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 1 answerLast reply by: Professor HovasapianSun Jan 20, 2013 11:01 PMPost by mateusz marciniak on January 20, 2013my cross product came out different. mine came out to be cos pheta + sin pheta + 0, and for the curl(F) i got 1, -1, 1. it could be my algebra mistake. would that still be considered a constant vector but the second term would be flipped because of the negative one?

### Stokes' Theorem, Part 1

Let F(x,y,z) be a vector field such that F = (z,y, − x) and P(t,u) be a parametrization such that P = (u,t,u + t).
Find curl(F)oP
• To find the composition curl(F)oP we first compute curl(F) = ∇×F.
• Then ∇×F = 0i + 2j + 0k or (0,2,0).
So curl(F)oP = (0,2,0)o(u,t,u + t) = (0,2,0). Note that curl(F) is a constant vector.
Let F(x,y,z) be a vector field such that F = (x2 + y − z,x + y2 − z,x + y − z2) and P(t,u) be a parametrization such that P = (u,t,u − t).
Find curl(F)oP
• To find the composition curl(F)oP we first compute curl(F) = ∇×F.
• Then ∇×F = 2i − 2j + 0k or (2, − 2,0).
So curl(F)oP = (2, − 2,0)o(u,t,u − t) = (2, − 2,0). Note that curl(F) is a constant vector.
Let F(x,y,z) be a vector field such that F = (xey,yez,zex) and P(t,u) be a parametrization such that P = (u,t,1).
Find curl(F)oP
• To find the composition curl(F)oP we first compute curl(F) = ∇×F.
• Then ∇×F = − yezi − zexj − xeyk or ( − yez, − zex, − xey).
So curl(F)oP = ( − yez, − zex, − xey)o(u,t,1) = ( − te, − eu, − uet).
Let F(x,y,z) be a vector field such that F = ( − x, − y,1) and P(f,q) be a parametrization such that P = (cosq,sinf,q + f).
Find [ curl(F)oP ] ×N
• To find the scalar product [ curl(F)oP ] ×N we first find curl(F)oP and N.
• Now, curl(F) = ∇×F = 0i − 0j + 0k or (0,0,0).
• So curl(F)oP = (0,0,0)o(cosq,sinf,q + f) = (0,0,0).
• Recall that N = [dP/df] ×[dP/dq]. We have [dP/df] = (0,cosf,1) and [dP/dq] = ( − sinq,0,1)
• Then N = (0,cosf,1) ×( − sinq,0,1) = cosfi − sinqj + cosfsinqk or (cosf, − sinq,cosfsinq)
Hence [ curl(F)oP ] ×N = (0,0,0) ×(cosf, − sinq,cosfsinq) = 0 + 0 + 0 = 0. Note that if curl(F) = (0,0,0) then [ curl(F)oP ] ×N = 0.
Let F(x,y,z) be a vector field such that F = (x,y,x + y) and P(f,q) be a parametrization such that P = (sinf,cosq,q − f).
Find [ curl(F)oP ] ×N
• To find the scalar product [ curl(F)oP ] ×N we first find curl(F)oP and N.
• Now, curl(F) = ∇×F = 1i − 1j + 0k or (1, − 1,0).
• So curl(F)oP = (1, − 1,0)o(sinf,cosq,q − f) = (1, − 1,0).
• Recall that N = [dP/df] ×[dP/dq]. We have [dP/df] = (cosf,0, − 1) and [dP/dq] = (0, − sinq,1)
• Then N = (cosf,0, − 1) ×(0, − sinq,1) = sinqi − cosfj − cosfsinqk or (sinq, − cosf, − cosfsinq)
Hence [ curl(F)oP ] ×N = (1, − 1,0) ×(sinq, − cosf, − cosfsinq) = sinq + cosf
Find a parametrization P(t,u) for 4x − 3y + 2z = 3 and compute N = [dP/dt] ×[dP/du].
• To parametrize 4x − 3y + 2z = 3 we let x = t, y = u and z = f(x,y). Solving for z gives z = [3/2] − 2x + [3/2]y and so P(t,u) = ( t,u,[3/2] − 2t + [3/2]u ).
Now, [dP/dt] = (1,0, − 2) and [dP/du] = ( 0,1,[3/2] ) so N = [dP/dt] ×[dP/du] = 2i − [3/2]j + k or ( 2, − [3/2],1 )
Find a parametrization P(t,u) for z = 3 − x2 − y2 and compute N = [dP/dt] ×[dP/du].
• To parametrize z = 3 − x2 − y2 we let x = t, y = u and z = f(x,y) and so P(t,u) = (t,u,3 − t2 − u2).
Now, [dP/dt] = (1,0, − 2t) and [dP/du] = (0,1, − 2u) so N = [dP/dt] ×[dP/du] = 2ti + 2tj + k or (2t,2u,1)
Find the circulation of the vector field F(x,y,z) around the curve P(t,u) given that F = (x,y,x + y + z) and P = ( [t/2],[u/2],t + u ) for 0 ≤ t ≤ 1 and 0 ≤ u ≤ 1.
• We can use Stoke's Theorem to find the circulation of a vector field F around a curve P, that is computing dtdu where S is the surface of the region P faces counterclockwise.
• To integrate ∫0101 [ curl(F)oP ] ×N dtdu we first compute [ curl(F)oP ] ×N.
• First, curl(F) = ∇×F = 1i − 1j + 0k or (1, − 1,0), so curl(F)oP = (1, − 1,0)o( [t/2],[u/2],t + u ) = (1, − 1,0).
• Second, N = ( [1/2],0,1 ) ×( 0,[1/2],1 ) = − [1/2]i − [1/2]j + [1/4]k or ( − [1/2], − [1/2],[1/4] )
Hence [ curl(F)oP ] ×N = (1, − 1,0) ×( − [1/2], − [1/2],[1/4] ) = 0 and ∫0101 [ curl(F)oP ] ×N dtdu = 0
Find the circulation of the vector field F(x,y,z) around the curve P(t,u) given that F = (z + y,z − x,x + y) and P = ( u2,t2,1 ) for 0 ≤ t ≤ 1 and 0 ≤ u ≤ 1.
• We can use Stoke's Theorem to find the circulation of a vector field F around a curve P, that is computing dtdu where S is the surface of the region P faces counterclockwise.
• To integrate ∫0101 [ curl(F)oP ] ×N dtdu we first compute [ curl(F)oP ] ×N.
• First, curl(F) = ∇×F = 0i − 1j + 0k or (0, − 1,0), so curl(F)oP = (0, − 1,0)o( u2,t2,1 ) = (0, − 1,0).
• Second, N = ( 0,2t,0 ) ×( 2u,0,0 ) = 0i − 0j − 4tuk or ( 0,0, − 4tu )
• Hence [ curl(F)oP ] ×N = (0, − 1,0) ×( 0,0, − 4tu ) = 0.
We now integrate ∫0101 [ curl(F)oP ] ×N dtdu = ∫0101 0 dtdu = 0
Find the circulation of the vector field F(x,y,z) around the curve P(t,u) given that F = ( − x, − y,xyz) and P = ( [1/t],[1/u],t − u ) for 0.5 ≤ t ≤ 1 and 0.5 ≤ u ≤ 1.
• We can use Stoke's Theorem to find the circulation of a vector field F around a curve P, that is computing dtdu where S is the surface of the region P faces counterclockwise.
• To integrate ∫0.510.51 [ curl(F)oP ] ×N dtdu we first compute [ curl(F)oP ] ×N.
• First, curl(F) = ∇×F = xzi − yzj + 0k or (xz, − yz,0), so curl(F)oP = (xz, − yz,0)o( [1/t],[1/u],t − u ) = ( 1 − [u/t],1 − [t/u],0 ).
• Second, N = ( − [1/(t2)],0,1 ) ×( 0, − [1/(u2)], − 1 ) = − [1/(u2)]i − [1/(t2)]j + [1/(t2u2)]k or ( − [1/(u2)], − [1/(t2)],[1/(t2u2)] )
• Hence [ curl(F)oP ] ×N = ( 1 − [u/t],1 − [t/u],0 ) ×( − [1/(u2)], − [1/(t2)],[1/(t2u2)] ) = − [1/(t2)] + [2/tu] − [1/(u2)].
• We now integrate ∫0.510.51 [ curl(F)oP ] ×N dtdu = ∫0.510.51 ( − [1/(t2)] + [2/tu] − [1/(u2)] ) dtdu
Thus ∫0.510.51 ( − [1/(t2)] + [2/tu] − [1/(u2)] ) dtdu = ∫0.51 ( − 1 − [1/(2u2)] − [2/u]ln( [1/2] ) ) du = − 1 − 2[ ln(0.5) ]2

*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 1

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
• Stokes' Theorem 0:25
• Recall Circulation-Curl Version of Green's Theorem
• Constructing a Surface in 3-Space
• Stokes' Theorem
• Note on Curve and Vector Field in 3-Space
• Example 1: Find the Circulation of F around the Curve 12:40
• Part 1: Question
• Part 2: Drawing the Figure
• Part 3: Solution

### Transcription: Stokes' Theorem, Part 1

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

In the last lesson, we discussed the divergence theorem in 3-space. We generalized the flux divergence theorem, a version of Green's Theorem into 3-space.0004

Now we are going to discuss Stoke's Theorem, which in some sense is a generalization, is a... we are going to kick up a circulation curl form of Green's Theorem into 3-space.0012

So, let us go ahead and get started. Okay. So, let us start off by recalling the circulation curl form of Green's theorem.0023

So, recall, circulation curl version of Green's theorem. It looked like this.0031

The integral over that of f(c) · c' dt is equal to a, this time we have the curl dy dx, so what this basically says is that the circulation of a given vector field around a closed curve, so again we are talking about a region, that way we are traversing it in a counter-clockwise direction, keeping the region contained to our left.0052

The circulation of a vector field around this curve, as we move it around it from beginning to end, it is equal to the integral of the curl of that vector field over the area enclosed by that curve. That is it.0087

The flux concerned that version, that part of the vector field.0102

The circulation concerns this part of the vector field. To what extent is the vector field actually rotating vs. diverging.0107

That is all this is. Okay. So let us go ahead and draw a picture here. So we have this nice 3-dimensional picture.0119

So, now, what I am going to do. So, this is the x axis, this is the y axis and this is the z axis, so I am just going to draw a little region here, slightly triangular, looks kind of like a pizza.0126

This is our region and we are traversing it in a counter-clockwise direction, so this theorem actually applies to this region.0137

Now, what happens if I actually cut this out, cut out this region, lift it off the xy plane, so that now it is floating in 3-space, in xyz space.0147

So, I take that little piece and if I just deform it and turn it into a surface, here is what it looks like.0159

So, let us now lift this region out of the xy plane, deform it to construct a surface in 3-space.0167

What you get is something that looks like this. So, let me go ahead and draw this and that, and that, make this one a little bit longer, so again, I am not going to label the axes.0205

So what you have is something that looks like that... This goes that way... that way... that way...0218

So what I have done is I have taken this triangular region, which was this triangular region right here, and I have literally lifted it up, I have just peeled it off the page, and I have sort of deformed it, and I have actually turned it into some sort of a surface.0233

Well, on that surface there is a point. This can be parameterized. This is a parameterized surface. Our parameterization is going to be p(t,u).0250

This point on that surface is going to be some... some point, so that is some p. Well, there is a vector field on that surface, right?0262

There is vector fields here, here, here, right? Because a vector field, that is defined in a 3-dimensional region, is... they're are the points on the surface, we can actually apply it to that vector field.0270

So, we have f(p). In other words, the composition. Once we have a point, we can go ahead and take that point and put it into f, so there is some vector emanating from that point on that surface.0283

Well, we can also form the normal vector. That is all that is going on here. Notice, we still have a boundary, and we can still traverse this boundary, this time in 3-space. We are no longer in 2-space.0302

We can still traverse this boundary in the counter clockwise direction, except now, this boundary is not containing a region in 2-space. It is actually the boundary of a surface. That is all that is happening here.0317

So, this surface, is still a 2-dimensional object. The surface is still a 2-dimensional object, because that is what a surface is. It is still a 2-dimensional object. The surface has a boundary... and this surface has a boundary.0337

Okay. Now, if we traverse... let me draw the region again actually, just so we have it on this page as well. So, we had some sort of a triangular region, so let us say it's that, it's like that, yea, something like that.0368

If we traverse this boundary again, this boundary again, and this is our surface... If we traverse the boundary in a counter-clockwise direction, keeping the region to our left... clockwise direction... then, Stoke's Theorem says the following.0392

Stoke's Theorem says the integral over the boundary of f(c(t)) · c' dt is equal to... okay, here is where you have to be careful... the curl of f(p) · n dt du.0423

Okay, here is what this says. Remember when we did this surface integrals just in the last couple of lessons, we formed -- let me do this in red -- this surface, the integral of a given vector field over a surface, it was equal to this integral except instead of the curl we just did f(p) · n dt du.0466

Now, what we are doing. Instead of using our vector field, f, we are actually going to take the curl of that vector field.0495

Well, the curl of a vector field in 3-dimensions is another vector field, because you remember the curl of a vector field is a vector.0502

So at any given point, you are going to get another vector. So if I have one vector field, well at all the points on that surface where there is some vector, I can form the curl. So I get the curl vector field.0510

Instead of f, we are just using the curl of f. So, instead of forming f(p), we are forming the curl of f(p). That is it.0522

This is a derived vector field from the original f. That is what the curl is. It is a type of derivative. SO, we are deriving a new vector field from f.0533

What Stoke's Theorem says, it is actually the exact same thing as Green's Theorem, except now instead of a 2-dimensional flat region like this, all I have done is I have a surface in 3-space.0545

That surface still has a boundary. If I calculate the circulation of a vector field around the boundary of a surface in 3-space, it happens to equal the integral of the curl of the vector field over the surface itself.0556

This is a surface integral, this is a line integral. That is all this is. I have just generalized the circulation curl form of Green's Theorem. Now, into 3-dimensional space. I hope that is okay.0575

Now, a couple of things to note. This time, this c(t) in Green's Theorem, we were in R2, we were in 2-dimensional space, so our parameterization for a curve had 2 coordinate functions.0592

But, now, notice, this surface is in 3-space. Well, this boundary is still a curve and we can parameterize it with a curve except now it has 3 coordinate functions because now it is a curve in 3-space.0606

Note. So, now, c(t) is a curve in 3-space, not 2-space. It is a curve in 3-space. So it has 3 coordinate functions.0620

You will see what we mean in a minute when we actually start doing some examples. Okay.0643

f is, or I should say and, okay, and f is now a vector field in 3-space, not 2-space, so f has 3 coordinate functions. So, it has 3 coordinate functions.0648

Interestingly enough, Stoke's Theorem is the most natural generalization yet it is the strangest and most difficult to wrap your mind around, I think, because you are not accustomed to seeing a surface integral with a curl in it. That is what makes Stoke's Theorem different.0681

If you do not completely understand it, do not worry about it. Just deal symbolically. You deal symbolically until you actually -- you know -- get your sea legs with this stuff, then it is not a problem.0698

Just, find p, find the curl of f, form the composite function, form the dot product, put it in the integrand, find your domain of a parameterization for the integration, and just solve the integral until you get a feel for it.0707

Normally -- you know -- historically, Stoke's Theorem is the one that gives kids the biggest problem, because it seems to be the most unnatural.0724

Yet strangely enough, it is actually the most natural generalization, all you have done is taken a 2-dimensional region, imagine taking a piece of paper, which has the paper itself, the boundary is the edge, and just bending it like this.0732

When you have bent it, now you have created a surface. Well it is still a surface. It is still a flat 2-dimensional object. There is still a boundary around it, and now there is a surface, and there is a relationship between the integral of something around the boundary, and the integral over the surface.0742

Okay. Now, let us go ahead and just solve some problems. I think that will be the best way to handle this. So example 1.0760

Let p(t,Θ) = tcos(Θ), tsin(Θ), and Θ for t > or = 0, < or = 1, and Θ > or = 0, and < or = pi/2.0779

We will let f of xyz, so you notice... okay... equal to z, x, and y. Okay.0804

our task is to find the circulation of f around the curve. Let us draw this out and see what it looks like.0822

Okay. This parameterization, t goes from 0 to 1, Θ goes from 0 to pi/2. This is the parameterization of a spiral surface.0843

In other words, imagine a parking ramp. It is a spiral surface if I take a... if I am looking from the top, and I take some track like this, and if I split this and I lift it up, what I end up is having a track that goes up like that. A spiral.0856

So, what it actually looks like is it is this surface right here, actually spiraling up. Θ goes from 0 to pi/2.0879

This is y, this is x, so this is just that part of it. Only half of it.0898

t goes from 0 to 1. So, this is a length -- just, 1, 1, that is all it is.0905

Now, here is what is interesting. The boundary actually has 4 parts. There is this curve, c1, there is this curve, c2, there is this curve, which is c3, and there is this curve, which is c4.0915

Therefore, if I actually wanted to solve this. If I wanted to find the circulation of f around this curve, I have to parameterize 4 curves, c1, c2, c3, and c4.0940

I can do that, it is not a problem. I mean I can parameterize these things if I need to. But I do not need to because I have Stoke's Theorem at my disposal.0952

So, I am going to go ahead and solve these surface integrals. The surface is actually easily parameterized. In fact, they gave us the parameterization.0958

You definitely want to use Stoke's Theorem in this case.0967

Circulation... is this.0971

So, c = c1, c2, c3, c4, so it is the integral around c of f(c(t)) · c' dt. That is the circulation integral.0982

c is composed of these four curves. If I go all the way around again, I am traversing it this way. Keeping the region to my left. Well, I do not want to solve this via circulation.1004

So, if we solve the integral, if we solve the line integral directly, we will have to parameterize 4 curves. Parameterize 4 curves... and solve 4 integrals, which we definitely do not want to do.1016

Okay. But we have Stoke's Theorem, which says that f(c) · c' dt = so this is the curve, this is the surface, the curl of f(p) · n dt du.1052

Now, let us go ahead and run it through. Let us figure out what each of these is, put it in... okay.1085

So, let us go ahead and find n first. Let me do this in blue. So n, so n, let us go ahead and find dp dt.1095

When I take the partial of the parameterization with respect to t, I get cos(Θ)sin(Θ), and I get 0.1108

When I take the partial of the parameterization with respect to Θ, I get -tsin(Θ), I get tcos(Θ), and I get 1.1118

Well, n is equal to dp dt cross dp du -- oh, sorry, I already have... I do not need t and u, I need dp dt dp dΘ... There we go.1133

So, it is going to be dp dt cross dp dΘ, and when I go ahead and run that particular one, I end up with the following. I end up with sin(Θ) -cos(Θ), and t. That is n.1151

Now, when I take the curl, so I have taken care of n, now I need curl of f, well, the curl of f, I am going to actually do this one explicitly... i, j, k, d dx, d dy, d dz, of z, x, and y.1174

When I expand along the top row, I get 1, 1, and 1. So, that is the curl of f. Well, I need the curl of f(p). I need that. The curl of f is just... is (1,1,1). It is a constant vector.1202

There is no composition here, so I can just leave it as 1, 1, and 1. Well, the curl of f(p) dotted with the vector n is equal to sin(Θ) - cos(Θ) + t.1235

So, the integral that we are looking for, our circulation is t goes from 0 to 1, Θ goes from 0 to pi/2, our curl of f of p · n is sin(Θ) - cos(Θ) + t, and we did Θ first, so it is dΘ dt, and our final answer is pi/4.1265

There you go. So a circulation of this vector field happens to equal pi/4.1295

It is positive. What this means is that this vector field is actually rotating on that surface. That is what we have done. That is what Stoke's Theorem does.1300

Okay. Thank you for joining us here at Educator.com, for our first part of the discussion of Stoke's Theorem.1312

We will see you next time for the second part of the discussion of Stoke's Theorem. Bye-bye.1317