Verify that Stokes' Theorem is true for the given vector field F and surface S . F(x, y, z) = -2

step 1 All right, let's go ahead and solve this problem. So we're given a vector f negative 2y z, comma

Verify that Stokes' Theorem is true for the given vector...

Verify that Stokes' Theorem is true for the given vector field $ \textbf{F} $ and surface $ S $. $ \textbf{F}(x, y, z) = -2yz \, \textbf{i} + y \, \textbf{j} + 3x \, \textbf{k} $, $ S $ is the part of the paraboloid $ z = 5 - x^2 - y^2 $ that lies above the plane $ z = 1 $, oriented upward

Verify that Stokes' Theorem is true for the given vector field $ \textbf{F} $ and surface $ S $.

$ \textbf{F}(x, y, z) = -2yz \, \textbf{i} + y \, \textbf{j} + 3x \, \textbf{k} $,
$ S $ is the part of the paraboloid $ z = 5 - x^2 - y^2 $ that lies above the plane $ z = 1 $, oriented upward

Transcript

00:02 All right, let's go ahead and solve this problem.

00:05 So we're given a vector f negative 2y z, comma, y, 3x, and the surface is the paraboloid.

00:14 Z is equal to 5 minus x squared minus y squared.

00:18 I just wrote it like this because it's usually going to be more convenient later.

00:22 And z is above 1 is how it's stated, but from the expression, we know that z is going to be between 1 and 5.

00:30 So it is going to be forming a cap with a bottom lid, which is a plane or a disk, at a height 1.

00:42 Okay? all right.

00:44 So we want to confirm stokes theorem again.

00:47 So we are going to have two approaches.

00:50 One is to take a look at the region c, which is the boundary of the circle right here, x square plus y squared is equal to two.

01:04 Is it two? let me double check.

01:08 We know that the radius is two.

01:09 Yeah, so the radius is going to be two.

01:12 And we're going to parameterize c as this.

01:22 We know that the angle is going to be useful, theta.

01:26 And then i will say that x is equal to cosine, y is equal to sine.

01:32 Now z is constant, it's at 1, so we know that this is equal to 1.

01:41 Okay, from here we just need to take the derivative, oh sorry, the radius is actually 2, so i need my 2's there, x is equal to 2 cosine, y is equal to 2 sine, and we take the derivative with respect to theta, so r sub theta, it is equal to negative 2 sine, comma, 2 cosine, derivative of 1 is, 0 let's put f in terms of theta it will be negative 4 sine theta 2 sine theta and 6 cosine theta now we take the dot product f dot r okay so you will have 8 sine squared plus 4 sine theta is what you will get but i like to put that as 2 sine of 2 theta because sign of 2 theta is equal to 2 sine theta cosine theta okay so what we're trying to evaluate here is through c we have 8 sine squared theta plus 2 um 2 sine of 2 theta d theta and the angle varies from 0 to 2 pi because it's one revolution this is a straightforward integration so we will get 8 pi i want you to do the calculation on your own.

03:40 So the other integral, the double integral here, we're supposed to get 8 pi as well.

03:45 So let's see if that is going to be the case.

03:49 So first i will calculate the curl.

04:01 I've already done the calculation.

04:03 However, you will get 0 ,000, negative 2y minus 3, comma, 2 .z.

04:15 All right.

04:16 Now i will choose the parameterization.

04:18 Of the surface s let's actually make it clear that i am using two parameters are of row comma theta there you go so x i choose row cosine row sine row sign for the y and for z we know that it is equal to five minus the sum of x squared plus y squared we know that x square plus y squared according to these choices of the parameters we're going to get our row squared.

05:01 So we will get 5 minus row squared.

05:11 Okay.

05:12 Let's take the partial derivatives with respect to row.

05:16 We will get cosine, comma, sine, comma, negative to row.

05:25 With respect to theta, you will get negative row sine, comma, row, co.

05:36 Negative 2 row.

05:38 Oh, this one is actually 0.

05:41 Excuse me.

05:45 There you go.

05:52 We will take the cross product because we want the normal vector.

06:07 It comes out quite nicely...