For more information, please see full course syllabus of Multivariable Calculus

For more information, please see full course syllabus of Multivariable Calculus

### Stokes' Theorem, Part 2

^{2}+ z

^{2}= 1 with − 1 ≤ x ≤ 1.

- Note that y
^{2}+ z^{2}= 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.

- Note that our region is a tetrahedron with vertices at the origin, (5,0,0), (0,5,0) and (0,0,5).

_{B}F(C(t)) ×C′(t) dt where B is the boundary.

- We can parametrize the circle through C(t) = (cos(t),sin(t)) as well as a line through x(t) = P + t(Q − P).

_{B}F(C(t)) ×C′(t) dt where B is the boundary.

- We can parametrize the three lines through x(t) = P + t(Q − P), but we would be adding three integrals.

- We can parametrize the parabola with C(t) = (t,t
^{2}) as well as a line through x(t) = P + t(Q − P).

_{B}F(C(t)) ×C′(t) dt where B is the boundary.

- 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.

- 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.

- 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.

_{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.

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

### Multivariable Calculus

### 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 - x ^{2} - y^{2}, 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 - x ^{2}y^{2}, 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 - x ^{2} - y^{2}, therefore the radius is going to have to be 2, so this is going to be x^{2} + y^{2} = 4, = 2^{2}, because the radius of this circle is 2.*0266

*I will write it this way. x ^{2} + y^{2} = 2^{2}. 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 - x ^{2} - y^{2}, 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 - x ^{2} - y^{2}. 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 x ^{2} + y^{2} = 4.*0583

*That means y = 4 - x ^{2} 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 - x ^{2}) all the way to +sqrt(4 - x^{2}).*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-x ^{2}) to + sqrt(4-x^{2}), 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 - x ^{2} - y^{2}, 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 - x ^{2} - y^{2}.*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 - x ^{2} - y^{2}.*0934

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

*So, it is going to be x ^{2} + y^{2} - 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 + 2x ^{2}y + 2y^{3}, 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 - x ^{2}) all the way to +sqrt(4 - x^{2}), - 2xy - 8y + 2x^{2}y + 2y^{3} - 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

1 answer

Last reply by: Professor Hovasapian

Mon Dec 21, 2015 7:28 PM

Post by Hen McGibbons on December 16, 2015

fantastic course. your ability to explain complex concepts in an understandable way is a powerful skill.

as a freshman in college, i can only think of 2 teachers that have this same ability. one of them was an ap us history teacher and the other was an intro psychology professor. I believe that multivariable calculus is even more difficult to grasp than those subjects, so I give you even more credit!

I'm curious, can you list all the subjects you have been trained to teach or have taught in the past? I know your expertise is not limited to mathematics because I saw that you have many chemistry videos.

thanks,

hen

1 answer

Last reply by: Professor Hovasapian

Sat Aug 17, 2013 8:24 PM

Post by Mikael Svedmyr on August 17, 2013

Thank you so much for a great course. I now have an understanding for these matters that I did not achive during the course in my university. However, I was wondering if you plan to add new lessons about Taylor's series in multivarible calculus.

Thank you again.

1 answer

Last reply by: Professor Hovasapian

Mon Jul 8, 2013 4:08 PM

Post by Josh Winfield on July 8, 2013

Results are in! MATH201 (Multivariate And Vector Calculus)

Mark 90 (High Distinction)

Thanks so Much Raffi, not only for helping me to get a good grade but for helping me too fully understand the maths its applications and it power!

Josh (Australia)

1 answer

Last reply by: Professor Hovasapian

Wed Jul 3, 2013 9:56 PM

Post by barry ben-ishai on June 29, 2013

Great Job teaching :-)

1 answer

Last reply by: Professor Hovasapian

Sun Apr 21, 2013 4:14 AM

Post by Riley Argue on April 18, 2013

Thanks