Hello again, Tom from everystepcalculus.com, everystepphysics.com, a calculus video on Green’s Theorem. Let’s do it. Index 8 to get to my menu. We’re gonna scroll down to the “G”‘s and choose Green’s theorem. Normally here’s the way problems are given, you know, something before the dy and something before the dy, and so I’m gonna ask you to enter those into this– in any problem, but you can see an example of what problem we’re gonna be doing right now. This is from Patrick JMT’s site, exactly the way he does it. Of course I have the program so that you can add other variables et cetera, so that’s what’s neat about programming and using the Titanium. So we’re going to enter alpha before we enter anything into these entry lines here, alpha X for the dx, and then alpha minus X squared times y squared. Green’s theorem turns a line integral into a double integral and I have found that line integrals are very difficult, I’d say it’s a waste of time, but to me as most calculus is. In fact, even, I was searching and searching line integrals and there’s not even SI units for the answer. You get some some sort of answer like 8192 over 5 or something. And there’s not even– so I don’t know what that computes and nobody else does. It’s called a line integral and it’s not a length, it’s not a width, it’s not an area… so I question that. But of course the Green’s Theorem makes it even a little bit more easy or difficult, depending on what you can do with integrals. But anyways, here’s my program on it. I say it’s okay once we enter that and of course they always give you a triangle. They give you a square, a curve, a unit circle, or two circles, or one circle. Calculus always makes something difficult out of something simple. If that’s simple. So anyways, this is a triangle, which we’re gonna choose. The object within the region is a right triangle. It has to be a right triangle because of the pythagorean theorem, which is x squared and y squared in there, which calculus can work with exponents like Newton designed it. And the first vertices given is (0,0), we’re gonna choose that. The next one is (0,y) and we have to enter something for the y, because even though these are all ones, you can’t enter something different. You can enter 5– and I have to make sure I understand that when I program. But this is alpha 1 and then that changes the one here and the last third vertices can’t be anything different than one. And we’re gonna enter the x-value for that. And that’s 1, 2, alpha 1. And so I show you what the vertices given are, if that’s correct then choose “okay”. And then something that other programs don’t give you, I draw it, because you’re supposed to draw it on your paper– adding the vertices and stuff like that, so here’s the way it’s drawn. And y equals x. Why does this equal x? Because this slope, if you graph it, is x. Okay, now what if this was 5 and 7? Well then, slope is rise over run so this would be 5– or, if the x was 5 and the y was 7, rise over run is 7/5. So it would be 7/5 x here. Right now its (1,1) x here. But you don’t see it because 1 divided by 1 is 1. So i exclude that part, but those are the kind of things that upset me– and this is the tough part to me, this was always the tough part– getting the vertices and how to put it in the integral. So here’s the formula for the Green’s theorem integral, here’s the region, and I show you this is ax over bx, you know, these are the limits. And so we add those automatically. 0 over 1 for the x, and x over 1 for the y. We’re gonna do the order of integration is gonna be dy first and then dx. So here’s the– we’re integrating this and we’re just gonna go through it quick and keep this video kind of short. At y equals 1 here’s the answer. At y equals x here’s the answer when you substitute that. I show you the substitutions in here. Here’s one substitute of that. Now, you’ll use parenthesis for this instead, I use quotes because that’s the way that the calculator operates, but you’re going to use parentheses. Two times parenthesis “one” parenthesis cubed. Times– and that’s substituting 1 for the y. And here we’re substituting x for the y so there would be parenthesis around this x here. And of course the upper limit minus the lower limit, look at the minus signs here and how it works out. I always screwed that up when I was in calculus. That’s the reason I programmed, so I wouldn’t screw up. And now we’re gonna do this integral here from 0 to 1. And… at x equals one it equals minus one fifth and zero. Subtract the upper limit from the lower. Or, lower from the upper limit. And the answer is -1/5. Pretty neat, huh? everystepcalculus.com, go to my site, subscribe and see other videos that I make. Have a good one.