375.79629166666666,"leg_in is hidden from the screen."],[375.79629166666666,"leg_out is hidden from the screen."],[375.79629166666666,"waypoint is hidden from the screen."],[375.79629166666666,"p_start is hidden from the screen."],[375.79629166666666,"p_end is hidden from the screen."],[375.79629166666666,"contours is hidden from the screen."],[375.79629166666666,"heading is hidden from the screen — left the board."],[375.79629166666666,"work is hidden from the screen — left the board."]]},{"start":376.3962916666666,"say":"Which hands you a corollary that falls straight out. Suppose the route is a closed loop, like this circle Gamma. You finish exactly where you started, so the start point and the end point are the same point.","live":["axes","point_2"],"does":[[376.3962916666666,"heading_closed is shown on the screen, written out."],[377.2692916666666,"point_2 is hidden from the screen."],[382.06829166666665,"loop is shown on the screen, drawn."],[385.1102916666666,"loop_dir is shown on the screen, written out."]]},{"start":389.4132916666666,"say":"So the endpoint term is f of P minus f of P. The two copies cancel, and the result is zero.","live":["axes","heading_closed","loop","loop_dir"],"does":[[390.55129166666666,"closed_endpoints is shown on the screen, written out."],[394.78929166666666,"closed_endpoints (the \"f(P)\" part) is slashed through — it cancels."],[394.78929166666666,"closed_endpoints (the \"f(P)#2\" part) is slashed through — it cancels."],[396.18229166666663,"closed_law is shown on the screen, written out."]]},{"start":397.6297916666666,"say":"Around any closed loop, a gradient field integrates to zero. You can see it coming without computing anything: closed curve, gradient field, nothing to do.","live":["axes","closed_endpoints","closed_law","heading_closed","loop","loop_dir"],"does":[[400.7642916666666,"A box is drawn around closed_law."]]},{"start":407.71479166666666,"say":"Same theorem. Region, boundary, done. Now the one that looks hardest and really isn't.","live":null,"does":[[409.86229166666664,"closed_law is indicated — a transient flash."],[413.7386458333333,"axes is hidden from the screen — left the board."],[413.7386458333333,"loop is hidden from the screen — axes left the board."],[413.7386458333333,"loop_dir is hidden from the screen — axes left the board."],[413.7386458333333,"closed_endpoints is hidden from the screen — left the board."],[413.7386458333333,"closed_law is hidden from the screen — left the board."],[413.7386458333333,"heading_closed is hidden from the screen — left the board."]]}]},{"title":"Green's Theorem","start":414.7803124999999,"end":584.8123958333333,"objects":{"axes":"an Axes (x_range=(0.0, 5.0), y_range=(0.0, 4.0), aspect=(5.0, 4.0))","cell":"a TileCell [yellow] drawn in axes (v=1, outline=((0.6, 1.5333333333333332), (0.825, 1.5333333333333332), (1.05,…, sample=(1.05, 1.9999999999999996))","edge_from_left":"a TileEdge [red] drawn in axes (edges=('v:1,1',), paths=(((1.5, 1.5333333333333332), (1.5, 1.7666666666666666), (1.5, 1…, induced='from_lower')","edge_from_right":"a TileEdge [green] drawn in axes (edges=('v:1,1',), paths=(((1.5, 2.4666666666666663), (1.5, 2.233333333333333), (1.5, 1.…, induced='from_higher')","field":"a VectorField [green] drawn in axes (function=<function>, at=((1.0, 1.0), (1.0, 2.0), (1.0, 3.0), (2.0, 1.0), (2.0, 2.0), (2…)","green_law":"a Math [text] that says \"$integral.double_R (frac(partial Q, partial x) - frac(partial P, partial y)) thin dif A = integral.cont_(partial R) P thin dif x + Q thin dif y$\"","heading":"a Heading that says \"Green's Theorem\"","heading_cancel":"a Heading that says \"Why The Interior Disappears\"","patch":"a Region [blue] labelled \"R\" drawn in axes (predicates=(<function <lambda> at 0x2b4ce85f6840>,), x_range=(0.6, 4.2), y_range=(0.6, 3.4))","rim":"a ParametricCurve [red] labelled \"partial R\" drawn in axes (function=<function>, breakpoints=(0.2857142857142857, 0.5, 0.7857142857142857))","rim_dir":"an Orientation [red] drawn in axes (path=((0.6, 0.6, 0.0), (0.796875, 0.6, 0.0), (0.99375, 0.6, 0.0), (1…, closed=True, arrows=6)","rim_t":"a VariableNumber (format_spec='.1f')","rim_traveller":"a Point [yellow] labelled \"0.0\" drawn in axes (location=(0.6, 0.6))","shared_edge":"a TileEdge [yellow] drawn in axes (edges=('v:1,1',), paths=(((1.5, 1.5333333333333332), (1.5, 1.7666666666666666), (1.5, 1…, of='tiling')","surviving":"a ParametricCurve [red] labelled \"partial R\" drawn in axes (function=<function>, breakpoints=(0.2857142857142857, 0.5, 0.7857142857142857))","surviving_dir":"an Orientation [red] drawn in axes (path=((0.6, 0.6, 0.0), (0.796875, 0.6, 0.0), (0.99375, 0.6, 0.0), (1…, closed=True, arrows=6)","tiling":"a Tiles [yellow] drawn in axes (v_cells=3, nodes=(((0.6, 0.6), (0.6, 0.8333333333333333), (0.6, 1.06666666666666…, of='patch')"},"beats":[{"start":414.7803124999999,"say":"So, the flat case. Green's theorem. This is the one that looks frightening the first time you meet it, and it is the same sentence again.","live":[],"does":[[414.7803124999999,"heading is shown on the screen, written out."],[417.1023124999999,"axes is shown on the screen, written out."]]},{"start":424.63331249999993,"say":"Our region is now genuinely two dimensional: a flat patch R in the plane, carrying a field of arrows. The rectangle keeps the boundary easy to read, but the theorem does not depend on that shape.","live":["axes","heading"],"does":[[428.3483124999999,"patch is shown on the screen, written out."],[430.39231249999995,"field is shown on the screen, written out."]]},{"start":437.4003124999999,"say":"Its boundary is the closed curve that runs all the way around the outside. We write it as del R. Orientation matters, exactly as the plus and minus mattered on the interval.","live":["axes","heading","patch","field"],"does":[[437.95731249999994,"rim is shown on the screen, drawn."],[444.02931249999995,"rim_dir is shown on the screen, written out."]]},{"start":449.18631249999993,"say":"Travel counterclockwise, so the region stays on your left. Starting here, one complete trip comes back to the point it left. That choice of direction is the same choice as the plus sign on the interval's right hand end.","live":["axes","heading","patch","field","rim","rim_dir"],"does":[[453.8363124999999,"rim_traveller is shown on the screen, written out."],[454.9273124999999,"rim_t ticks to 1.0."],[455.3363124999999,"rim_traveller is redrawn as the numbers it depends on change."]]},{"start":463.1198124999999,"say":"Here is the statement. The double integral over every point of the region lives on the left. One line integral around its boundary lives on the right. Inside on the left, edge on the right.","live":["axes","heading","patch","field","rim","rim_dir","rim_traveller"],"does":[[463.9443124999999,"axes moves to a new place on the board."],[463.9443124999999,"green_law is shown on the screen, written out."],[468.3093124999999,"green_law (the \"integral.double_R (frac(partial Q, partial x) - frac(partial P, partial y)) thin dif A\" part) is emphasized."],[471.89731249999994,"green_law (the \"integral.cont_(partial R) P thin dif x + Q thin dif y\" part) is emphasized."],[471.89731249999994,"green_law (the \"integral.double_R (frac(partial Q, partial x) - frac(partial P, partial y)) thin dif A\" part) is no longer emphasized."],[473.0233124999999,"green_law (the \"integral.cont_(partial R) P thin dif x + Q thin dif y\" part) is no longer emphasized."]]},{"start":476.28231249999993,"say":"Read the left side as the total swirl at all points inside R. Read the right side as the total push along one counterclockwise trip around del R. Every arrow inside contributes to the first count; the rim supplies the second.","live":["green_law","axes","heading","patch","field","rim","rim_dir","rim_traveller"],"does":[[477.1533124999999,"green_law (the \"integral.double_R (frac(partial Q, partial x) - frac(partial P, partial y)) thin dif A\" part) is emphasized."],[481.7733124999999,"green_law (the \"integral.cont_(partial R) P thin dif x + Q thin dif y\" part) is emphasized."],[481.7733124999999,"green_law (the \"integral.double_R (frac(partial Q, partial x) - frac(partial P, partial y)) thin dif A\" part) is no longer emphasized."],[487.1263124999999,"green_law (the \"integral.cont_(partial R) P thin dif x + Q thin dif y\" part) is no longer emphasized."]]},{"start":492.6948124999999,"say":"Green's theorem says those two numbers are equal. Total swirl inside equals total push around the rim. That is the whole theorem.","live":null,"does":[[495.06331249999994,"green_law is indicated — a transient flash."],[502.48181249999993,"axes moves to a new place on the board."],[502.48181249999993,"green_law is hidden from the screen — left the board."],[502.48181249999993,"heading is hidden from the screen — left the board."]]},{"start":503.08181249999996,"say":"Now for the reason. Chop R into little tiles. Each tile contributes a walk around its own boundary.","live":["axes","patch","field","rim","rim_dir","rim_traveller"],"does":[[503.08181249999996,"heading_cancel is shown on the screen, written out."],[505.2993124999999,"rim is hidden from the screen."],[505.2993124999999,"rim_dir is hidden from the screen."],[505.2993124999999,"rim_traveller is hidden from the screen."],[505.2993124999999,"field is hidden from the screen."],[505.2993124999999,"patch is hidden from the screen."],[506.5183124999999,"tiling is shown on the screen, written out."]]},{"start":511.7278124999999,"say":"Choose this tile. Its contribution includes this little edge. Adding the tiles means reading that edge once from each tile that touches it.","live":["axes","heading_cancel","tiling"],"does":[[512.7153125,"cell is shown on the screen, written out."],[515.4893124999999,"shared_edge is shown on the screen, written out."],[516.5813125,"cell is hidden from the screen."]]},{"start":521.8133124999999,"say":"One tile walks the shared edge upward. Its neighbour walks the very same edge downward. The geometric edge is one object, but its two induced directions are opposites.","live":["axes","heading_cancel","tiling","shared_edge"],"does":[[522.1613124999999,"edge_from_left is shown on the screen, written out."],[525.2963124999999,"edge_from_right is shown on the screen, written out."]]},{"start":533.8143124999999,"say":"Those two contributions cancel. The same happens at every shared edge, so the entire interior grid deletes itself.","live":["axes","heading_cancel","tiling","shared_edge","edge_from_left","edge_from_right"],"does":[[535.4163124999999,"edge_from_left is hidden from the screen."],[535.4163124999999,"edge_from_right is hidden from the screen."],[538.1793124999999,"shared_edge is hidden from the screen."],[540.4783124999999,"tiling is hidden from the screen."]]},{"start":542.8428124999999,"say":"Only edges with no neighbour remain: the outside of R. Their induced walk is the surviving boundary, with the region on its left. That is Green's theorem.","live":["axes","heading_cancel"],"does":[[546.0933124999999,"surviving is shown on the screen, drawn."],[549.5653124999999,"surviving_dir is shown on the screen, written out."],[554.6268124999999,"axes moves to a new place on the board."],[554.6268124999999,"heading_cancel is hidden from the screen — left the board."]]},{"start":555.2268124999999,"say":"The equality now reads from the region to its surviving boundary. The derivative is integrated over R on the left; the original field is integrated over del R on the right.","live":["axes","surviving","surviving_dir"],"does":[[555.2268124999999,"heading is shown on the screen, written out."],[555.8543124999999,"green_law is shown on the screen, written out."],[557.5023124999999,"patch is shown on the screen, written out."],[562.0993125,"green_law (the \"integral.double_R (frac(partial Q, partial x) - frac(partial P, partial y)) thin dif A\" part) is emphasized."],[563.5393124999999,"field is shown on the screen, written out."],[565.6643124999999,"green_law (the \"integral.cont_(partial R) P thin dif x + Q thin dif y\" part) is emphasized."],[565.6643124999999,"green_law (the \"integral.double_R (frac(partial Q, partial x) - frac(partial P, partial y)) thin dif A\" part) is no longer emphasized."],[566.4188125,"green_law (the \"integral.cont_(partial R) P thin dif x + Q thin dif y\" part) is no longer emphasized."]]},{"start":567.0188125,"say":"If a line integral around a closed curve is horrible, read the equality backward. Swap the loop integral to the left and the double integral to the right. Often that reversal is the whole difference between an integral you can do and one you cannot.","live":["green_law","axes","heading","patch","field","surviving","surviving_dir"],"does":[[573.2653124999999,"green_law becomes \"$integral.cont_(partial R) P thin dif x + Q thin dif y = integral.double_R (frac(partial Q, partial x) - frac(partial P, partial y)) thin dif A$\"."],[574.8673124999999,"green_law (the \"integral.cont_(partial R) P thin dif x + Q thin dif y\" part) is emphasized."],[576.7833125,"green_law (the \"integral.cont_(partial R) P thin dif x + Q thin dif y\" part) is no longer emphasized."],[576.7833125,"green_law (the \"integral.double_R (frac(partial Q, partial x) - frac(partial P, partial y)) thin dif A\" part) is emphasized."],[577.8853124999999,"green_law (the \"integral.double_R (frac(partial Q, partial x) - frac(partial P, partial y)) thin dif A\" part) is no longer emphasized."],[583.7707291666666,"axes is hidden from the screen — left the board."],[583.7707291666666,"patch is hidden from the screen — axes left the board."],[583.7707291666666,"field is hidden from the screen — axes left the board."],[583.7707291666666,"surviving is hidden from the screen — axes left the board."],[583.7707291666666,"surviving_dir is hidden from the screen — axes left the board."],[583.7707291666666,"green_law is hidden from the screen — left the board."],[583.7707291666666,"heading is hidden from the screen — left the board."]]}]},{"title":"The Big Picture","start":584.8123958333333,"end":746.22325,"objects":{"blob":"a Polygon [blue] labelled \"R\" drawn in trio (vertices=((6.0, 0.8), (8.2, 0.9), (8.5, 2.1), (7.3, 2.8), (5.8, 2.1)), fill_opacity=0.22)","blob_boundary":"a ParametricCurve [red] drawn in trio (function=<function>, breakpoints=(0.2, 0.4, 0.6, 0.8))","blob_direction":"an Orientation [red] drawn in trio (path=((6.0, 0.8, 0.0), (6.171875, 0.8078125, 0.0), (6.34375, 0.81562…, closed=True, arrows=4)","end_a":"a Point [red] labelled \"a\" drawn in trio (location=(0.5, 1.7))","end_b":"a Point [red] labelled \"b\" drawn in trio (location=(2.2, 1.7))","heading_master":"a Heading that says \"One Line\"","heading_next":"a Heading that says \"One More Dimension\"","heading_table":"a Heading that says \"All Three At Once\"","ladder":"a Table [text] that says \"Theorem Region Boundary Calculus $[a, b]$ $a$ and $b$ Line integrals $C$ $P$ and $Q$ Green $R$ $partial R$\" (rows=(('Theorem', 'Region', 'Boundary'), ('Calculus', '$[a, b]$', '$…, header=True)","master":"a Math [text] that says \"$integral_Omega dif omega = integral_(partial Omega) omega$\"","path":"a ParametricCurve [blue] labelled \"C\" drawn in trio (function=<function>)","path_end":"a Point [red] labelled \"Q\" drawn in trio (location=(4.9, 2.1))","path_start":"a Point [red] labelled \"P\" drawn in trio (location=(3.0, 1.0))","segment":"a Line [blue] drawn in trio (start=(0.5, 1.7), end=(2.2, 1.7))","sheet":"a Surface [blue] drawn in space (function=<function>, u_range=(-1.3, 1.3), v_range=(-1.3, 1.3))","sheet_rim":"a ParametricCurve [red] labelled \"partial S\" drawn in space (function=<function>, breakpoints=(0.25, 0.5, 0.75))","sheet_rim_marks":"an Orientation [red] drawn in space (path=((-1.3, -1.3, 0.35499999999999987), (-1.1375, -1.3, 0.454023437…, closed=True, arrows=4)","space":"an Axes3D (x_range=(-1.8, 1.8), y_range=(-1.8, 1.8), z_range=(0.0, 2.2))","trio":"a Figure (x_range=(0.0, 9.0), y_range=(0.0, 3.4), aspect=(9.0, 3.4))"},"beats":[{"start":584.8123958333333,"say":"Put the three theorems side by side with the three regions we started from. For each theorem, name its region and then its complete boundary.","live":[],"does":[[584.8123958333333,"heading_table is shown on the screen, written out."],[584.8123958333333,"trio is shown on the screen, written out."],[590.6173958333333,"trio moves to a new place on the board."],[590.6173958333333,"ladder is shown on the screen, written out."]]},{"start":594.0153958333333,"say":"For the fundamental theorem of calculus, the region is the interval from a to b. Its boundary is the two endpoints a and b.","live":["trio","heading_table"],"does":[[595.6063958333333,"ladder is shown on the screen, written out."],[597.3013958333333,"segment is shown on the screen, written out."],[600.4943958333333,"end_a is shown on the screen, written out."],[600.4943958333333,"end_b is shown on the screen, written out."]]},{"start":603.0328958333333,"say":"For line integrals, the region is the curve C. Its boundary is the point P where the curve starts and the point Q where it ends.","live":["trio","heading_table","segment","end_a","end_b"],"does":[[603.5903958333333,"ladder is shown on the screen, written out."],[605.1223958333333,"path is shown on the screen, drawn."],[607.3753958333333,"path_start is shown on the screen, written out."],[609.6043958333333,"path_end is shown on the screen, written out."]]},{"start":611.6323958333334,"say":"For Green's theorem, the region is the patch R. Its boundary is the closed loop del R, walked with the region on the left.","live":["trio","heading_table","segment","end_a","end_b","path","path_start","path_end"],"does":[[612.1543958333333,"ladder is shown on the screen, written out."],[614.0003958333333,"blob is shown on the screen, written out."],[616.5783958333333,"blob_boundary is shown on the screen, drawn."],[618.0643958333333,"blob_direction is shown on the screen, written out."]]},{"start":620.6493958333333,"say":"Now read the two right hand columns. This one names the region being integrated over. This one names its boundary, always one dimension smaller. Every row is saying the same thing.","live":["trio","heading_table","segment","end_a","end_b","path","path_start","path_end","blob","blob_boundary","blob_direction"],"does":[[624.5623958333333,"ladder (the \"column=2\" part) is emphasized."],[627.8133958333333,"ladder (the \"column=2\" part) is no longer emphasized."],[627.8133958333333,"ladder (the \"column=3\" part) is emphasized."],[631.0753958333333,"ladder (the \"column=3\" part) is no longer emphasized."],[633.1538958333333,"trio moves to a new place on the board."],[633.1538958333333,"heading_table is hidden from the screen — left the board."],[633.1538958333333,"ladder is hidden from the screen — left the board."]]},{"start":633.7538958333333,"say":"And that shared structure is the line I showed you at the start. Here it is again, and this time it should read as ordinary English.","live":["trio","segment","end_a","end_b","path","path_start","path_end","blob","blob_boundary","blob_direction"],"does":[[633.7538958333333,"heading_master is shown on the screen, written out."],[635.5653958333334,"master is shown on the screen, written out."]]},{"start":642.5508958333334,"say":"Omega names the region, whatever dimension it has. Del Omega names its boundary, always one dimension smaller.","live":["trio","segment","end_a","end_b","path","path_start","path_end","blob","blob_boundary","blob_direction","master","heading_master"],"does":[[642.8993958333333,"master (the \"Omega\" part) is emphasized."],[646.4633958333333,"master (the \"Omega\" part) is no longer emphasized."],[646.4633958333333,"master (the \"partial Omega\" part) is emphasized."],[649.6553958333333,"master (the \"partial Omega\" part) is no longer emphasized."]]},{"start":651.1493958333333,"say":"The symbol d means take the appropriate derivative: an ordinary derivative on an interval, a gradient along a curve, or curl over a patch. It is the operation that turns the original quantity into the inside quantity.","live":null,"does":[[653.5413958333334,"master (the \"dif omega\" part) is emphasized."],[661.1223958333333,"master (the \"dif omega\" part) is no longer emphasized."]]},{"start":666.0058958333333,"say":"The left side integrates that derivative over the region. The right side integrates the original quantity over the boundary. Read at three different dimensions, this one line is all three theorems.","live":null,"does":[[666.5513958333333,"master (the \"integral_Omega dif omega\" part) is emphasized."],[669.9073958333333,"master (the \"integral_(partial Omega) omega\" part) is emphasized."],[669.9073958333333,"master (the \"integral_Omega dif omega\" part) is no longer emphasized."],[673.5753958333333,"master (the \"integral_(partial Omega) omega\" part) is no longer emphasized."],[677.8253958333333,"heading_master is hidden from the screen — left the board."],[677.8253958333333,"trio is hidden from the screen — left the board."],[677.8253958333333,"segment is hidden from the screen — trio left the board."],[677.8253958333333,"end_a is hidden from the screen — trio left the board."],[677.8253958333333,"end_b is hidden from the screen — trio left the board."],[677.8253958333333,"path is hidden from the screen — trio left the board."],[677.8253958333333,"path_start is hidden from the screen — trio left the board."],[677.8253958333333,"path_end is hidden from the screen — trio left the board."],[677.8253958333333,"blob is hidden from the screen — trio left the board."],[677.8253958333333,"blob_boundary is hidden from the screen — trio left the board."],[677.8253958333333,"blob_direction is hidden from the screen — trio left the board."]]},{"start":679.0253958333333,"say":"And notice what that line does not say. It never says how many dimensions the region has. So take one more: a surface in space, a sheet of it hanging in the air.","live":["master"],"does":[[679.0253958333333,"heading_next is shown on the screen, written out."],[679.0253958333333,"space is shown on the screen, written out."],[688.7773958333333,"sheet is shown on the screen, written out."]]},{"start":691.2933958333333,"say":"Let me turn it round, because a curved sheet and a flat one look exactly alike until something moves. Now you can see it is bowed upward, like a cloth held up by its four corners.","live":["master","space","heading_next","sheet"],"does":[[692.1293958333333,"space turns in its own slot."]]},{"start":702.8188958333333,"say":"Its boundary is the closed curve running round the rim, walked so that the surface stays on your left. A region, and an edge one dimension smaller. The same two things as every other picture today.","live":null,"does":[[705.8023958333333,"sheet_rim is shown on the screen, drawn."],[706.4643958333334,"sheet_rim_marks is shown on the screen, written out."],[709.5993958333333,"master (the \"Omega\" part) is emphasized."],[710.6443958333333,"master (the \"Omega\" part) is no longer emphasized."],[710.6443958333333,"master (the \"partial Omega\" part) is emphasized."],[712.9543958333333,"master (the \"partial Omega\" part) is no longer emphasized."]]},{"start":716.2478958333334,"say":"Read that same line over a surface like this one, and what you get is the theorem that actually carries Stokes' name. Read it one dimension higher again and it is the divergence theorem. Nobody had to invent a new idea for either of them.","live":["master","space","heading_next","sheet","sheet_rim","sheet_rim_marks"],"does":[[717.0603958333334,"master is emphasized."],[727.5563958333332,"master is no longer emphasized."]]},{"start":731.0358958333334,"say":"So there it is. That line is Stokes' theorem. Omega is whatever region you have, del Omega is its edge, and adding up the change inside always leaves you something that was settled on that edge.","live":null,"does":[[734.4143958333333,"A box is drawn around master."],[735.4243958333333,"master (the \"Omega\" part) is emphasized."],[738.1873958333333,"master (the \"Omega\" part) is no longer emphasized."],[738.1873958333333,"master (the \"partial Omega\" part) is emphasized."],[740.7073958333333,"master (the \"partial Omega\" part) is no longer emphasized."],[745.1815833333333,"heading_next is hidden from the screen — left the board."],[745.1815833333333,"master is hidden from the screen — left the board."],[745.1815833333333,"space is hidden from the screen — left the board."],[745.1815833333333,"sheet is hidden from the screen — space left the board."],[745.1815833333333,"sheet_rim is hidden from the screen — space left the board."],[745.1815833333333,"sheet_rim_marks is hidden from the screen — space left the board."]]}]}]},"durationSeconds":746,"chapters":[{"title":"One Line, Three Theorems","startSeconds":0,"narration":"Let me tell you what Stokes' theorem actually says. It is one sentence. Not a page of conditions, not a machine you turn a handle on. One sentence, which then gets written down three times, in three different numbers of dimensions, and picks up a different name each time. Before the sentence, put its three regions side by side. A chunk of the number line, a path through the plane, and a two dimensional patch. These regions have different dimensions, but each has an edge one dimension smaller. Start with the chunk of line. Its edge is two dots, one at each end. Those two endpoints are the complete boundary of the interval from a to b. The path has the same kind of boundary. Its edge is the point P where the journey starts, together with the point Q where it stops. The patch is different. Its edge is not a pair of points. It is the closed curve running all the way around the outside, walked in one direction. Now look at the pattern. The edge of a piece of line is a couple of points. The edge of a path is its two endpoints. The edge of a blob is a curve. In every case the edge is one dimension smaller, and here is the sentence all three theorems share: add up how fast something changes inside a region, and the answer depends only on its edge. And here is that sentence written in symbols. This is Stokes' theorem. The whole of it, on one line. You are not supposed to be able to read that yet. I only want you to have seen it once. Omega is the region, whatever it happens to be. Del Omega is its edge. And the line says: what you get by integrating inside is decided on the edge. And you have almost certainly met that line three times already, under three different names. The fundamental theorem of calculus. The fundamental theorem for line integrals. And Green's theorem. So let's take them one at a time. One dimension, then a curve through the plane, then a flat patch, and watch the same line come back every time."},{"title":"The Fundamental Theorem of Calculus","startSeconds":128.0810625,"narration":"Start in one dimension, where you have known this theorem the longest. Here is the question it actually answers. You know how fast something is changing at every moment: you're driving, and you can watch the speedometer the whole way. How far did you go? The fundamental theorem of calculus says this. If you integrate the derivative of f from a to b, you don't get anything complicated. You get f at b, minus f at a. Read that out loud in words. Add up all the little changes over the whole interval, and you get the net change from one end to the other. The speedometer tells you the odometer. Let's make that concrete, because a number you can check beats a formula you can only nod at. Take f of x equal to x squared over four. Its derivative is x over two, which is this blue line. Fix the left end at one, and let the right endpoint b start at two. The left hand side is the shaded area from one to b. Now slide b out to four. The region grows with it, and the running total up in the corner counts all the way to three point seven five. Now check the other side. f of four is four. f of one is a quarter. Four minus a quarter is three point seven five. Same number, obviously. But here is the thing I really want you to notice. The right hand side never looks inside the interval at all. It only asks two questions. What is f at the right end, and what is f at the left end? Everything in the middle cancels. So here is the piece of language you need. The interval from a to b is the region. Its boundary is just the two endpoints. That word, boundary, is the whole lecture. The boundary is not the number b minus a. It is a formal sum of two points: plus b at the right hand end, together with negative a at the left hand end. Those signs are exactly why the answer is f of b minus f of a. That sign is orientation, the direction in which you travel. Watch the traveller move from the negative endpoint to the positive endpoint. Keep that direction in mind. It comes straight back in Green's theorem."},{"title":"The Same Theorem, On A Curve","startSeconds":270.32129166666664,"narration":"Now go up a dimension. Same sentence, new region. This time the region is not a chunk of the number line. It is a route, drawn through the plane, from a point P to a point Q. And instead of a function of one variable, there are now two. Picture it as a hill: f of x, y is the height of the ground above the point x, y. These grey rings are its contour lines. A contour is the set of places at one particular height. So put a walker on the ring where f is nine, and send it the whole way round. It climbs nothing at all, because every point of that ring is at the same height. Here is theorem one again, so you can see what is coming. Integrate the derivative across the region, and read the answer off the two ends. And here is theorem two. Integrate the gradient of f along the curve C, and what you get is f at Q minus f at P. Look at those two lines together. It is the same sentence twice. Read the left side as, add up every little bit of climbing as you walk. Read the right side as, the height where you finished minus the height where you started. It is the odometer story again, on a hill. And that hands you something genuinely useful. The route does not matter. Take hold of it here, in the middle, and bend it wherever you like, keeping P and Q where they are. Up over the shoulder of the hill. Or down round the bottom instead. The climb you total up is the same number every single time, because the right hand side never mentions the route at all. Because the formula only ever looks at the endpoints. There is that word again. The boundary of a curve is its two ends, and the answer depends on nothing else. Which hands you a corollary that falls straight out. Suppose the route is a closed loop, like this circle Gamma. You finish exactly where you started, so the start point and the end point are the same point. So the endpoint term is f of P minus f of P. The two copies cancel, and the result is zero. Around any closed loop, a gradient field integrates to zero. You can see it coming without computing anything: closed curve, gradient field, nothing to do. Same theorem. Region, boundary, done. Now the one that looks hardest and really isn't."},{"title":"Green's Theorem","startSeconds":414.7803124999999,"narration":"So, the flat case. Green's theorem. This is the one that looks frightening the first time you meet it, and it is the same sentence again. Our region is now genuinely two dimensional: a flat patch R in the plane, carrying a field of arrows. The rectangle keeps the boundary easy to read, but the theorem does not depend on that shape. Its boundary is the closed curve that runs all the way around the outside. We write it as del R. Orientation matters, exactly as the plus and minus mattered on the interval. Travel counterclockwise, so the region stays on your left. Starting here, one complete trip comes back to the point it left. That choice of direction is the same choice as the plus sign on the interval's right hand end. Here is the statement. The double integral over every point of the region lives on the left. One line integral around its boundary lives on the right. Inside on the left, edge on the right. Read the left side as the total swirl at all points inside R. Read the right side as the total push along one counterclockwise trip around del R. Every arrow inside contributes to the first count; the rim supplies the second. Green's theorem says those two numbers are equal. Total swirl inside equals total push around the rim. That is the whole theorem. Now for the reason. Chop R into little tiles. Each tile contributes a walk around its own boundary. Choose this tile. Its contribution includes this little edge. Adding the tiles means reading that edge once from each tile that touches it. One tile walks the shared edge upward. Its neighbour walks the very same edge downward. The geometric edge is one object, but its two induced directions are opposites. Those two contributions cancel. The same happens at every shared edge, so the entire interior grid deletes itself. Only edges with no neighbour remain: the outside of R. Their induced walk is the surviving boundary, with the region on its left. That is Green's theorem. The equality now reads from the region to its surviving boundary. The derivative is integrated over R on the left; the original field is integrated over del R on the right. If a line integral around a closed curve is horrible, read the equality backward. Swap the loop integral to the left and the double integral to the right. Often that reversal is the whole difference between an integral you can do and one you cannot."},{"title":"The Big Picture","startSeconds":584.8123958333333,"narration":"Put the three theorems side by side with the three regions we started from. For each theorem, name its region and then its complete boundary. For the fundamental theorem of calculus, the region is the interval from a to b. Its boundary is the two endpoints a and b. For line integrals, the region is the curve C. Its boundary is the point P where the curve starts and the point Q where it ends. For Green's theorem, the region is the patch R. Its boundary is the closed loop del R, walked with the region on the left. Now read the two right hand columns. This one names the region being integrated over. This one names its boundary, always one dimension smaller. Every row is saying the same thing. And that shared structure is the line I showed you at the start. Here it is again, and this time it should read as ordinary English. Omega names the region, whatever dimension it has. Del Omega names its boundary, always one dimension smaller. The symbol d means take the appropriate derivative: an ordinary derivative on an interval, a gradient along a curve, or curl over a patch. It is the operation that turns the original quantity into the inside quantity. The left side integrates that derivative over the region. The right side integrates the original quantity over the boundary. Read at three different dimensions, this one line is all three theorems. And notice what that line does not say. It never says how many dimensions the region has. So take one more: a surface in space, a sheet of it hanging in the air. Let me turn it round, because a curved sheet and a flat one look exactly alike until something moves. Now you can see it is bowed upward, like a cloth held up by its four corners. Its boundary is the closed curve running round the rim, walked so that the surface stays on your left. A region, and an edge one dimension smaller. The same two things as every other picture today. Read that same line over a surface like this one, and what you get is the theorem that actually carries Stokes' name. Read it one dimension higher again and it is the divergence theorem. Nobody had to invent a new idea for either of them. So there it is. That line is Stokes' theorem. Omega is whatever region you have, del Omega is its edge, and adding up the change inside always leaves you something that was settled on that edge."}]}}