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### Stokes' Theorem, Part 2

Describe the boundary of the surface above the xy - plane and bounded by y2 + z2 = 1 with − 1 ≤ x ≤ 1.
• Note that y2 + z2 = 1 is a cylinder of radius 1 which goes along the x - axis.
• The xy - plane cuts this cylinder in half, and the restriction − 1 ≤ x ≤ 1 cuts the cylinder into a measureable section.
Hence the boundary of the surface is a square with vertices at ( − 1, − 1), ( − 1,1), (1,1) and (1, − 1).
Describe the boundary of the surface above the first octant and below the plane x + y + z = 5.
• Note that our region is a tetrahedron with vertices at the origin, (5,0,0), (0,5,0) and (0,0,5).
Our boundary is therefore a triangle.
The boundary below describes the surface S of a region. Determine which integral from Stoke's Theorem is most convenient for the region.
Since we can parametrize a circle through C(t) = (cos(t),sin(t)), Stokes's theorem works best for the integral ∫B F(C(t)) ×C′(t) dt where B is the boundary.
The boundary below describes the surface S of a region. Determine which integral from Stoke's Theorem is most convenient for the region.
• We can parametrize the circle through C(t) = (cos(t),sin(t)) as well as a line through x(t) = P + t(Q − P).
We would be adding two integrals, but Stokes's theorem works best for the integral ∫B F(C(t)) ×C′(t) dt where B is the boundary.
The boundary below describes the surface S of a region. Determine which integral from Stoke's Theorem is most convenient for the region.
• We can parametrize the three lines through x(t) = P + t(Q − P), but we would be adding three integrals.
Since we can parametrize f(x,y), Stokes's theorem works best for the double integral dt where S is the region the surface lays in.
The boundary below describes the surface S of a region. Determine which integral from Stoke's Theorem is most convenient for the region.
• We can parametrize the parabola with C(t) = (t,t2) as well as a line through x(t) = P + t(Q − P).
We would be adding two integrals, but Stokes's theorem works best for the integral ∫B F(C(t)) ×C′(t) dt where B is the boundary.
The boundary below describes the surface S of a region. Determine which integral from Stoke's Theorem is most convenient for the region.
• Although we may be able to compute two integrals (as well as represent the boundaries parametrically) it is easier to compute the double integration for the area.
Hence, Stokes's theorem works best for the double integral dt where S is the region the surface lays in.
The boundary below describes the surface S of a region. Determine which integral from Stoke's Theorem is most convenient for the region.
In this scenario it is best to compute the double integration for the area, Stokes's theorem works best for the double integral dt where S is the region the surface lays in.
The boundary below describes the surface S of a region. Determine which integral from Stoke's Theorem is most convenient for the region.
• For ellipses, computing the double integral for the area may be tedious at first, but we can always use polar coordinates to simplify our calculations.
Hence, Stokes's theorem works best for the double integral dt where S is the region the surface lays in.
The boundary below describes the surface S of a region. Determine which integral from Stoke's Theorem is most convenient for the region.
• In order for Stoke's Theorem to apply, our orientation must be counterclockwise.
• Recall that we can always change the orientation of a curve as ∫C F = − ∫C F . Hence our parametrization of our circle still applies as before.
Therefore, Stokes's theorem works best for the integral ∫B F(C(t)) ×C′(t) dt where B is the boundary.

*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.

### Stokes' Theorem, Part 2

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
• Example 1: Calculate the Boundary of the Surface and the Circulation of F around this Boundary 0:30
• Part 1: Question
• Part 2: Drawing the Figure
• Part 3: Solution
• Example 2: Calculate the Boundary of the Surface and the Circulation of F around this Boundary 13:11
• Part 1: Question
• Part 2: Solution

### Transcription: Stokes' Theorem, Part 2

Hello and welcome back to educator.com and multivariable calculus.0000

In the last lesson, we started talking about Stoke's Theorem, we discussed a little bit of its background. We did an example.0004

Today, we are going to continue on with some more examples of Stoke's Theorem.0008

Again, Stoke's Theorem is the one that seems to give kids the biggest problem, and yet interestingly enough, it is the one that generalizes most naturally.0013

It is just the circulation curl form of Green's Theorem kicked up one dimension. Let us just jump in and do some examples. Okay.0022

So, this time -- let us see here -- We will do example 1. This time we will define our surface first, so let s be the surface z = 4 - x2 - y2, and z > or = 0.0032

They give it to us in the form of an equation, we have to find out, we have to draw it out, find out what it is, and let f = z - yx + z - x -y.0061

Okay, in this particular case, we want you to find a couple of things. We want you to find the ds, which is the boundary, remember this symbol is the boundary of the surface, and calculate the circulation of f around this boundary.0085

Okay. Let us go ahead and draw out what this looks like. 4 - x2y2, this is a paraboloid that is inverted, and it starts up at 4.0121

When x is 0, y is 0, z = 4. So, what you have is basically something that goes down like this, goes down like this, and goes around like this.0138

That is it. It is a paraboloid that is just upside down. z > or = 0 means it is everything above the xy plane.0150

So, we have this surface right here, okay, so this is the surface. This is s. The boundary of this surface is this right here, this curve. It is the circle in the xy plane.0160

So, let us see here. Let us see how we can, so two, this is two, and it would be two this way, 4, so it looks like... so this is our boundary curve, and that is what is really, really important here.0180

So, imagine this is kind of like some hat, the bottom of the hat, the circle that the hat makes, that is the boundary of the surface.0202

Imagine if I took some circle, and if I took the piece of material that is in between that circle, and if I just deformed it, if I pulled it and stretched it, so now I have this tent like thing, right?0214

Well, the boundary itself has not changed. I have deformed the surface. I have not changed anything else, so as it turns out, this circle is the boundary of the flat region, or if it is the boundary of any deformation that I make to that flat region.0226

This is like a hat, that is the boundary. So, this -- oops -- Δs.0241

So, that is going to equal the following. It is of course a circle in the xy plane, so it is going to be a function of x and y.0251

It is going to be a circle, and let us see what its radius is going to be. Well, let us see.0261

If this is going to be... if this is 4 - x2 - y2, therefore the radius is going to have to be 2, so this is going to be x2 + y2 = 4, = 22, because the radius of this circle is 2.0266

I will write it this way. x2 + y2 = 22. That is the boundary curve, okay?0287

If I wanted to find the circulation, I could find the circulation of this vector field directly by integrating around this boundary curve, which I can parameterize by 2cos(Θ), 2sin(Θ), this is just a curve in the xy plane.0292

But I am not going to do that. I am going to go ahead and solve this via Stoke's Theorem.0310

We want to integrate over the surface, not the boundary of the surface. That is the whole idea. We are getting some practice with Stoke's theorem, but that is the boundary. I hope that makes sense. Okay.0314

So, let us use, let us see... so let us calculate over s the curl of the vector field of the parameterization dotted n dt du.0327

Let us actually calculate the flux of the curl vector field. That is what we are doing. The surface integral, in other words. Let us calculate the surface integral of the curl. Let us use Stoke's Theorem.0355

Well, let us go ahead and see what our... parameterization, surface... okay, so let us go ahead and write down actually, you know what, let us start on another page. So let us write down what our parameterization is, because we need a parameterization.0370

Well, they said that z is equal to 4 - x2 - y2, so they are giving us the surface as a function of x and y.0387

We know that our parameterization is going to equal x,y, and z itself, which is 4 - x2 - y2. That is our parameterization for our surface.0398

Well, we want to go ahead and form dp dx. dp dx is going to equal (1,0,-2x).0409

We want to form dp dy. That is going to equal (0,1,-2y).0419

We want to form n, which is equal to the cross product of these two, so that is going to equal, I am going to go ahead and write this one out actually, i, j, k... (1,0) - 2x (0,1) - 2y, and when I go ahead and remember cross product, we do a determinant, symbolic calculation.0430

When I do that, I get that n is actually equal to 2x, 2y, and 1. That is the normal vector to this surface. In other words, if I have this upside down parabola, at some point there is some vector sticking out at that point. That is what this is. It is the normal vector to this surface.0455

So, we have n. Okay. Now let us go ahead and find the curl of the vector field.0475

So the curl of f, that is also a symbolic determinant calculation. This time I am going to use capital D notation. Oops, d2, d3.0481

I have z - y, I have x + z, and I have -x - y. Okay, so at this point you should be reasonably comfortable with calculating... expanding determinants along the first row, and finding what the curl of a vector field is.0498

When I do this, I get the following. I get (-2,2,2). Okay.0515

Now, we want to find the following. We want to find the curl of f of the parameterization.0524

Well the curl of f of the parameterization, in this case the curl of f is a constant vector, so I do not have to do anything with the parameterization and the curl.0536

In the next example, I actually will do that. So, this is just (-2,2,2). Well, of course the next step is to find the curl of f of p dotted with n.0545

When I do that, so I do this, dot it with this and I end up with -4x + 4y + 2. That is going to be my integrand. Therefore my final integral looks like this.0562

Okay. Let us go ahead and draw out this region. Remember we said we had a circle, right? A circle of radius 2. So, this is x2 + y2 = 4.0583

That means y = 4 - x2 all under the radical sign + or -, and x... so x is going to go from -2 to 2.0601

y is going to go from -sqrt(4 - x2) all the way to +sqrt(4 - x2).0618

The curl of f(p) dotted with n is my integrand. That is -4x + 4y + 2, and this is y and this is x, so we do y first... x.0627

When I put this into my mathematical software, I get 8pi. That is it. Positive circulation, this thing is actually rotating. It is rotating in a counter-clockwise direction. That is what is happening, that is it. Nice and straight forward and simple.0644

Find the region, find the parameterization, in this particular case find the boundary and then you can decide if you are actually going to do the circulation integral, or if you are going to do the surface integral.0661

In this case we decided to do Stoke's Theorem, so we decided to integrate over the surface. We had the parameterization because we know what that is. We are given what the surface was. Then we just filled in the rest.0674

We found n, we found the curl of f, we found the curl of f of p, we took the dot product of the two things, and we figured out what the region was, -2 to 2, -sqrt(4-x2) to + sqrt(4-x2), and you just solve the integral.0687

Okay. So, little note here. Now, in this example and the example from the previous lesson, in the previous two examples... the curl of f was a constant vector.0706

So, we did not have to form, at least explicitly, this thing... the curl of f of p... we did not have to form the composite function, which is what the integrand requires.0746

The double integral, that is what is requires. Okay. Now, let us go ahead and do an example where we do have to do that. Where we do have to run through the entire process.0766

So, example number 2 here. I think I am going to do this one in red ink. So, example number 2.0781

We are going to let s be the same surface. Let s be the same surface as before, in other words the upside down paraboloid, be the same surface which is z = 4 - x2 - y2, z > or = 0.0800

This time we will let our vector field f, we will let it equal to xy, yz, and xz. Okay.0824

So, we have our parameterization, which is what we always want to start off with, xy is x and y and of course the function itself. 4 - x2 - y2.0838

Okay. We already know what n is, we calculated it from before, so let us go ahead and write down what that is.0851

So, n = (2x,2y,1).0856

So, now let us go ahead and calculate the curl of f. The curl of this particular vector field, well let us go ahead and form our symbolic determinant i,j,k d dx, d dy, and d dz.0867

We have xy, yz, and we have xz. When we expand this determinant along the first row, we get the following. We get (-y,-z,-x). So this is not a constant vector. Okay.0887

Now, we are going to form the curl of f(p), we are going to form the composite function. This is the curl of f, we are going to put the parameterization into that. The curl of f(p).0906

Wherever I see y, I put the second coordinate function. Wherever I see z, I put the first coordinate function, wherever I see x, I put the third coordinate function.0923

Okay... equals -y, okay, let us write down what the parameterization is so we remind ourselves -- I am going to do this one in blue, we have it on the same page -- the parameterization of xy is equal to x and y, 4 - x2 - y2.0934

So, -y, that just stays -y. -z, well -z is - this thing. This is z.0955

So, it is going to be x2 + y2 - 4 and of course, -x, that is -x, that stays the same... -x. this is our curl of f(p).0969

Now, we have to form the curl of f(p) dotted with n. Well, let us go ahead and remind ourselves what n is. N = (2x, 2y, 1).0985

We are going to take the dot product of this thing, and this thing. When we do that, here is what we get.1007

The dot product of this and this is this × that. So it is going to be -2xy, and it is going to be + this × that.1015

It is going to be... I am going to write it the way I have it here... -8y + 2x2y + 2y3, and then + this × that, - x.1026

There you go. This is our integrand, right here. Now, we are ready to solve the integral.1051

This one I am going to write out entirely. So, the integral over s of the curl of f(p) dotted with n dt du = -2 to 2, -sqrt(4 - x2) all the way to +sqrt(4 - x2), - 2xy - 8y + 2x2y + 2y3 - x, dy dx.1061

When I put that into my mathematical software, I get 0. That is it. In this particular case, this particular vector field is not this vector field.... xy, yz, xz, it is not rotating around that surface. It is not circulating around the boundary. That is what this says.1112

Very, very simple. I find n, I have p, I find the curl of f, I form the composite function, curl of f(p), that is a vector.1134

I am going to dot that vector with vector n. That gives me a scalar. That is this thing. That is my integrand. I take my integrand, I take a look at my domain for my parameter, in this case -2 to 2, this is this to this for y, and I solve it.1147

I get a particular number. That is all that is going on here. In case you lose your way, just work symbolically. As long as you have this, you can just run down the list and find what you need.1163

That is what is important. So, I hope you have an opportunity to practice this as much as possible.1176

Again, there is no one thing that you can always do over and over and over again. The process is the same, but the problem is going to come in different ways.1182

Sometimes you are given a parameterization, sometimes you are not. Sometimes you are... it is described one way, sometimes it is described another way.1193

Sometimes you will have to solve the line integral, sometimes you will have to solve the surface integral. We want to get you to be able to stand back from these problems and think about what is going on.1200

These are reasonably complicated problems, because this is serious mathematics. This is not - you know - arithmetic here. This is serious mathematics. This is happening on a daily basis.1212

So, you should be very, very proud of your ability to actually solve these problems, because this is pretty high end mathematics.1222

Thank you for joining us here at educator.com, we will see you next time.1230