Hi everyone. Welcome to our lecture on Green's theorem. This theorem is a very important theorem. This theorem starts off with a definition. A simple closed curve C in the xy-plane is positively oriented. Remember, simple and closed means that there's no crossings, and closed means that it starts and ends at the same place. Like a circle or some variation of it. A simple closed curve C in the xy-plane is positively oriented if it is traversed in the counterclockwise direction. This is important because we're going to start to introduce integrals and remember the orientation affects if there's a negative sign or not. It doesn't have to be a circle. You can certainly think of any body of water, any lake. There's your curves C and this is in the positive direction. Be careful. Counterclockwise is positive and any other direction that would be the negative direction as we evaluate these integrals. Now, let's state the Green's theorem. Let C be a positively oriented, piecewise smooth. I can be a curve that could be just a bunch of curves glued together maybe and I'm going in the counterclockwise direction or so. I want it to be closed. Now I can have any polygon I want going counterclockwise in the xy-plane. Let D be the interior of C. Since it's closed, there's clearly an interior and an exterior. The set of points inside is in fact D. If P and Q have continuous partial derivatives on D. Remember P is a function of x and y, and Q also is a function of x and y. We could take partials, everything's defined, we're talking about this region inside. Then, and here comes the statement of Green's theorem, the work done by a vector field Fds. Another way to write this is as Pdx plus Qdy. Remember, at the end of the day this is a dot product, so you can write it out as such. This is actually the statement of Green's theorem, that this value, we have a third way to compute this now. Remember, we had the fundamental theorem for line integrals. We had the brute force method to parameterize the curve and evaluate it. Now we have a new way when it's closed, again, keyword here being closed. I can do this as the double integral over the interior of the curve of dq dx minus dp dy da. This is mildly amazing. Again, let this sink in. This relates the line integral of a vector field to the double integral along the interior. I'm relating the boundary of an object to its interior. This is a way that relates a one-dimensional object, a curve, to its two-dimensional interiors. We're crossing the boundary among dimensions of objects. We're exploring their relationship and this is it. I remember this formula as quicks p, quicks dq dx p. That's just my little mnemonic that I remember, and of course, it's a minus sign to evaluate these things. This gives us a new way, when the curve is closed, to evaluate the integral. Sum notation, since C is the boundary of D we use this notation to discuss boundary. But just to remind you, we're integrating in the positive direction of C as well and that C is closed. We can also use a circle along the integral sign, you'll see this once in a while. Instead of just the plain integral sign to stress the fact that C is closed. When C is closed, we have this really powerful theorem at our disposal and so to emphasize that, sometimes we'll do an integral sign with a circle in the middle of it to just stress that C is closed. Put this formula down with your ever-growing list of formula and let's go ahead and do some examples to see it in action. Let's compute the line integral of the vector field. I'll write it as x y dx plus x squared y cubed dy. Now, remember that this is a vector field. I'm handing you a vector field with the dot product already worked out. This is really x y, x squared y cubed. The curve that I want, the simple closed curve that I want, will be a triangle with one vertex at the origin, one vertex at (1,0), and will go all the way up to (1,2). Picture not quite drawn to scale but it's a triangle nonetheless. We're going to of course go in the positive direction, the counterclockwise direction. I want you to realize that there are three parts to this curve. It's piecewise smooth, it's simple closed, so Green's theorem is going to apply. We're going to think of this as three line segments as we start from the base and work our way around. The curve itself, the whole triangle, is really made up of three parts. We're going to do this two ways. We're going to do it the long way, we're going to do it the short way and the short way will show the power of Green's Theorem. Each one of these is a line segment. Way number one is going to be with brute force. We're going to parameterize the curves C1, C2, and C3. C1, of course, it's defined by, remember our formula for a line segment. If you want the general formula, if you don't want to go and look it up, it's one minus t times the points of p and then plus tq. We're going from t is zero to one. Using that formula, I'm going to write this quickly. x is going to be t for the bottom curve and y, well that's just zero for the entire time. Again, each one of these is going to have its own integral so I'll write the bounds zero to one. The second curve, this should be C2, sorry. This is the curve that goes up. For this one, x is fixed at one, y is on the run so we can think of it as 2t. I'm going to go from zero to t to one. Last but not least, the third curve is going to be given by one minus t, two minus 2t again. This is coming from using the formula and then plugging in the values of the points (1,2) and (0,0). Of course T goes from zero to one, one more time. Great. Now, the line integral of this closed curve is really three separate line integrals, so as I continue the brute force method, just a friendly reminder what that is, I take the parameterization, I substituted into the vector field, and I dot it with the derivative vector of each piece. I have three curves, I'm going to do it three times. Can get a little rough and we'll go quickly, you can check me as I go here. The first curve, because the y component is zero, and there's a y in both these components, I get the zero vector, dot the derivative, which is one comma zero, and that's just zero dt. That's nice. That means since the function is zero, or y is zero, so I'm going to get zero back there. For the second curve, C2, the one that goes up on where x is one, this becomes, is little more involved here. This becomes xy, so 2t comma and then x squared y cubed. Vectors one and y cubed becomes 8t cubed, and then I dot this with the derivative vector of the second component here, so the derivative of one is zero, derivative of 2t is just two, and I'll put that all together in a second. Then the last curve, C3, the one that comes down the hypotenuse of this right triangle. This one gets a little more involved as well. This is going to be, probably going to run out of room on this one, but xy, this becomes the product of the vector that has the product one minus t parentheses, two minus 2t, so xy. Then x squared y cubed comma, we have one minus t squared to minus 2t cubed. That closes up the vector and all of that is dotted with the derivative vector, which is minus one, minus two dt. The first one, integral is the easiest one, the rest got nasty and no fun. I'll leave it to you to work this out. This is not difficult, and it's just not super pleasant to do, but it's definitely something that you can easily do or throw into a computer to do this. When you work this out, you do get two thirds. One should realize though it's annoying and this is a triangle, you can imagine if I had a square, or any other polygon, I hec or an octagon. As the shape gets more complicated, so does the integral, just the number of computations increases as well. Now we take advantage of the fact that it's a simple closed curve, and let's do way number 2, and of course this will be Green's theorem. What I'm going to do now is to compute this line integral of the vector field, I'll write the formula down one more time. To compute the line integral, I can instead compute the double integral of over the region of quick speed. Partial derivative q with respect to x minus the partial derivative of p with respect to y, dA. This is Green's Theorem. Remember p is the function in the first component of the vector field, and q is the function in the second component of the vector field. Now I just have to take partial derivatives. Dq dx becomes 2xy cubed and then minus dpdy, which is just x. All of that is done dA at the moment, when I set this double integral up, we're going to go dydx. 2xy cubed minus xdydx. Let's work our way outside in the bounds on x. As x moves over this triangle, I go from zero to one. As y moves over the triangle, I start at zero, but then my upper bound is a line, the equation of that line, of course is 2x, and I have a nice double integral. Now you have to ask yourself, what's a little easier? Would you rather do many multiple single integrals that get gross or one single, double integral? This one is not that bad, and again, I leave the work for you to work out here, but again, you get two-thirds. I care more about the setup at this point than working through the calculations. To integrate these things, they're all easy polynomials, they're not too bad. Both ways will give you the same answer. If you ever do it both ways and you get a different answer, something went wrong, go back and check your work as you go through this. Let's do another example. Let's compute the line integral about the curve. I'll say what the curve is in a second, but let's do y cubed dx minus x cubed dy. We'll let c be the smooth, close will be the circle. Now is a curve here. I want to do, center it zero, zero. Will have a radius be two. Let's go in the positive direction, of course, by default. I hope you agree that this satisfies all the conditions of Green's Theorem. To parameterize a circle, this is something we know. We can go ahead and parameterize it, but let's see if we can use Green's theorem instead. Our vector field, if we wanted to score that out, is y cubed comma negative x cubed. This is our function p, this is our function q here. Just be careful that minus sign on the x cubed, that does come along with it. By Green's theorem, be ashamed not to actually use the color green here. This is Green's theorem. I'm allowed to now look at the double integral over the interior of, I'll write the formula quick speed, dqdx minus dpdy, all dA. Let's just work out the function of the partial derivatives first, and then we'll worry about the bounds to work this out. We have dqdx. Remember our q function is negative x cubed. It's partial derivative just becomes negative 3x squared, and then the partial derivative with respect to y becomes minus three y squared dA. Now we can factor out the -3. It's just a constant in front, you get the double integral of x squared plus y squared dA. We're going to make a substitution on this one. Since we have a circle, we're going to convert to polar coordinates. In that particular case, we're going to go from 0 to 2 Pi, [inaudible] our angle around and 0 to 2 x squared plus y squared, that's just r squared. Then we do r, dr, d theta. Whenever you convert to polar, you have to throw in this extra r for technical reasons. Now you can work that out. That's just rq there's a function. So it's minus 3, there's no theta, so this just 2 Pi and you get the integral from 0 to 2 of r cubed dr. That's of course minus 6 Pi times r to the fourth over 4 from 0 to 2, and that all works out to be 24, well, negative 24 Pi. This tells you something by the way, because it's not 0, right? If we had a conservative vector field over a smooth closed curve, because it's not 0, this also tells me that the vector field is not conservative. So just getting a non-zero answer here is another way to test if the vector field is conservative. We can also use this theorem in the reverse direction. We have a double integral, we can convert it to a line integral instead. We do this often when we have areas, we have another way to find area. So the area of a region is the double integral of the region 1dA. This is the way to find the size, just like arc length when you integrate 1, you get the size of the arc. When you do the double integral of 1dA, then you get the area of the region. If you think of this though as dq/dx minus dp/dy, you have to come up with, you have to create the right functions p and q, whose partial derivatives work and when their difference, is in fact just 1, there is more than one answer here. If you think about these for a while, you can come up with p of xy is 0, and perhaps q of xy is x. Again, convince yourself that dq dx is 1 minus 0. That is in fact 1. Or you can do p of xy is minus y and q of x, y is 0. That will also work. That'll give you 1 again or one more, p of xy is minus 1.5 y and q of xy is positive 1.5 x. Again, convince yourself if I take a partial of q, I get 1.5 and then minus, minus a half becomes plus a half and a half plus a half is of course 1. We'd like the third way, just so we have a non-zero component, gives more information. So I can rewrite the area formula for a region. So I started with the double integral over that region of the function 1dA. Let's write this. We'll use the double integral of d dq dx minus dp dy, all dA. We're going to use p as minus 1.5y and q as 1.5x. Now we're going to use Green's theorem in reverse. So Green's theorem says, this is reading it right to left now, that this can be equated to the line integral around the boundary of d. So c would be our boundary of d of the vector field with functions <P,Q>.<dx, dy>, which is just the same as f dot. We call this ds or dr doesn't matter. But s dot, it's differential. Now you can see I can rewrite this. Again, if I use the first one, then p is 0 and I just get back the q component, which is x.dy. We can write it as the curve of x.dy. We can also write it if we use the second component for all the same reasons, minus y, dx or If we use the third component, the one we like the most, we get 1.5. Just factor that out, front of xdy minus ydx. My advice would be use whatever's easiest to integrate after plugging in the parameterization for c. But if you have multiple tools, you come at this with the easiest one you can. But the key here is you can get the area of a region by studying its boundary. Let's compute this area for our friend, the circle. We'll give it our generic radius r, we'll center it at the origin and let's find the area formula for a circle. Yet again, we'll use the parameterization of the circle, of course, we'll have c be the boundary of the interior of the circle. This will be defined by its usual parameterization. So r cosine t, r sine of t, r is fixed here, r is the fixed radius. We're going one lap around, so 0 to t to 2 Pi. We'll find that the area of this circle will be given by, we'll use the first one just for it's being easy, x.dy. It's our formula from the last slide. Now we're going to use the parameterization, so remember what x is. We are going from zero to 2Pi, X is R cosine T, Dy the derivative of the Y component becomes positive R cosine of T Dt, and we have from zero to 2Pi. Remember, R is a constant, so there's two R's that just comes out front R-squared, and you get good old cosine squared of T Dt. This is the good old double-angle formula that we have to use whenever we have cosine, this becomes one half, one plus cosine of 2t Dt. We can pull out the one half to get R squared over two. We can integrate each piece to get T plus sine of 2t over two evaluated from zero to 2Pi. Again, R is a constant, so we're plugging these values in for T, R squared over two, you get 2Pi plus sine of 4Pi over two. Then minus, if you plug in zero, you just get minus zero, so we'll leave it here. Sine of 4Pi, remember 4Pi is one lap around is 2Pi, so two laps around would be 4Pi. That just lands you at zero again, and you get R squared over two times 2Pi, the two's cancel, and perhaps not surprisingly, you get Pi R squared. The area of a circle, here's just another way to show you that all the theory is pointing to the same result. Let's do one more example, I want to find the line integral of Y plus E to the square root of X, this will be Dx plus 2x plus cosine of Y squared Dy. I want C to be the boundary of the region enclosed by two parabolas. We're going to do Y equals X squared, so that's our usual one that we know and love. We're going to do X equals Y squared, this is the parabola turned sideways. It normally go around, opening to the right, then the parabola opens upward, but I want this region of these two curves here. That's my parabola, so X equals Y squared. The curve will be oriented in positive directions. I have two pieces, C1 and C2, together they make the whole curve, and my piecewise, and I guess my two parts will be Q_ x and Y are given by each of the components. Remember, they are the pieces that make my vector valued function or my vector field. Let's say you get a question like this, and they ask you to work this out. Remember, this is the work done by this vector. The first thing you can ask yourself, if you want a quick answer is you can ask yourself, is the vector field conservative, so is F conservative, is it a gradient vector field? Why don't we ask that question first? Because over a simple closed curve, remember the start and end point would be the same, if that's true, the answer is zero. Friendly reminder, our test to do this would check is the partial derivative of P with respect to Y equal to the partial derivative of Q with respect to X. We can take this partial derivative with respect to Y is just one, for p and the second one is just two. As two equal to one, not conservative. We can't walk away here and just say it's zero. I really don't want to parameterize the curve. Remember our option now is I want to parameterize the curve, and plug in my parametrization. This E to the square root of X is going to get gross, I really don't want to deal with that. Cosine or Y squared, these are notoriously difficult functions to solve. If you want to ruin your weekend, go parameterize this curve, plug it in and see if you can do with the direct way, it's going to be very difficult. However, this curve is simple closed, satisfies the conditions of Green's Theorem. Let's work out this line integral of this vector field using Green's theorem, which says I can compute the double integral over the interior of the region, quick speed, Dq Dx minus Dp Dy. Remember, it comes up to nice. The partial derivative with respect to X_q, we said was just two, and Dp Dy was just one, and I end up with just the double integral of one Da. Now remember, this is just the area of the shape. Unfortunately, the shape is not a common shape where we know sometimes you get lucky, and it's a square, or a rectangle, or a circle, or some other shape where the area formula I just know. Unfortunately here we don't have that, so we're going to have to work out the double integral, but our function is one, it's a nice shape to work with. The point of intersection, if you evaluate these two things, you get the point of intersection is one. Let's go from zero to one, Dx. Then the Y, if you draw a line in the Y direction, we're going from the parabola Y equals X squared to, I'm going to rewrite this as Y is the square root of X, of one Dy. I'm going Dy Dx on this integral, now we have a relatively straightforward double integral to workout. This becomes zero to one. Integrate one, you get Y, take the difference square root of X minus X squared Dx, and then we integrate this. Remember this is X to the one half, so when I take it's anti-derivative, I get X_3 over two times two thirds minus X cubed over three from 0-1, that's not too bad. That's just two thirds minus one third, which is just one third. Green's theorem actually makes this integral possible, without Green's theorem, again, if you tried to do the other way doing parameterize the curve, and brute force, it's never going to work. You're going to get this integral that's just impossible to do. You're at best you're going to get approximation methods are something really challenging. Use Green's theorem when you can. It's a beautiful theorem. It really turns some problems into easier double integrals versus multiple line integrals. Just look for it, use it when you can. Once in a while you're lucky, the vector field is conservative, and you get zero right away, and you can walk away. Excellent job with these difficult examples. Keep this formula handy, and good luck on the practice problems, will see you next time.
