Lec #18 Ch 16.9 The Divergence Theorem

What I want everyone to think. Because when I do it directly, when I find the flow directly, it's a little annoying, right? I mean, this

theorem

makes it so smooth, so elegant, so easy. So my problem now is, what if we had been there? But now I want to point out that it has an open bottom. So what I mean is don't do the flow from the bottom. Now it is an open surface. Well. Now it is no longer the entire surface. Upper hemisphere. Focused on the origin. Radius two, facing outward. Everything is facing outwards. So I just want to remove the flow from this.

Well, I guess I'll have to go back and do it directly. Or I can do something. Any suggestions? What are you saying? -Change the limits of the integral φ? - Then I'll do that. But that would mean using Gauss in some way. I just have to change that, do you agree? So I'll do that right now. Where would it go from? Zero to? - π/2 - π/2. So far, so good. Let me change them all. Change that. Change that. Change this. That changes this answer. What is the negative cosine of π/2 and then minus the negative cosine of one? This becomes what? - One one.

So this just has to be divided why? A two? I don't know what 1536π over 10 is. What has that become? I just took that last answer and split the two. What is that 153.6? Or I can leave it like this, right? 1536π/10 or 153.6π, right? Well. What would that be? What is 12 times two times 32? That would simply be it. Let me do that. I'll write it that way once it's simplified a bit. 12 times two times 32. 768π/5. Well. Then I understood it. But that's not the answer, is it? Because what does this include? That's removing the flow here. So, according to Gauss's

theorem

, we include this.

But I want to use Gauss's theorem because doing it directly takes forever. It is a great pain. Parameterizing a surface is a pain. Any suggestions what I can do with this answer? To get that answer? What I can do? Seeing how to make the flow from a simple disk is quite simple. So let's go back and leave that answer right there. Let's go back and find the flow out of that disk and subtract it from this. Then we will have the answer. Makes sense? That's what we'll do. And I'm telling you it would be a lot faster than doing what?

Parameterization of this surface by the other means. Because the flow of a flat disc is easy. What is the unit vector that goes in this direction? I have to go this way because originally it had an outward flow, right? What would the unit vector be like that? Zero zero? Negative! So what will we do? Leave this here. We'll take this answer and subtract this one. I can delete this. Take all this out. So everyone has done a great job. Now let's say, draw an arrow right there. What would be the outward flow? Because you can say it would be downwards.

From this album. Let me draw a picture of that. It's a disk. It is in the xy plane. What is the radius? The radius is two, right? So this is a disk with a radius of two. And put your normal vectors up. So I have to do this. Point rx cross ry dydx. But you all just told me. Do you know what is it. What is this? < 0, 0, -1dydx. Scalar product. What is F composed of that surface? What is z equal to there? Z is zero. So I'll write all this but for every z I have to put a what? - Zero. - A zero.

So here it comes. < 4x^3 + 4y^3, 4y^3 + 0 and 4(0) + 4x^3 . It looks a little sloppy, but look what you're punctuating it with. Just a <0, 0, -1, what do you get? Just a -4x^3 dydx. Now do you want to continue being Cartesian or do something else? We cannot make a shortcut, it has variables. Is it a double integral and you want to make it polar? Okay, let's go polar. Go polar woohoo. Alright, what's that in polar? That would be, how much is x? Oh, I have to put my automatic what? R dr dθ. Do you remember what x was?

Rcosθ. Good? Okay, I'll leave that r in there. Rcosθ. So you have it, now you go from zero to 2π and what is the radius? Zero to two. Well, remember this was a hemisphere with radius two, right there. Let me correct this again. Negative four, zero to 2π, zero to two, r^4cos^3(θ) dr dθ. And we would take this answer. And what would we do with this solution? From this? Take this and subtract that. Makes sense? Take that answer right there. Subtract this answer if you have it. Now, if you want to calculate this, you can do it.

Cosine cubed might be annoying now. You have to integrate that. You have to use it as a trigonometric substitution. What is sin squared equal to? Cosine squared is one minus what? Sin squared. We can do it. But this is how we're going to eliminate that integral. But we will do it. And I have enough space there on that side. We will calculate everything. We'll deal with that cosine cubed. At the end, take this, subtract this answer. If this is negative, I will subtract it, but a subtraction with a negative will add. And we'll find the flow through that, cool?

I'm telling you, this would still be a lot easier. Than do it the other way around. Alright. Let's find that. I'm going to finish this. I have enough space here. Negative four, zero to 2π, cosine cubed, can I write it as what? Cosine squared is one minus sin squared, multiplied by another cosθ, right? There's the dθ and, oh, that r will be a piece of cake. Take that and multiply the integral from zero to two of r^4. First we will do the easy r. What is the integral of r^4? - R^5 out of five. - R^5 out of five.

I get 32 ​​fifths. That was easy. It's this guy. Can you make a u-sub for that? What would you leave the same? First I used a trigonometric identity. So I can get that cosine squared times the cosine, which gives me the cosine cubed. What would you leave the same? - Cosθ? I will let u be equal to sinθ. And I'm very much here. Is it okay for you to see that? Oh well, sure, thumbs up, okay? U is equal to sinθ. Du is equal to cosθ dθ. Look what happens. Wow, you're gone. I'll keep that negative four in there.

You obtained the integral from zero to 2π. You got a (1 - u^2)du, not bad. So doing the integral of a cosine cubed isn't so bad. What is the integral of this with respect to u? U less what? Do you cube over three? Now put it back, what were you? U is cosine, no, u was not cosine. You were what? - Sin - theta - Minus sin^3(θ) over three. And we are going to go from zero to 2π. I'll get this answer and multiply it by 32 fifths. So I want to make sure you don't forget about those 32 fifths.

Do all that, multiply by 32 fifths of r^4. Take this and subtract it from that. Let's solve this. Insert a 2π here, what do you get? Zero minus zero. Nice. Minus, now put zeros, what do you get? - Zero. - Put a stand right here. Oh my God. This is interesting. What is everyone going to get? What do we have to subtract from that? To get the flow out of it? - Zero? - I know. And then why does that happen? It's just the nature of it. This was where? This was where z is equal to zero. And when you get here, you get zero.

So the answer to this is to say: taking this, subtract zero and you have it. Cool? Awesome. Any questions about the

divergence

theorem? And just to wrap this up. When is that

divergence

theorem amazing? When is it cool? It's great for closed surfaces. Spheres, cubes, cylinders when discs are included. A paraboloid but it also includes the upper part as a disk that finishes it. A cone but there is a disk that closes the surface that way. Cool? And super work. Tomorrow, what we will do tomorrow is one last section. But I skipped it. This is 16.9 and will be 16.8.

Stokes' theorem. And what we're going to do is with this sheet we're just going to solve those last four problems. Because they're set up nice and easy for that, cool? Great job! Have a nice day