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Lecture Comments (3)

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Post by Francis Tremblay on July 28, 2015

This lecture doesn't work with the chrome browser, but it works with safari.

1 answer

Last reply by: Professor Hovasapian
Mon Aug 5, 2013 11:02 AM

Post by michael Boocher on August 5, 2013

Video and voice seem out of sync on this one

Divergence Theorem in 3-Space

Let S1:z = 0 and S2:x2 + y2 + z2 = 1. Explain why (or why not) S1∪S2 creates a closed surface.
  • A closed surface is one which contains volume.
  • Since x2 + y2 + z2 = 1 is a circle of radius 1 and z = 0 is the xy - plane, our result is the upper semicircle above the xy - plane.
This surface does contains volume and hence S1∪S2 is a closed surface.
Let S1:y = 0 and S2: − x2 − y2 + 2 = z. Explain why (or why not) S1∪S2 creates a closed surface.
  • A closed surface is one which contains volume.
  • Since - x2 − y2 + 2 = z is a paraboloid opening downwards and y = 0 is the xz - plane, our result is an ever extending half paraboloid.
This surface does not contain volume (it opens down infinitely) and hence S1∪S2 is not a closed surface.
Let S1:z = √{x2 + y2} , S2:x = 0 and S3:y = 0. Explain why (or why not) S1∪S2∪S3 creates a closed surface.
  • A closed surface is one which contains volume.
  • Since z = √{x2 + y2} is a cone opening upwards, x = 0 is the yz - plane and y = 0 is the xz - plane, our result is an ever extending quarter of a cone.
This surface does not contain volume (it up down infinitely) and hence S1∪S2∪S3 is not closed surface.
Let S1:x2 + y2 = 1, S2:z = 1 and S3:z = − 1. Explain why (or why not) S1∪S2∪S3 creates a closed surface.
  • A closed surface is one which contains volume.
  • Since x2 + y2 = 1 is a cylinder of radius one on the z - axis, and z = 1 along with z = − 1 are planes which cut our cylinder into a section, our result is a small cylinder.
This surface does contain volume and hence S1∪S2∪S3 is a closed surface.
Let S1:x + y + z = 1 and S2:x − y + z = 1. Explain why (or why not) S1∪S2 creates a closed surface.
  • A closed surface is one which contains volume.
  • Since x + y + z = 1 and x − y + z = 1 are planes they either intersect at a line, are parallel or are equal.
This surface does not contain volume (it creates no definite shape) and hence S1∪S2 is not a closed surface.
Find the flux of the vector field F(x,y) = (x,y,x + y + z) on the region described by the rectangular prism 0 ≤ x ≤ 2, 0 ≤ y ≤ 1 and 0 ≤ z ≤ 1.
  • Since our surface is closed, we can apply the divergence theorem to obtain the flux by evaluating dxdydz where V is the volume the region encloses.
  • Now, div(F) = ∇×F = 1 + 1 + 1 = 3 so that dxdydz = ∫010102 3 dxdydz
Integrating yields ∫010102 3 dxdydz = ∫0101 6 dydz = ∫01 6 dz = 6
Find the flux of the vector field F(x,y) = ( − x, − y,x + y + z) on the region described by the tetrahedron bounded by the first octave and the plane x + y + z = 1.
  • Since our surface is closed, we can apply the divergence theorem to obtain the flux by evaluating dxdydz where V is the volume the region encloses.
  • By observing the xy - plane, we can find the bounds for x and y. We also note that z is bounded by the xy - plane and the plane x + y + z = 1.
  • The intervals of integration for our tetrahedron are x ∈ [0,1], y ∈ [0, − x + 1] and z ∈ [0,1 − x − y].
  • Now, div(F) = ∇×F = − 1 + ( − 1) + 1 = 1 so that dzdydx = ∫010 − x + 101 − x − y 1 dzdydx
Integrating yields ∫010 − x + 101 − x − y 1 dzdydx = ∫010 − x + 1 ( 1 − x − y ) dydz = ∫01 ( [1/2]x2 − 2x + [3/2] ) dz = [2/3]
Find the flux of the vector field F(x,y) = (z + x2,z + x,x + y) on the region described by the cylinder x2 + y2 = 4 bounded by the first octave and the plane z = 2.
  • Since our surface is closed, we can apply the divergence theorem to obtain the flux by evaluating dxdydz where V is the volume the region encloses.
  • By observing the xy - plane, we can find the bounds for x and y. We also note that z is bounded by the xy - plane and the plane z = 2.
  • The intervals of integration for our cylinder are x ∈ [0,2], y ∈ [ 0,√{4 − x2} ] and z ∈ [0,2].
  • Now, div(F) = ∇×F = 2x + 0 + 0 = 0 so that dzdydx = ∫020√{4 − x2} 02 2x dzdydx
Integrating yields ∫020√{4 − x2} 02 2x dzdydx = ∫020√{4 − x2} 4x dydz = ∫01 4x√{4 − x2} dz = [32/3]
Find the flux of the vector field F(x,y) = (xez,yez,exy) on the sphere (x − 1)2 + (y − 1)2 + z2 = 1.
Use rectangluar coordinates, do not integrate.
  • Since our surface is closed, we can apply the divergence theorem to obtain the flux by evaluating dzdydx where V is the volume the region encloses.
  • Our region is a sphere centered at (1,1,0), by observing the xy - plane, we can find the bounds for x and y. We also note that z is bounded by z = ±√{1 − (x − 1)2 − (y − 1)2} .
  • The intervals of integration for our sphere are x ∈ [0,2], y ∈ [ − √{1 − (x − 1)2} ,√{1 − (x − 1)2} ] and z ∈ [ − √{1 − (x − 1)2 − (y − 1)2} ,√{1 − (x − 1)2 − (y − 1)2} ].
Now, div(F) = ∇×F = ez + ez + 0 = 2ez so that dzdydx = ∫02 − √{1 − (x − 1)2} √{1 − (x − 1)2} − √{1 − (x − 1)2 − (y − 1)2} √{1 − (x − 1)2 − (y − 1)2} 2ez dzdydx
Find the flux of the vector field F(x,y) = (zcos(x),zsin(y),tan(z)) on the ellipsoid [((x + 2)2)/4] + [((y − 1)2)/9] + z2 = 1.
Use rectangluar coordinates, do not integrate.
  • Since our surface is closed, we can apply the divergence theorem to obtain the flux by evaluating dzdydx where V is the volume the region encloses.
  • Our region is an ellipsoid centered at ( − 2,1,0), by observing the xy - plane, we can find the bounds for x and y. We also note that z is bounded by z = ±√{1 − [((x + 2)2)/4] − [((y − 1)2)/9]} .
  • The intervals of integration for our ellipsoid are x ∈ [ − 4,0], y ∈ [ 1 − 3√{1 − [((x + 2)2)/4]} ,1 + 3√{1 − [((x + 2)2)/4]} ] and z ∈ [ − √{1 − [((x + 2)2)/4] − [((y − 1)2)/9]} ,√{1 − [((x + 2)2)/4] − [((y − 1)2)/9]} ].
Now, div(F) = ∇×F = − zsin(x) + zcos(y) + sec2(z) so that dzdydx = ∫ − 401 − 3√{1 − [((x + 2)2)/4]} 1 − 3√{1 − [((x + 2)2)/4]} − √{1 − [((x + 2)2)/4] − [((y − 1)2)/9]} √{1 − [((x + 2)2)/4] − [((y − 1)2)/9]} ( − zsin(x) + zcos(y) + sec2(z) ) dzdydx

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Divergence Theorem in 3-Space

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Divergence Theorem in 3-Space 0:36
    • Green's Flux-Divergence
    • Divergence Theorem in 3-Space
    • Note: Closed Surface
    • Figure: Paraboloid
    • Example 1
    • Example 2
    • Recap for Surfaces: Introduction
    • Recap for Surfaces: Surface Area
    • Recap for Surfaces: Surface Integral of a Function
    • Recap for Surfaces: Surface Integral of a Vector Field
    • Recap for Surfaces: Divergence Theorem

Transcription: Divergence Theorem in 3-Space

Hello and welcome back to educator.com and multi variable calculus.0000

Today's topic is going to be the divergence theorem in 3-space. When we did Green's theorem, we had -- remember -- two different forms of Green's theorem, we had the circulation curl which was the original form, and then we had the flux divergence.0004

We talked about what each integral meant, what flux meant, what divergence meant, well, the divergence theorem in 3-space is just the flux divergence form of Green's theorem, which is a 2-dimensional theorem kicked up one dimension.0017

That is exactly what we are doing. We are just moving up one dimension at a time. So, that is it. Let us just jump in and see what we can do here.0030

Okay. So, let us just remind ourselves of the flux divergence version of Green's theorem was as follows: so, I will write Green's -- oops, let us make it a little bit clearer here -- Green's flux divergence theorem... I will just write Green's flux divergence.0039

It said this in terms of symbols. That over Δ a, we have f(c) · n dt = a divergence of f dy dx. This was a symbolic representation.0067

It said that this flux integral = to this divergence integral, hence flux divergence.0087

In words, this is what this is saying. I am not going to write it out. I was going to write it out, but it is available in the previous lesson, so I will just say what it is and it is really, really important to sort of hear this over and over again.0093

It says the flux of a vector field across a closed curve, so again, remembering Green's theorem, we are talking about a closed curve, we are talking about traversing this curve in such a way, normally counter-clockwise, so that you are keeping the actual region that you are containing with this curve to the left. 0106

That is the important part. You want to keep the region to your left. So, you are traversing it this way. It is a closed curve. Okay.0124

So, the flux of a vector field across a closed curve, in other words, the extent to which the vector field is actually flowing across is equal to the integral of the divergence of that vector field over the area that is enclosed by the curve.0132

So, once again, the flux of a vector field across a closed curve is equal to the integral of the divergence of that vector field, over the area enclosed by that curve. 0149

Green's theorem expresses a relationship between a region and the boundary of that region. That is what the fundamental theorem of calculus is.0159

It expresses a relationship between a region, and the boundary of a region. In single variable calculus, your region was just this line, this interval along the x axis.0167

Your boundary were the points a and b. Here in 2-dimensions, your region is some 2-dimensional region, the boundary is a closed curve.0178

Now what we are going to do is we are going to kick up one dimension. We are going to go to a region in 3-space, a volume, and we are going to talk about the integral over that volume, and the integral over the boundary of that volume, which happens to be a surface, a closed surface.0186

There is nothing here that you actually have not done, it is just a new dimension. Notationally, it is actually the same.0200

So, another name for this is the divergence theorem, so we just generalize and call it the divergence theorem in 3-space.0207

Okay, so I am going to go ahead and write the notation for the divergence theorem in 3-space, and then we will go ahead and start talking about it.0215

So, this generalizes to 3-space as follows, so now it looks like this... Okay... Over s of f of p · n dt du = the triple integral over v of the divergence of f, and this time it is over the volume element, dy dx and dz.0225

So, let us stop and think about what this says. This right here, this integral, this is the surface integral that we just learned.0269

It says the flux of the vector field across this closed surface is equal to the integral of the divergence of that vector field over the volume enclosed by that surface.0276

If you want to think of a sphere, that is probably the most natural thing... geometric object that you know of in 3-space.0292

So, what you have is this 3-dimensional sphere. Well, you have the region inside. That is the volume and then you have the surface outside, the sphere itself, the shell. That is this.0302

So, the divergence theorem is expressing a relationship between those two. It says if you can either solve this integral, or you can solve this integral, it is up to you depending on what the problem demands.0314

That is exactly what it is. One of the choices you are going to have to make is how do I want to solve this. I want the surface integral, but is it easier to just solve the integral as a triple integral, volume integral, if the surface is not too easy to parameterize or not too easy to contain.0324

Okay. So, this p right here, this is the parameterization. Remember, any time we do any of these integrals, it is based on some parameterization. This is the parameterization p and this n, that was the cross product of dp dt cross dp du, right? The two vectors of the surface.0341

So, if you do not remember, go back one lesson, and it will discuss this very, very carefully. Okay. So, in words, what this says is the flux of the vector field across a closed surface is equal to the integral of the divergence of that vector field over the volume enclosed by that surface. 0363

Again, we are establishing a relationship between the region and the boundary of that region, the closed boundary of that region.0380

Okay. So, let us see what we have got. Meanings... called the meanings of the two integrals... right... let us see... what shall we do, oh, yes, absolutely. 0388

Let us make sure we understand what we mean by closed surface... so we know what we mean by closed curve, one end of the curve happens to be the same as the... the beginning of the curve is the same as the end of the curve. It is closed, it contains something. 0406

The same thing with a closed surface, it is a completely intuitive idea. It just has to be closed, so let us do a slightly different version of a closed surface so you know exactly what we mean when we say closed surface.0417

So, this is a special note... to apply this theorem s must be a closed surface. Now, you are not always going to get something nice like a sphere. You might get several different things. You might have to form your surface as a union of multiple surfaces. We will do an example in a second.0430

In other words, there must be some volume that is actually contained by this surface. That is another intuitive way of thinking about it.0465

Some volume... contained by the surface, and again, any time we do any of these. I have not made it explicit in a lot of the lessons, but we are assuming all of the necessary notions of smoothness, and differentiability, and things like that.0485

So, that just goes without saying. What you really have to watch out for is this stuff, remember, we are talking about a parameterization.0504

Somehow or other, you need to parameterize that surface. That is what is important, because the integral demands that there be a parameterization. That is the p. From the parameterization, you derive the n.0512

Okay. So, let us talk about this particular surface right here. A paraboloid, so let me draw myself something like this... a paraboloid that is going around the z axis, this is y, this is z, and this is x so we have this right here.0525

Well, if I take just the paraboloid, that is not a closed surface. The reason is that the top is open, okay? Yes, all around it is closed, but the top is open.0544

In other words, there is not volume that is actually contained, it just keeps going up like that. But, if I specify, so if I say z = x2 + y2, as my surface, I cannot apply the theorem yet because it is not closed. The top is not closed.0555

But, if I add another constraint, if I say z = 3, now, I am saying that there is this plane on top of that cup, so now I have closed it off.0570

So, now it is a paraboloid, everything along the paraboloid, and it is the top where z = 3. Now I have a completely closed surface. That is what is important.0585

In this particular case, your surface over which you are going to integrate is actually equal to a union of two surfaces. This one is surface one, we will call the paraboloid surface 1, and this top here, the plane, z = 3, is surface 2.0598

So if you actually decide to solve these surface integrals for this particular integral, you are actually going to end up solving 2 integrals.0617

You have to solve the surface integral over this surface, and you have to solve the surface integral over this surface.0625

What is nice is the divergence theorem allows you to say okay, well, okay, instead of doing that, it is probably easier if I just integrate over the volume that is enclosed by that surface, so you will only have to do one integral.0633

Again, the problem is going to have you decide which approach is better, but it is nice to have all of the tools in the tool box.0645

So, again, very, very important. You need to be discussing a closed surface.0655

Now, you are probably noticing something. As far as this closed surface is concerned, it is not exactly smooth. You have this paraboloid, and then you have this thing on top of it, and right there around the edge -- around this -- this is not a smooth surface.0660

There is a kink there. You know, from both sides, it is not going to be differentiable. That is okay. The theorem still applies.0676

You can actually have a surface that is made up of a bunch of separate pieces and those pieces can actually meet like this. They do not have to meet smoothly, they can meet like this where there is a kink -- you know -- where there is a sharp corner or a sharp edge.0685

That is actually fine if as long as the number of these sharp corners, these kinks or whatever it is, these edges, as long as they are finite it is not a problem... the theorem actually still holds.0699

The integral does not notice. So it is absolutely fine, and it is really, really great that it is that way. Otherwise it would be really, really hard to apply this theorem.0711

Fortunately, we do not have to worry about a finite number of edges, when we put a bunch of surface together.0722

You might think of it as a piece-wise continuous surface. Something like that. Okay.0729

Okay. Let us go ahead and do an example, and I think that is always the best way to go... I hope that this little intuitive discussion was helpful.0734

So, let us go ahead and find the flux of the vector field f, which happens to equal (x,y3, and... you know what, I am actually going to specify the variables in x, y, and z here just to make sure you guys realize that we are talking about a vector field that is a function of 3 variables.0744

So, f(x,y,z) = (x,y3,z5), right? That is our vector field over the unit cube in the first octant.0770

So, again the problem can give you the surface, can give you the volume in any number of ways. There is no one way it is going to give it to you... sometimes it will be direct and explicit and you do not have to do anything... other times you have to extract and draw it out. Figure out what is going on.0792

In this case, they say over the unit cube in the first octant. So, in the first octant that just means x = 0 to 1, y is 0 to 1, z is 0 to 1, that cube.0805

So, let us go ahead and draw that out, so we know what we are looking at. So, this is that way, that is that way, so this is z, this is y, this is x, so we go 1, 1, 1, so we are looking at this cube right here.0815

Probably not the best cube drawer... but... there we go, that is the unit cube in the first octant. All of these are length one.0833

We want to find the flux of this vector field across this surface. In other words, this vector field is defined for every point (x,y,z), therefore it is certainly defined over this entire surface... over all of the points that make up that surface. 0842

There are vectors that are emanating from all over there. We want to figure out what the flux is. Well, take a look at this thing. You have a cube. 0860

A cube is made of 6 sides, so basically I have to integrate this vector field 6 times, over 6 different surfaces. 0870

Not only that, but I have to parameterize each surface, the top, the bottom, the front, the back, this side and this side. 0877

Well, I do not want to solve 6 different integrals. Fortunately I have the divergence theorem that allows me to take the triple integral over the volume. I just have to solve 1 integral. That is what makes this really, really, fantastic.0883

Okay. So, let us go ahead and write out the formula, which is always what we do. Always a great idea, so when you do these problems go ahead and do the formula, it is a great way to familiarize yourself with the notation and sort of start to keep it in your head.0896

So, the integral looks like this. The flux is f(p) · n dt du, that is what we are looking for.0911

The divergence theorem actually says that this is equal to the triple integral of v, over the divergence of f, dx dy dz. That is nice. 0924

I do not want to do this. I do not want to solve all 6 integrals, I do not want to find a parameterization. I do not want to take... I do not want to find the normal vector. I do not want to take dot products. I could be here for the next 45 minutes doing this.0935

I just want to be able to take the divergence, and then integrate over the volume. This is a really easy volume to integrate over. It is a cube. It is the perfect volume.0946

Okay. Let us go ahead and do it. So, the divergence of f, let me write f again so we have it on this page.0955

We have x, we have y3, and we have z5. Well, the divergence of f, you remember, it is df1 dx + df2 dy + df3 dz, so let me write that in regular partial derivative notation.0962

It is df1 dx + df2 dy + df3 dz. It is a scalar. This is f1, this is f2, and this is f3. Let me go ahead and change colors here... I am going to go ahead and go to blue. 0981

So, the divergence of f, now, the derivative with respect to x is 1, so it is 1... the derivative with respect to y is going to be 3y2, plus the third one with respect to z... + 5z4.0997

There you go. That is our divergence. So, we have the first part of the integral. Now, x, now we will do our upper lower limits of integration.1011

Well, x is going to run from 0 to 1, because we are doing it over the unit cube, y is going to run from 0 to 1, and z is going to run from 0 to 1. Could not be easier.1023

Let me just draw this out one more time. We are looking at this cube right here, so x goes from 0 to 1, and y goes from 0 to 1, and z goes from 0 to 1. That is it. That will give us the entire volume contained in there. 1034

So, our integral is... yea, that is fine, we can do it here... our integral... we are doing our divergence integral... the integral from 0 to 1, the integral from 0 to 1, the integral from 0 to 1... The divergence is 1 + 3y2 + 5z4.1051

Of course we did -- in this case it does not really matter -- let us do z, y, and x, therefore it will be dz dy dx, working our way out.1074

When we go ahead and either solve this by hand... this one you can do by hand, it is not a problem, or just put it into your math software, you are going to get an integral of 3. 1085

That means... so the integral of 3, it is a positive divergence. What that means is that this vector field is actually moving... if this were a fluid field, in other words if this were a liquid, the liquid would be flowing out of a cube. That is what this means.1095

If this were negative, it would mean the fluid was flowing into the cube. In this case you have fluid that is flowing across the surface. It is moving away from the center of the cube. That is what this means.1113

Okay. Let us see. Let us try another example. Example 2.1127

This time we will let f, our vector field, be equal to... they get a little bit more complicated, x, y, z, x2, y2, z2, and x3, y3, and z3. Okay.1140

Okay. We want to find the flux of this vector field... okay, let us see... and what surface do we want... well, let us -- oops -- let r be the region... so this time the region is going to be defined by several equations in x, y, and z... be the region z > or = 0, x > or = 0, y > or = 0, and the relationship between x, y, and z, is x + y + z = 1.1159

When you put all of that together, what you have is this following region. It is the pyramidal region in the first octant. Let me draw it over here... give myself some room on the left.1204

So, there is that, there is that, there is that. Basically what you have is this, this is y, this is x, and this is z.1223

So what you have is that little pyramid right there. This pyramid made up of the yz plane, the zx plane, the xy plane, and this plane, that surface... that is the x + y + z = 1 plane.1235

You basically have a surface, a volume that is contained by 4 surfaces. You have 1, 2, 3, and 4. The plane.1252

So, if you wanted the flux of this vector field, you have to sort of solve 4 integrals. You have to find 4 different parameterizations, you have to solve 4 integrals. 1262

Fortunately, we do not have to do that, because we have the divergence theorem. We can just solve 1 volume integral. This volume is very, very easily extracted.1272

Okay. So, let me actually label these. This is going to be surface 1, so we know this is going to be surface 2, and this is going to... the back is going to be surface 3, and of course this plane right here, this thing, we call that surface 4.1281

So, again, our surface is the union of all of these surfaces. This 3 union s4.1302

If we wanted the integral over this, we would have to do 1, 2, 3, 4 integrals. We do not want to do 4 integrals, we want to do 1. The divergence theorem is the perfect tool for this. Excellent.1309

Let us go ahead and calculate the divergence. Let me rewrite f so we have it on the page.1319

So, we have x, y, z... whoof, I have got to slow down here, I am getting kind of crazy and excited... this is great stuff... x2, y2, z2... x3, y3, and z3.1325

So, the divergence of f is equal to... well, so this is f1, this is f2, this is f3, so df1 dx is going to be yz, df2 dy is going to be 2x2yz2, and as always... I hope that you are confirming this for me because I do have a tendency to make lots of arithmetic mistakes, and differentiation and integration mistakes... and the derivative with respect to third one, z, it is going to be 3x3y3z2, if I am not mistaken. 1343

Now we have the divergence, so we have the integrand, right? We are looking for this. The divergence of f, dx, dy, dz. We have just found the divergence, so we have taken care of that, the integrand.1383

Now we need to figure out the upper and lower limits of integration for this particular region.1397

Okay. x is going to go from 0 to 1. In other words it is going to go from here... to here.1403

y -- well, the relationship between y and x -- y is going to be this height right here. It is going to go from 0 all the way to -x + 1, that is what this line is right here.1411

I take the xy plane, this line is -x + 1, or 1 - x. So, y is going to go from 0 to -x + 1.1428

I have integrated in the x direction, I have integrated in the y direction, now I am going to integrate in the z direction.1438

Well, this one is easy. z goes from 0 all the way to -- well, here is the relationship, jus solve this for z. I am going to go from 0, which is its base all the way to the plane here. This surface... -x, -y, +1. That is it.1444

So, again, all of these integrals ultimately come down to 2 things. You need to be able to extract the integrand, you need to be able to figure out what the upper and lower limits of integration is, and you are done. The rest is very, very easy. 1468

That is what this problem... all of these problems are about. 1479

So, our flux, our integral is equal to 0 to 1, 0 to -x + ..., we will do y next... 0 to -x - y + 1, and we have our divergence, which is this thing right here.1485

We have yz + 2x2yz2 + 3x3y3z2, and we did dz dy dx, because this is z, this is y, and this is x. Working our way out.1510

When I put this into mathematical software, I get some odd number, but it is a number... 4733/554,000, you know what, no commas, I will just write out the numbers, how is that. 5 5 4 4 0 0.1533

Again we have a positive divergence. It is a very, very small number, but it is a positive divergence. This vector field is actually -- if this were a fluid field, the fluid is leaving this volume -- it is actually flowing across the surface. 1554

That is what is happening. That is what we are measuring. We are measuring the extent to which the vector field is moving out or moving in. That is all we are doing.1569

okay. So, again, what all of these integral theorems are doing is they are establishing relationships between line integrals and area integrals... surface integrals, volume integrals, they are establishing a relationship between a region in a particular dimensional space, and the boundary of that region. 1578

This is very, very, very profound. It is the reason it is called the fundamental theorem of calculus. It is absolutely fundamental. You have established a region, you have established a relationship between a region and its boundary. 1601

There is no reason to believe that such a relationship should exist, but not only does it exist, it exists in any number of dimensions. That is extraordinary.1613

There is nothing really new here. We have just kicked up one dimension. The only difficulty is the mechanics, you know?1621

Being able to extract the region, the upper and lower limits of integration, usually finding the divergence is a very, very easy thing. 1629

Finding the curl is actually not that difficult, just slightly more notationally intensive, but again also very, very simple.1639

So, that is it. Sometimes if you are looking for a particular integral, you will actually have to solve the surface integral. Maybe you are going to deal with a region where the double integral, or the triple integral, once you find the divergence, is going to be too difficult. 1645

Though, in that case, you have to do the double integral directly. You might have to integrate over one surface, maybe two surfaces, maybe three, four, five surfaces.1658

Again, this is where mathematical software comes in very, very handy. 1668

Okay. So, let us do a recap for services so that we know what it is that we have in our toolbox. So a recap for surfaces.1670

We did a similar one for curves, a recap for surfaces. Now, we will let p(t(u)), which is equal to some first function, some second function, some third coordinate function, be a parameterized surface.1683

This is of course, what we want. A parameterized surface, we are always dealing with a parameterization.1707

Now, the normal vector to that surface at a given point is given by the cross product of the partial of p with respect to t cross the partial of p with respect to u. There are going to be two normals. 1714

One is going to be pointing away from the surface, and the other is going to be pointing into the surface. We want the one that points away, but the relationship is just a negative sign, that is it.1728

So, this is the vector normal to the surface at a given point. Now, we defined 3 different things. We defined surface area, and our symbol for it was just dΣ, and the mathematics for it was the norm(n) dt du.1737

This is the symbol, this is the actual mathematics. You find n, you take the norm of n, you put it in here, you find dt, you find du, the domain of the parameterization, and you solve the double integral. That is the surface area.1778

We also defined the surface integral of a function. In other words the integral of a function over a given surface. Okay.1791

That one was... symbolically, it was just fdΣ, and that equals exactly what you think. Just stick an f in here... f(p), remember in each case we are always taking the parameterization, we are always taking the f of the parameterization p times the norm dt du.1809

We are just basically, all we are doing is we are adding things to the integrand. That is all we are doing. 1836

Then of course we had the surface integral of a vector field. The surface integral of a vector field.1841

So that is vf, and that is... the symbol is f · ds... that is f(p) -- you know what, let me write a little bit more for this one, let me erase this and actually specify which vector field... do some coordinate functions of a vector field.1851

So, f, which is equal to first coordinate function, second coordinate function, third coordinate function... so that is going to be... in symbols... it is going to be f · ds, that is the symbolic representation, the integral over the surface, equal to the integral over the surface of f(p) dotted with n dt du.1886

Notice, do not mistake these two. This is the composition function. That is a little o, f(p), also you can write if you want... f(p)... if you like it that way. I actually prefer it like that just to remind myself what I am doing. 1919

This is a dot product right here... and n, n is the same, that is this. Okay. So, surface area.1936

The integral of a function over a surface, the integral of a vector field over a surface, and last but not least... the divergence theorem, which is what we did today -- the divergence theorem.1945

Okay. Let us see, I am going to write this one in two ways. The divergence theorem says that the flux integral, the surface integral, in other words the f · ds is equal to over s is equal to the integral of the divergence over the volume... divergence of f... I am going to write dv. That is a differential volume element.1968

I am also going to write it as dx dy dz. So, I am going to rewrite this. f · ds is equal to this over v, divergence of f dx dy dz.1998

That is it. It is just some little differential cube. That is it. dv is the actual differential volume element. It is dx × dy × dz. That is all it is. These are the same thing, you will see it both ways.2013

So there you have it. That is the divergence theorem in 3-space. The flux of a vector field across a closed surface is equal to the integral of the divergence of that vector field over the volume enclosed by that surface.2026

Often your surface will be reasonably complicated and you do not want to solve this integral. You want to end up solving this integral, because it will end up being much easier. That is the whole idea. 2039

Thank you so much for joining us here at educator.com, we will see you next time for a discussion of Stoke's Theorem. Bye-bye.2048