  Raffi Hovasapian

Stokes' Theorem, Part 1

Slide Duration:

Section 1: Vectors
Points & Vectors

28m 23s

Intro
0:00
Points and Vectors
1:02
A Point in a Plane
1:03
A Point in Space
3:14
Notation for a Space of a Given Space
6:34
Introduction to Vectors
9:51
14:51
Example 1
16:52
18:24
Example 2
21:01
Two More Properties of Vector Addition
24:16
Multiplication of a Vector by a Constant
25:27
Scalar Product & Norm

30m 25s

Intro
0:00
Scalar Product and Norm
1:05
Introduction to Scalar Product
1:06
Example 1
3:21
Properties of Scalar Product
6:14
Definition: Orthogonal
11:41
Example 2: Orthogonal
14:19
Definition: Norm of a Vector
15:30
Example 3
19:37
Distance Between Two Vectors
22:05
Example 4
27:19
More on Vectors & Norms

38m 18s

Intro
0:00
More on Vectors and Norms
0:38
Open Disc
0:39
Close Disc
3:14
Open Ball, Closed Ball, and the Sphere
5:22
Property and Definition of Unit Vector
7:16
Example 1
14:04
Three Special Unit Vectors
17:24
General Pythagorean Theorem
19:44
Projection
23:00
Example 2
28:35
Example 3
35:54
Inequalities & Parametric Lines

33m 19s

Intro
0:00
Inequalities and Parametric Lines
0:30
Starting Example
0:31
Theorem 1
5:10
Theorem 2
7:22
Definition 1: Parametric Equation of a Straight Line
10:16
Definition 2
17:38
Example 1
21:19
Example 2
25:20
Planes

29m 59s

Intro
0:00
Planes
0:18
Definition 1
0:19
Example 1
7:04
Example 2
12:45
General Definitions and Properties: 2 Vectors are Said to Be Paralleled If
14:50
Example 3
16:44
Example 4
20:17
More on Planes

34m 18s

Intro
0:00
More on Planes
0:25
Example 1
0:26
Distance From Some Point in Space to a Given Plane: Derivation
10:12
Final Formula for Distance
21:20
Example 2
23:09
Example 3: Part 1
26:56
Example 3: Part 2
31:46
Section 2: Differentiation of Vectors
Maps, Curves & Parameterizations

29m 48s

Intro
0:00
Maps, Curves and Parameterizations
1:10
Recall
1:11
Looking at y = x2 or f(x) = x2
2:23
Departure Space & Arrival Space
7:01
Looking at a 'Function' from ℝ to ℝ2
10:36
Example 1
14:50
Definition 1: Parameterized Curve
17:33
Example 2
21:56
Example 3
25:16
Differentiation of Vectors

39m 40s

Intro
0:00
Differentiation of Vectors
0:18
Example 1
0:19
Definition 1: Velocity of a Curve
1:45
Line Tangent to a Curve
6:10
Example 2
7:40
Definition 2: Speed of a Curve
12:18
Example 3
13:53
Definition 3: Acceleration Vector
16:37
Two Definitions for the Scalar Part of Acceleration
17:22
Rules for Differentiating Vectors: 1
19:52
Rules for Differentiating Vectors: 2
21:28
Rules for Differentiating Vectors: 3
22:03
Rules for Differentiating Vectors: 4
24:14
Example 4
26:57
Section 3: Functions of Several Variables
Functions of Several Variable

29m 31s

Intro
0:00
Length of a Curve in Space
0:25
Definition 1: Length of a Curve in Space
0:26
Extended Form
2:06
Example 1
3:40
Example 2
6:28
Functions of Several Variable
8:55
Functions of Several Variable
8:56
General Examples
11:11
Graph by Plotting
13:00
Example 1
16:31
Definition 1
18:33
Example 2
22:15
Equipotential Surfaces
25:27
Isothermal Surfaces
27:30
Partial Derivatives

23m 31s

Intro
0:00
Partial Derivatives
0:19
Example 1
0:20
Example 2
5:30
Example 3
7:48
Example 4
9:19
Definition 1
12:19
Example 5
14:24
Example 6
16:14
20:26
Higher and Mixed Partial Derivatives

30m 48s

Intro
0:00
Higher and Mixed Partial Derivatives
0:45
Definition 1: Open Set
0:46
Notation: Partial Derivatives
5:39
Example 1
12:00
Theorem 1
14:25
Now Consider a Function of Three Variables
16:50
Example 2
20:09
Caution
23:16
Example 3
25:42
Section 4: Chain Rule and The Gradient
The Chain Rule

28m 3s

Intro
0:00
The Chain Rule
0:45
Conceptual Example
0:46
Example 1
5:10
The Chain Rule
10:11
Example 2: Part 1
19:06
Example 2: Part 2 - Solving Directly
25:26
Tangent Plane

42m 25s

Intro
0:00
Tangent Plane
1:02
Tangent Plane Part 1
1:03
Tangent Plane Part 2
10:00
Tangent Plane Part 3
18:18
Tangent Plane Part 4
21:18
Definition 1: Tangent Plane to a Surface
27:46
Example 1: Find the Equation of the Plane Tangent to the Surface
31:18
Example 2: Find the Tangent Line to the Curve
36:54
Further Examples with Gradients & Tangents

47m 11s

Intro
0:00
Example 1: Parametric Equation for the Line Tangent to the Curve of Two Intersecting Surfaces
0:41
Part 1: Question
0:42
Part 2: When Two Surfaces in ℝ3 Intersect
4:31
Part 3: Diagrams
7:36
Part 4: Solution
12:10
Part 5: Diagram of Final Answer
23:52
Example 2: Gradients & Composite Functions
26:42
Part 1: Question
26:43
Part 2: Solution
29:21
Example 3: Cos of the Angle Between the Surfaces
39:20
Part 1: Question
39:21
Part 2: Definition of Angle Between Two Surfaces
41:04
Part 3: Solution
42:39
Directional Derivative

41m 22s

Intro
0:00
Directional Derivative
0:10
Rate of Change & Direction Overview
0:11
Rate of Change : Function of Two Variables
4:32
Directional Derivative
10:13
Example 1
18:26
Examining Gradient of f(p) ∙ A When A is a Unit Vector
25:30
Directional Derivative of f(p)
31:03
33:23
Example 2
34:53
A Unified View of Derivatives for Mappings

39m 41s

Intro
0:00
A Unified View of Derivatives for Mappings
1:29
Derivatives for Mappings
1:30
Example 1
5:46
Example 2
8:25
Example 3
12:08
Example 4
14:35
Derivative for Mappings of Composite Function
17:47
Example 5
22:15
Example 6
28:42
Section 5: Maxima and Minima
Maxima & Minima

36m 41s

Intro
0:00
Maxima and Minima
0:35
Definition 1: Critical Point
0:36
Example 1: Find the Critical Values
2:48
Definition 2: Local Max & Local Min
10:03
Theorem 1
14:10
Example 2: Local Max, Min, and Extreme
18:28
Definition 3: Boundary Point
27:00
Definition 4: Closed Set
29:50
Definition 5: Bounded Set
31:32
Theorem 2
33:34
Further Examples with Extrema

32m 48s

Intro
0:00
Further Example with Extrema
1:02
Example 1: Max and Min Values of f on the Square
1:03
Example 2: Find the Extreme for f(x,y) = x² + 2y² - x
10:44
Example 3: Max and Min Value of f(x,y) = (x²+ y²)⁻¹ in the Region (x -2)²+ y² ≤ 1
17:20
Lagrange Multipliers

32m 32s

Intro
0:00
Lagrange Multipliers
1:13
Theorem 1
1:14
Method
6:35
Example 1: Find the Largest and Smallest Values that f Achieves Subject to g
9:14
Example 2: Find the Max & Min Values of f(x,y)= 3x + 4y on the Circle x² + y² = 1
22:18
More Lagrange Multiplier Examples

27m 42s

Intro
0:00
Example 1: Find the Point on the Surface z² -xy = 1 Closet to the Origin
0:54
Part 1
0:55
Part 2
7:37
Part 3
10:44
Example 2: Find the Max & Min of f(x,y) = x² + 2y - x on the Closed Disc of Radius 1 Centered at the Origin
16:05
Part 1
16:06
Part 2
19:33
Part 3
23:17
Lagrange Multipliers, Continued

31m 47s

Intro
0:00
Lagrange Multipliers
0:42
First Example of Lesson 20
0:44
Let's Look at This Geometrically
3:12
Example 1: Lagrange Multiplier Problem with 2 Constraints
8:42
Part 1: Question
8:43
Part 2: What We Have to Solve
15:13
Part 3: Case 1
20:49
Part 4: Case 2
22:59
Part 5: Final Solution
25:45
Section 6: Line Integrals and Potential Functions
Line Integrals

36m 8s

Intro
0:00
Line Integrals
0:18
Introduction to Line Integrals
0:19
Definition 1: Vector Field
3:57
Example 1
5:46
Example 2: Gradient Operator & Vector Field
8:06
Example 3
12:19
Vector Field, Curve in Space & Line Integrals
14:07
Definition 2: F(C(t)) ∙ C'(t) is a Function of t
17:45
Example 4
18:10
Definition 3: Line Integrals
20:21
Example 5
25:00
Example 6
30:33
More on Line Integrals

28m 4s

Intro
0:00
More on Line Integrals
0:10
Line Integrals Notation
0:11
Curve Given in Non-parameterized Way: In General
4:34
Curve Given in Non-parameterized Way: For the Circle of Radius r
6:07
Curve Given in Non-parameterized Way: For a Straight Line Segment Between P & Q
6:32
The Integral is Independent of the Parameterization Chosen
7:17
Example 1: Find the Integral on the Ellipse Centered at the Origin
9:18
Example 2: Find the Integral of the Vector Field
16:26
Discussion of Result and Vector Field for Example 2
23:52
Graphical Example
26:03
Line Integrals, Part 3

29m 30s

Intro
0:00
Line Integrals
0:12
Piecewise Continuous Path
0:13
Closed Path
1:47
Example 1: Find the Integral
3:50
The Reverse Path
14:14
Theorem 1
16:18
Parameterization for the Reverse Path
17:24
Example 2
18:50
Line Integrals of Functions on ℝn
21:36
Example 3
24:20
Potential Functions

40m 19s

Intro
0:00
Potential Functions
0:08
Definition 1: Potential Functions
0:09
Definition 2: An Open Set S is Called Connected if…
5:52
Theorem 1
8:19
Existence of a Potential Function
11:04
Theorem 2
18:06
Example 1
22:18
Contrapositive and Positive Form of the Theorem
28:02
The Converse is Not Generally True
30:59
Our Theorem
32:55
Compare the n-th Term Test for Divergence of an Infinite Series
36:00
So for Our Theorem
38:16
Potential Functions, Continued

31m 45s

Intro
0:00
Potential Functions
0:52
Theorem 1
0:53
Example 1
4:00
Theorem in 3-Space
14:07
Example 2
17:53
Example 3
24:07
Potential Functions, Conclusion & Summary

28m 22s

Intro
0:00
Potential Functions
0:16
Theorem 1
0:17
In Other Words
3:25
Corollary
5:22
Example 1
7:45
Theorem 2
11:34
Summary on Potential Functions 1
15:32
Summary on Potential Functions 2
17:26
Summary on Potential Functions 3
18:43
Case 1
19:24
Case 2
20:48
Case 3
21:35
Example 2
23:59
Section 7: Double Integrals
Double Integrals

29m 46s

Intro
0:00
Double Integrals
0:52
Introduction to Double Integrals
0:53
Function with Two Variables
3:39
Example 1: Find the Integral of xy³ over the Region x ϵ[1,2] & y ϵ[4,6]
9:42
Example 2: f(x,y) = x²y & R be the Region Such That x ϵ[2,3] & x² ≤ y ≤ x³
15:07
Example 3: f(x,y) = 4xy over the Region Bounded by y= 0, y= x, and y= -x+3
19:20
Polar Coordinates

36m 17s

Intro
0:00
Polar Coordinates
0:50
Polar Coordinates
0:51
Example 1: Let (x,y) = (6,√6), Convert to Polar Coordinates
3:24
Example 2: Express the Circle (x-2)² + y² = 4 in Polar Form.
5:46
Graphing Function in Polar Form.
10:02
Converting a Region in the xy-plane to Polar Coordinates
14:14
Example 3: Find the Integral over the Region Bounded by the Semicircle
20:06
Example 4: Find the Integral over the Region
27:57
Example 5: Find the Integral of f(x,y) = x² over the Region Contained by r= 1 - cosθ
32:55
Green's Theorem

38m 1s

Intro
0:00
Green's Theorem
0:38
Introduction to Green's Theorem and Notations
0:39
Green's Theorem
3:17
Example 1: Find the Integral of the Vector Field around the Ellipse
8:30
Verifying Green's Theorem with Example 1
15:35
A More General Version of Green's Theorem
20:03
Example 2
22:59
Example 3
26:30
Example 4
32:05
Divergence & Curl of a Vector Field

37m 16s

Intro
0:00
Divergence & Curl of a Vector Field
0:18
Definitions: Divergence(F) & Curl(F)
0:19
Example 1: Evaluate Divergence(F) and Curl(F)
3:43
Properties of Divergence
9:24
Properties of Curl
12:24
Two Versions of Green's Theorem: Circulation - Curl
17:46
Two Versions of Green's Theorem: Flux Divergence
19:09
Circulation-Curl Part 1
20:08
Circulation-Curl Part 2
28:29
Example 2
32:06
Divergence & Curl, Continued

33m 7s

Intro
0:00
Divergence & Curl, Continued
0:24
Divergence Part 1
0:25
Divergence Part 2: Right Normal Vector and Left Normal Vector
5:28
Divergence Part 3
9:09
Divergence Part 4
13:51
Divergence Part 5
19:19
Example 1
23:40
Final Comments on Divergence & Curl

16m 49s

Intro
0:00
Final Comments on Divergence and Curl
0:37
Several Symbolic Representations for Green's Theorem
0:38
Circulation-Curl
9:44
Flux Divergence
11:02
Closing Comments on Divergence and Curl
15:04
Section 8: Triple Integrals
Triple Integrals

27m 24s

Intro
0:00
Triple Integrals
0:21
Example 1
2:01
Example 2
9:42
Example 3
15:25
Example 4
20:54
Cylindrical & Spherical Coordinates

35m 33s

Intro
0:00
Cylindrical and Spherical Coordinates
0:42
Cylindrical Coordinates
0:43
When Integrating Over a Region in 3-space, Upon Transformation the Triple Integral Becomes..
4:29
Example 1
6:27
The Cartesian Integral
15:00
Introduction to Spherical Coordinates
19:44
Reason It's Called Spherical Coordinates
22:49
Spherical Transformation
26:12
Example 2
29:23
Section 9: Surface Integrals and Stokes' Theorem
Parameterizing Surfaces & Cross Product

41m 29s

Intro
0:00
Parameterizing Surfaces
0:40
Describing a Line or a Curve Parametrically
0:41
Describing a Line or a Curve Parametrically: Example
1:52
Describing a Surface Parametrically
2:58
Describing a Surface Parametrically: Example
5:30
Recall: Parameterizations are not Unique
7:18
Example 1: Sphere of Radius R
8:22
Example 2: Another P for the Sphere of Radius R
10:52
This is True in General
13:35
Example 3: Paraboloid
15:05
Example 4: A Surface of Revolution around z-axis
18:10
Cross Product
23:15
Defining Cross Product
23:16
Example 5: Part 1
28:04
Example 5: Part 2 - Right Hand Rule
32:31
Example 6
37:20
Tangent Plane & Normal Vector to a Surface

37m 6s

Intro
0:00
Tangent Plane and Normal Vector to a Surface
0:35
Tangent Plane and Normal Vector to a Surface Part 1
0:36
Tangent Plane and Normal Vector to a Surface Part 2
5:22
Tangent Plane and Normal Vector to a Surface Part 3
13:42
Example 1: Question & Solution
17:59
Example 1: Illustrative Explanation of the Solution
28:37
Example 2: Question & Solution
30:55
Example 2: Illustrative Explanation of the Solution
35:10
Surface Area

32m 48s

Intro
0:00
Surface Area
0:27
Introduction to Surface Area
0:28
Given a Surface in 3-space and a Parameterization P
3:31
Defining Surface Area
7:46
Curve Length
10:52
Example 1: Find the Are of a Sphere of Radius R
15:03
Example 2: Find the Area of the Paraboloid z= x² + y² for 0 ≤ z ≤ 5
19:10
Example 2: Writing the Answer in Polar Coordinates
28:07
Surface Integrals

46m 52s

Intro
0:00
Surface Integrals
0:25
Introduction to Surface Integrals
0:26
General Integral for Surface Are of Any Parameterization
3:03
Integral of a Function Over a Surface
4:47
Example 1
9:53
Integral of a Vector Field Over a Surface
17:20
Example 2
22:15
Side Note: Be Very Careful
28:58
Example 3
30:42
Summary
43:57
Divergence & Curl in 3-Space

23m 40s

Intro
0:00
Divergence and Curl in 3-Space
0:26
Introduction to Divergence and Curl in 3-Space
0:27
Define: Divergence of F
2:50
Define: Curl of F
4:12
The Del Operator
6:25
Symbolically: Div(F)
9:03
Symbolically: Curl(F)
10:50
Example 1
14:07
Example 2
18:01
Divergence Theorem in 3-Space

34m 12s

Intro
0:00
Divergence Theorem in 3-Space
0:36
Green's Flux-Divergence
0:37
Divergence Theorem in 3-Space
3:34
Note: Closed Surface
6:43
Figure: Paraboloid
8:44
Example 1
12:13
Example 2
18:50
Recap for Surfaces: Introduction
27:50
Recap for Surfaces: Surface Area
29:16
Recap for Surfaces: Surface Integral of a Function
29:50
Recap for Surfaces: Surface Integral of a Vector Field
30:39
Recap for Surfaces: Divergence Theorem
32:32
Stokes' Theorem, Part 1

22m 1s

Intro
0:00
Stokes' Theorem
0:25
Recall Circulation-Curl Version of Green's Theorem
0:26
Constructing a Surface in 3-Space
2:26
Stokes' Theorem
5:34
Note on Curve and Vector Field in 3-Space
9:50
Example 1: Find the Circulation of F around the Curve
12:40
Part 1: Question
12:48
Part 2: Drawing the Figure
13:56
Part 3: Solution
16:08
Stokes' Theorem, Part 2

20m 32s

Intro
0:00
Example 1: Calculate the Boundary of the Surface and the Circulation of F around this Boundary
0:30
Part 1: Question
0:31
Part 2: Drawing the Figure
2:02
Part 3: Solution
5:24
Example 2: Calculate the Boundary of the Surface and the Circulation of F around this Boundary
13:11
Part 1: Question
13:12
Part 2: Solution
13:56
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• ## Related Books 1 answer Last 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

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