Green's theorem proof (part 1)
Part 1 of the proof of Green's Theorem. Created by Sal Khan.
Want to join the conversation?
- why does sal use the integral with the circle in it sometimes, and the normal one with a sub c other times?(8 votes)
- when it has th circle it means we want to integrate around a CLOSED loop :)
other than that , if there is no circle there will be another path to be followed which is not a closed path .(44 votes)
- î·î=1 ?
Is this expression true?
That is, when one has the x-unit-vector and the dot-product with another x-unit-vector (as at2:50), does that always equal one? Because it always seems to disappear when this happens.(6 votes)
- yes it is true.
the dot product is evaluated as the product of the magnitude of each vector times the cosine of the angle between them.(23 votes)
- So at around7:45Sal decides to swap the inside of the integral by multiplying with negative one. This isn't really an obvious thing to do but it is essential to get the right result for Green's Theorem. Does anybody know why he took that step?(7 votes)
- I think you need to do this because of the direction of the curve. The general form given in both these proof videos, that Green's theorem is dQ/dX- dP/dY assumes that your are moving in a counter-clockwise direction. If you were to reverse the direction and go clockwise, you would switch the formula so that it would be dP/dY- dQ/dX.
It might help to think about it like this, let's say you are looking at the unit circle, but only look at the part in the first quadrant. If you move counter-clockwise, you are going from larger to smaller x values, so when you integrate, it is going to be the smaller x minus the larger x (it's the inverse of how we generally view integrals, going from smaller to larger values). The y values on the other hand, are going from smaller values to larger. If you integrate all the same way, you have to subtract how you integrated along the x axis because it is going in a decreasing direction.
Of course, you could argue that this argument really only holds if you look at the first quadrant, but nevertheless I think it is a somewhat useful way to visualize it.
In short, he did this because of the direction of the curve.(7 votes)
- At around12:30, you start talking about how the integral we've taken is the volume under the surface. But that's only true if P is a scalar being treated as a "z" component. Didn't we say earlier that P is a vector with a component only in the X direction? How is it going up?(7 votes)
- he did say that p (lowercase) was a vector field with a component only in the x direction. However, he also said that the x component of this vector field p (lowercase) was a scalar function P(x,y) (capital). When he turned it into a double integral, the partial with respect to y of P (capital) became the inside piece, and one way to visualize double integrals is volume. p is a vector field. P is a scalar field. Neither of them really go up, but that is a way to visualize integrals of scalar functions, whether they go up or not. You can think of a scalar field in 2 dimensions as a surface in 3 dimensions with the value of the scalar being the z-value.(5 votes)
- But why does it end up being a volume? The integral in the beginning is supposed to be just an area right? I get the math and intuition behind it, but isn't something wrong considering we started out with something in 2D and ended up with something in 3D?(4 votes)
- At3:45he says that he considers the minimum and maximum values of x and states that he can split this into two function of y on x. Why is this legal to do? The way he draws it I think it's legal on this example, but in general those half loops may not be functions of x at all (such as when the curve takes multiple y values for a given x value).(6 votes)
- I am also a little dubious on this point, but I believe that it's possible to divide the curve into as many parts as necessary to make them all functions of x. As long as it eventually closes the loop and doesn't intersect with itself, the proof, while more complicated, will still work out the same way.(5 votes)
- Why is partial of P with respect to y the height of curve??(3 votes)
- Partial P with respect to y is the function inside the integral. When you have an integral of some function f(x), the f(x) is usually represented graphically as the height above the x-axis. In this case dP/dy is f(y) and is the "height" above the y-axis. Later on, you will have f(x,y) being the "height" on the z-axis above the x,y-plane. Etc. this continues into any number of dimensions. (e.g. f(x,y,z) will be the height in some 4th-dimension above the x,y,z-"hyperplane")(5 votes)
- I'm currently working my way through Khan academy in the hopes of pursuing a Physics degree soon. Currently on the Algebra 2 course and I just skipped ahead to see that knowledge jump I would need to obtain. I understand almost none of this but I can't wait until all the hard work pays off and I can!(4 votes)
- Hi Nick, keep it up and before long you will understand everything you see here. Math has a delightful way of clicking together in deep, surprising ways the more you learn. I found the YouTube channel 3blue1brown especially helpful to see some initial connections in math -- not to mention his work is beautiful in a purely aesthetic sense. Wishing you happy studies as you go forward!(3 votes)
- How do we know that P vector field is not a conservative vector field?(2 votes)
- At10:00why has P(x,y) been written as an anti-derivative of the differential of P w.r.t y ? Is there a physical meaning to this ?(2 votes)
Let's say we have a path in the xy plane. That's my y-axis, that is my x-axis, in my path will look like this. Let's say it looks like that; trying to draw a bit of an arbitrary path, and let's say we go in a counter clockwise direction like that along our path. And we could call this path-- so we're going in a counter clockwise direction --we could call that path c. And let's say we also have a vector field. And our vector field is going to be a little unusual; I'll call it p. p of xy. It only has an i component, or all of its vectors are only multiples of the i-unit vector. So it's capital P of xy times the unit vector i. There is no j component, so if you have to visualize this vector field, all of the vectors, they're all multiples of the i-unit vector. Or they could be negative multiples, so they could also go in that direction. But they don't go diagonal or they don't go up. They all go left to right or right to left. That's what this vector field would look like. Now what I'm interested in doing is figuring out the line integral over a closed loop-- the closed loop c, or the closed path c --of p dot dr, which is just our standard kind of way of solving for a line integral. And we've seen what dr is in the past. dr is equal to dx times i plus dy times the j-unit vector. And you might say, isn't it dx, dt times dt? Let me write that: can't dr be written as dx, dt times dti plus dy, dt times dtj? And it could, but if you imagine these differentials could cancel out, and you're just left with the dx and a dy, and we've seen that multiple times. And I'm going to leave it in this form because hopefully, if we're careful, we won't have to deal with the third parameter, t. So let's just look at it in this form right here with just the dx's and the dy's. So this integral can be rewritten as the line integral, the curve c-- actually let me do it over down here. The line integral over the path of the curve c of p dot dr. So we take the product of each of the coefficients, let's say the coefficient of the i component, so we get p-- I'll do that in green, actually do that purple color --so we get p of xy times dx plus-- well there's no 0 times j times dy; 0 times dy id just going to be 0 --so this our line integral simplified to this right here. This is equal to this original integral up here, so we're literally just taking the line integral around this path. Now I said that we play our cards right, we're not going to have to deal with the third variable, t; that we might be able just solve this integral only in terms of x. And so let's see if we can do that. So let's look at our minimum and maximum x points. That looks like our minimum x point. Let's call that a. Let's call that our maximum x point; let's call that b. What we could do is we can break up this curve into two functions of x. y is functions of x. So this bottom one right here we could call as y1 of x. This is just a standard curve; you know when we were just dealing with standard calculus, this is just you can imagine this is f of x and it's a function of x. And this is y2 of x. Just like that. So you can imagine two paths; one path defined by y1 of x-- let me do that in a different color; magenta --one path defined by y1 of x as we go from x is equal to a to x is equal to b, and then another path defined by y2 of x as we go from x is equal to b to x is equal to a. That is our curve. So what we could do is, we could rewrite this integral-- which is the same thing as that integral --as this is equal to the integral-- we'll first do this first path --of x going from a to b of p of x. And I could to say p of x and y, but we know along this path y is a function of x. So we say x and y1 of x. Wherever we see a y we substitute it with a y1 of x, dx. So that covers that first path; I'll do it in the same color. We could imagine this is c1. This is kind of the first half of our curve-- well it's not exactly the half --but that takes us right from that point to that point. And then we want to complete the circle. Maybe I'll do that, and I'll do that in yellow. That's going to be equal to-- sorry we're going to have to add these two --plus the integral from x is equal to b to x is equal to a of-- do it in that same color --of p of x. And now y is going to be y2 of x. Wherever you see a y, you can substitute with y2 of x along this curve. y2 of x, dx. This is already getting interesting and you might already see where I'm going with this. So this is the curve c2. too I think you appreciate if you take the union of c1 and c2, we've got our whole curve. So let's see if we can simplify this integral a little bit. Well one thing we want to do, we might want to make their end points the same. So if you swap a and b here, it just turns the integral negative. So you make this into a b, that into an a, and then make that plus sign into a minus sign. And now we can rewrite this whole thing as being equal to the integral from a to b of this thing, of p of x and y1 of x minus this thing, minus p of x and y2 of x, and then all of that times dx. I'll write it in a third color. Now, I'm going to do something a little bit arbitrary, but I think you'll appreciate why I did this by the end of this video, and it's just a very simple operation. What I'm going to do is I'm going to swap these two. So I'm essentially going to multiply this whole thing by negative 1, or essentially multiply and divide by negative 1. So I can multiply this by negative 1 and then multiply the outside by negative 1, and I will not have changed the integral; I'm multiplying by negative 1 twice. So if I swap these two things, if I multiply the inside times negative 1, so this is going to be equal to-- do the outside of the integral, a to b. If I multiply the inside-- I'll do a dx out here --if I multiply the inside of the integral by negative 1, these two guys switch. So it becomes p of x of y2 of x. And then you're going to have minus p of x and y1 of x. My handwriting's getting a little messy. But I can't just multiply just the inside by minus 1. I don't want to change the integral, so I multiplied the inside by minus 1, let me multiply the outside by minus 1. And since I multiplied by minus 1 twice, these two things are equivalent. Or you could say this is the negative of that. Either way, I think you appreciate that I haven't changed the integral at all, numerically. I multiplied the inside and the outside by minus 1. And now the next step I'm going to do, it might look a little bit foreign to you, but I think you'll appreciate it. It might be obvious to you if you've recently done some double integrals. So this thing can be rewritten as minus the integral from a to b of-- and let me do a new color --of the function p of x, y evaluated at y2 of x minus-- and let me make it very clear; this is y is equal to y2 of x --minus this function evaluated at y is equal to y1 of x. And of course all of that times dx. This statement and what we saw right here-- this statement right here --are completely identical. And then if we assume that a partial derivative of capital P with respect to y exists, hopefully you'll realize-- and I'll focus on this a little bit because I don't want to confuse you on this step. Let me write the outside of this integral. So this is going to be equal to-- and this is kind of a neat outcome, and we're starting to build up to a very neat outcome, which will probably have to take the next video to do --so we do the outside dx. If we assume that capital P has a partial derivative, this right here is the exact same thing. This right here is the exact same thing as the partial derivative of P with respect to y, dy, the antiderivative of that from y1 of x to y2 of x. I want to make you feel comfortable that these two things are equivalent. And to realize they're equivalent, you'll probably just have to start here and then go to that. We're used to seeing this; we're used to seeing a double integral like this, and then the very first step we say, OK to solve this double integral we start on the inside integral right there, and we say, OK let's take the antiderivative of this with respect to y. So if you take the antiderivative of the partial of p with respect to y, you're going to end up with p. And since this is a definite integral, the boundaries are going to be in terms of x, you're going to evaluate that from y is equal to y2 of x, and you're going to subtract from that y is equal to y1 of x. Normally we start with something like this, and we go to something like this. This is kind of unusual that we started, we kind of solved, we started with the solution of the definite integral, and then we slowly built back to the definite integral. So hopefully you realize that this is true, that this is just we're kind of going in a reverse direction than we normally do. And if you do realize that, then we've just established a pretty neat outcome. Because what is this right here? Let me go back, let me see if I can fit everything. I have some function-- and I'm assuming that the partial of P with respect to y exists --but I have some function defined over the xy plane. You know, you could imagine we're dealing in three dimensions now. We'll draw a little bit neater. So that's y, that's x, that's z, so this, you could imagine, is some surface; it just happens to be the partial of P with respect to x. So it's some surface on the xy plane like that. And what are we doing? We're taking the double integral under that surface, around this region. The region's boundaries in terms of y are defined by y2 and y1 of x. So you literally have that curve. That's y2 on top, y1 on the bottom. And so we're essentially taking the volume above. So if you imagine with the base is-- the whole floor of this is going to be the area inside of this curve, and then the height is going to be the function partial of P with respect to y. It's going to be a little hard for me to draw, but this is essentially some type of a volume, if you want to visualize it that way. But the really neat outcome here is if you call this region r, we've just simplified this line integral. And this was a special one. It only had an x-component, the vector field, but we've just simplified this line integral to being equivalent to-- maybe I should write this line integral because that's what's the really neat outcome. We've just established that this thing right here, which is the same as our original one, so let me write that. The closed line integral around the curve c of p of xy, dx, we've just established that that's the same thing as the double integral over the region r-- this is the region r --of the partial of P with respect to y. And we could write dy, dx, or we could write da, whatever you want to write, but this is the double integral over that region. The neat thing here is using a vector field that only had an x-component, we were able to connect its line integral to the double integral over region-- oh, and I forgot something very important. We had a negative sign out here. So this was a minus sign out here. Or we could even put the minus in here, but I think you get the general idea. In the next video, I'm going to do the same exact thing with the vector field that only has vectors in the y-direction. And then we'll connect the two and we'll end up with Green's theorem.