Lec #19 Pt.1 Ch 16.8 Stokes' Theorem

And that's X equals two. Is this the X axis, is this the Y axis? And that's the Z axis. Okay? Great, now. If I go polar, what letter is automatic? A. You already paid them. What has this become? They are four? Is X R cosine data or just cosine data? It's R cosine data. So this is what I'm understanding here. Yes, it's still easier than going Cartesian. Can I split that? What do you all think? Can I split that thing? - Yes, I love it. You know it. I can factor R. Although some of you don't believe in integrated integrals.

I'll avoid it if I can. So I'm going to do it. Zero pi over two. From the cosine data plus the sine data. And multiply this answer to the integral from zero to two. I have to factor all four, right? I'll put all four right there. And R factored with R becomes R squared, right? Are you okay with factoring? Oh God. That's really good, you see. I factored R with R, making it R squared. There are four that were discarded, and IPut it here. I could have put all four here. That becomes four times R cubed over three.

From zero to two. And four times eight over three is it? 32 thirds. And now. What are these? What is the integral of our cosine? Addition. Comprehensive data of sin? Negative cosines data. We have to go from zero to pi over two. So sin over pi over two is? One zero. Less? Zero. Less? One. Take this number. And multiply it. At 32 thirds. What is that number there? Are two? That's a two. Two times 32 thirds? We have just found the work done by the vector field to move a particular around the limit curve. The entire road closed. Super work.

I know some of you are still copying, so I'll wait. Stokes'

theorem

is quite clear. But we do it at the end because this is the only

theorem

that connects the line integral with the surface integral. Line and surface. Well, the last two problems we are going to do. We will use the opposite side. Because they're going to say, use Stokes to do that. So we'll go back and do that. Which is pretty good. We finished this entire course by going back to doing what we did in 16 two. Which I like. Everyone can check that out.

Remember when we had to find line integrals? It was just F point DR. What's that? So we'll just do this. Just there. The only thing is that RT means parameterized curve. It's not funny? Look at all the work he was doing. We had all these things with the F curl. And we had to parameterize the surface. What I'm saying is skip all that. We will only focus on the curve. A little bit of C. This will be a large surface. Look for the curve, the boundary curve with the surface. Now, it could be limited from above.

It is not always limited below. It depends. It depends on the image they give us. Hey. From the practice problems, I found out that I will do number 10 and then number nine. Oh, is everyone done copying? I'll keep going. Busy? I'll wait. I'll wait. There's a lot up there. Sure? - Yes. Well, I'll delete this and attack the problem hard, because we'll attack it another way. Alright. Use Stokes' theorem to evaluate. Curly. From F. Punto DS. Alright. That way. What is the vector field? They have. By the way, they call this the loop flow of F.

Through the surface S. Use Stokes' theorem to evaluate the loop flow F. For the vector field F, across the surface S. Okay, then, where F of XYZ is equal to 2YZI plus XGK. Alright. What is S? S is the part of the circular paraboloid. Z equals X squared plus Y squared. Below Z equals four. With upward orientation. Again, can you help me take a photo? Use Stokes' theorem. To evaluate this curl. I'll put an S there. F point DS curl. Where F is 2YZ plus X cubed. And S part of the circular paraboloid Z plus X squared plus Y squared under Z equals four.

With upward orientation. Let's draw a picture, but our focus will just be the limit curve, whatever that is. We'll draw the surface, but our focus will be. Alright, a paraboloid just opens up like that, like a windsock, okay. Four notches, one two three four. There goes the windsock. It's open. And cuts Z4 in a circle. On a circular path. In Z it is equal to four. So close to Z equals four. They said with upward orientation. So the upward orientation means normal vectors off this surface like this. Which means the path will move counterclockwise. So I just want to see it from above.

That's what I want you to think. And when the trajectory of that curve, think of a particle moving along that curve. By the way, do you agree that the yellow curve is the limiting curve? Look, sometimes it's underneath. If this were turned down, it would have a downward limiting curve like in the XY plane. But this opens like this, that is the limit curve. But I want to point out, you know, where are we going to move? Because that's important to us when we do the line integral. Am I doing it that way or that way?

We will move counterclockwise as seen from above. And the reason is they said upward orientation. Upward orientation, that means, as long as, I'll say it again, like the book says. Let me read the book. He goes. This means that if you walk in the positive direction around the curve, your head points in the direction of the normal vector M. In that case, your surface will always be to your left. And indeed, if he walked like that. The surface is still to my left, right? The surface is still to my left. OK? Alright. So. We have it, but what's so good about this problem?

How are we going to attack it? I just want to break this all down here, here's how we attack the problem. How that. Which is very interesting. Who cares about the surface? I just want that curve. I just want the curve. So let me find it. And I'm going to take this off so we can grab it right there. What is this curve? You have to help me parameterize it. Can you help me parameterize that curve? R of X and Y? No, it's just R of T, right? R of T. And we haven't done this in a while.

Do you remember how to parameterize a circle? And it should put three components in place. So let me make sure I move this around and I'll just say we're going to parameterize this curve. Just there. That there. R of T. Same. I'll get my three components. Do you want to use cosine and sin? Because it's a circle. What is the radius? - Two. - Good. You all know that the radius is two. Now, how did they know this? The set in this, equal to that. Looking at the equation of that circle, I'll go to the left.

They realize that when this is true, X squared plus Y squared equals four. That means the radius of the circle is two. So to parameterize this circle, they say it's two cos T for the X component. Two sin T. For the Y component. What the heck is the Z component? Man, you're good. I know it's complicated, just from experience. It is, so I want to make sure everyone sees this. I'm not going to put a zero there. That curve is in the air, it is equal to four. Now this is a Z axis? The Y axis and the X axis.

The positive Z axis, positive Y, positive X. That curve goes up through that Z equal to four. Common sense. This is Stokes' theorem, we are talking about three natural spaces, right? So we're going to say it's a four. Hey, I want to point out, maybe one of the practices, I don't remember. But what if that curve was down in the XY plane? So guess what this number has to be? Zero, very good. Cool. But that's a four, so I'll leave it. Let's do the line integral. What is my lens? See, isn't this strange for everyone? Here I only have one integral.

It's been a while since we did that. From zero to two pi. Alright, I'll put F here. 2YZ here. What's happening here? Component Y. Zero. And what is the Z component? X cubed. And then. Right outside. What's happening here? What I have to do, F composed by RT. Point R prime of T. DT. And go from zero to 2 pi. That's what I'll do. So that's D, what am I going to substitute for this Y to Z? That will go right there. This will go right there. What is going to happen to the X cubed? That will go there.

And then we're going to put points on what? R prime of T. So what is R prime? What is R prime of T? The tangential factor. Or we can simply say tangent factor, what is it? Two negatives without T? Two cos T and which one is the previous one? Okay, I'll put it here. So be careful. When you do a line integral, this is always R prime of T. So now I'm going to substitute everything in. I wrote it like that, but now I'll replace it. What becomes Y, what becomes Z, and what becomes X cubed, right?

Then I'll do it. Ready? What is two times this times this? What is two times this times this? OK. Let's say here I go now. Two times two sin T. Times four. Count the zero, comma, what is that last component? The last component is X cubed, which would be two cosines of T, all cubed. What is two times two times four? 16. So I get 16 sin T, zero two cos T cubed. Spot. This. Cool? What's the point part? I get negative 32. Sin squared T. Do you know how to do integral sin squared? You need, very well. You should know that you need a trigonometric power, yes, you call it the power of the under.

It's half minus half cosine two T. So I'm going to put that in here. However, I will put the negative 32 here. A half. Minus half. Cosine of. Two Ts. That's the identity reduction of sin squared. I just put the negative 32 out there. So what is the integral of a half? I only get half a T, right? OK. This becomes. Negative. Sin of 2T but for a sub U, two get another two, which become a fourth. I want to point out that you didn't really solve the calculations for sub U. Is everyone okay? Can you see that?

Yeah, because you've done U sub so often, that this, I actually let you match the 2T, and then I got half, and I put half in the other half, I made another one. So I just want to make a note, this is my U sub. This here came out for a U sub. In case you want to figure all that out. Didn't anyone need to see that U submarine? Sure? I let U equal 2T. Yeah, I thought everyone would understand. And I'm going to plug in what? Zero two pi, okay. What do we get for this answer?

I get negative 32. Times? I just put a big parenthesis, we'll do this, minus this.