Stokes' Theorem - MATH 311, Calculus IIIbanach.millersville.edu/~bob/math311/Stokes/main.pdf · Stokes’ Theorem MATH 311, Calculus III J. Robert Buchanan Department of Mathematics

Stokes’ TheoremMATH 311, Calculus III

J. Robert Buchanan

Department of Mathematics

Summer 2011

J. Robert Buchanan Stokes’ Theorem

Background (1 of 2)

Recall: Green’s Theorem,∮C

M(x , y) dx + N(x , y) dy =

∫∫R

(∂N∂x− ∂M∂y

)dA

where C is a piecewise smooth, positively oriented, simpleclosed curve in the xy -plane enclosing region R.

Define the vector field F(x , y) = 〈M(x , y),N(x , y),0〉.Note that ∇× F = 〈0,0, ∂N

∂x −∂M∂y 〉.

The component of ∇× F along k is

(∇× F) · k =

(∂N∂x− ∂M∂y

)k · k =

∂N∂x− ∂M∂y


Background (2 of 2)

Thus we have developed the vector form of Green’s Theorem∮C

M(x , y) dx + N(x , y) dy =

∫∫R

(∂N∂x− ∂M∂y

)dA∮

CF · dr =

∫∫R(∇× F) · k dA

Today we will generalize this result to three dimensions.


Curve Orientation

C

n

S

-1.0-0.5

0.0

0.5

1.0

x

-1.0

-0.5

0.0

0.5

1.0

y

0.0

0.5

1.0

1.5

2.0

z

Using the right-hand rule, curve C has positive orientation if ithas the same orientation as the right hand’s fingers when theright thumb points in the direction of the normal n to surface S.Otherwise C has negative orientation.


Stokes’ Theorem

Theorem (Stokes’ Theorem)Suppose that S is an oriented, piecewise-smooth surface withunit normal vector n, bounded by the simple closed,piecewise-smooth boundary curve ∂S having positiveorientation. Let F(x , y , z) be a vector field whose componentshave continuous first partial derivatives in some open regioncontaining S. Then,∮

∂SF(x , y , z) · dr =

∫∫S(∇× F) · n dS.


Interpretation

Recalling that the unit tangent vector is

T(t) =r′(t)‖r′(t)‖

⇐⇒ r′(t) = ‖r′(t)‖T(t)

and that differential arc length is defined as

ds = ‖r′(t)‖dt ,

sometimes Stokes’ Theorem is written as∫∫S(∇× F) · n dS =

∮∂S

F(x , y , z) · dr =

∮∂S

F · T ds

where T is the unit tangent in the direction of ∂S.

Interpretation: The line integral of the tangential component ofF is equal to the flux of the curl of F. This integral is the averagetendency of the flow of F to rotate around path ∂S.


Example (1 of 4)

Evaluate∮∂S

F · dr where F(x , y , z) = −y2i + x j + z2k and ∂S

is the intersection of the plane y + z = 2 and the cylinderx2 + y2 = 1.

-1

0

1

x

-1

0

1

y

1

2

3

z


Example (2 of 4)

Note that ∇× F = (1 + 2y)k.There are many surfaces which have ∂S as a boundary,choose the elliptical disk bounded by ∂S in the planey + z = 2.

The unit normal to this surface is n =1√2〈0,1,1〉.


Example (3 of 4)

-1

0

1

x

-1

0

1

y

1

2

3

z

S

C

n

-1.0-0.5

0.00.5

1.0 x

-1.0-0.5

0.00.5

1.0y

1.0

1.5

2.0

2.5

3.0

z


Example (4 of 4)

According to Stokes’ Theorem∮∂S

F · dr =

∫∫S(∇× F) · n dS

=

∫∫S(1 + 2y)k · 1√

2〈0,1,1〉dS

=

∫∫S

1√2(1 + 2y) dS

=

∫∫R(1 + 2y) dA

=

∫ 2π

0

∫ 1

0(1 + 2r sin θ)r dr dθ

=

∫ 2π

0

∫ 1

0r dr dθ

= π


Example (1 of 4)

Evaluate∫∫

S(∇× F) · ndS where F(x , y , z) = yzi + xzj + xyk

and S is the part of the sphere x2 + y2 + z2 = 4 that lies insidethe cylinder x2 + y2 = 1 and above the xy -plane.

-2

-1

0

1

2

x

-2

-1

0

1

2

y

0.0

0.5

1.0

1.5

2.0

z


Example (2 of 4)

Surface S is bounded by a circle formed by the intersectionof the sphere of radius 2 and the cylinder of radius 1.We can describe ∂S using the vector-valued function

r(t) = 〈cos t , sin t ,√

3〉,

with 0 ≤ t ≤ 2π.


Example (3 of 4)

-2

-1

0

1

2

x

-2

-1

0

1

2

y

0.0

0.5

1.0

1.5

2.0

zS

n

C

-1.0

-0.5

0.0

0.5

1.0

x

-1.0

-0.5

0.0

0.5

1.0

y

2.0

2.5

3.0

z


Example (4 of 4)

According to Stokes’ Theorem∫∫S(∇× F) · n dS

=

∮∂S

F · dr

=

∫ 2π

0F(cos t , sin t ,

√3) · 〈− sin t , cos t ,0〉dt

=√

3∫ 2π

0(cos2 t − sin2 t) dt

=√

3∫ 2π

0cos 2t dt

= 0


An Identity

Show that∮

C(f ∇f ) · dr = 0.

∮C(f ∇f ) · dr =

∫∫S

(∇× (f ∇f )) · n dS

=

∫∫S

(∇×

⟨f∂f∂x, f∂f∂y, f∂f∂z

⟩)· n dS

=

∫∫S

0 · n dS

= 0


Interpretation of the Curl (1 of 3)

Let P = (x0, y0, z0) be any point in a vector field F and let Sa bea circular disk of radius a > 0 centered at P. Let Ca be theboundary of Sa.

n

P

Ca

Sa



Average value of (∇× F) · n on surface Sa:

[(∇× F) · n]|Pa=

1πa2

∫∫Sa

(∇× F) · n dS

where πa2 is the area of Sa and point Pa ∈ Sa by the IntegralMean Value Theorem.

By Stokes’ Theorem∮Ca

F · dr =

∫∫Sa

(∇× F) · n dS

= πa2 [(∇× F) · n]|Pa



Average value of (∇× F) · n on surface Sa:

[(∇× F) · n]|Pa=

1πa2

∫∫Sa

(∇× F) · n dS

where πa2 is the area of Sa and point Pa ∈ Sa by the IntegralMean Value Theorem.

By Stokes’ Theorem∮Ca

F · dr =

∫∫Sa

(∇× F) · n dS

= πa2 [(∇× F) · n]|Pa



[(∇× F) · n]|Pa=

1πa2

∮Ca

F · dr

lima→0+

[(∇× F) · n]|Pa= lim

a→0+

1πa2

∮Ca

F · dr

[(∇× F) · n]|P = lima→0+

1πa2

∮Ca

F · T ds

If F describes the flow of a fluid then∮

CF · T ds is the

circulation around C, the average tendency of the fluid tocirculate around the curve.


Remarks

[(∇× F) · n]|P attains its maximum when ∇× F is parallelto n.We can define rot F = (∇× F) · n. This quantity is therotation of the vector field at a point.F is an irrotational vector field if and only if ∇× F = 0.


Example

Suppose F = 〈3y ,4z,−6x〉. Find the direction of the maximumvalue of (∇× F) · n.

Since ∇× F = 〈−4,6,−3〉, the maximum of (∇× F) · n willoccur in the direction of

n =〈−4,6,−3〉‖〈−4,6,−3〉‖

=1√61〈−4,6,−3〉.


Example

Suppose F = 〈3y ,4z,−6x〉. Find the direction of the maximumvalue of (∇× F) · n.

Since ∇× F = 〈−4,6,−3〉, the maximum of (∇× F) · n willoccur in the direction of

n =〈−4,6,−3〉‖〈−4,6,−3〉‖

=1√61〈−4,6,−3〉.


Irrotational Vector Fields

TheoremSuppose that F(x , y , z) has continuous partial derivativesthroughout a simply connected region D, then ∇× F = 0 in D if

and only if∮

CF · dr = 0 for every simple closed curve C in D.


Proof (1 of 2)

Suppose ∇× F = 0.According to Stokes’ Theorem∮

CF · dr =

∫∫S(∇× F) · n dS =

∫∫S

0 dS = 0.


Proof (2 of 2)

Suppose∮

C F · dr = 0 for every simple closed curve C.If at some point P, (∇× F) 6= 0, then by continuity thereexists a subregion of D on which (∇× F) 6= 0.In the subregion choose a circular disk whose normal n isparallel to ∇× F.∮

CF · dr =

∫∫S(∇× F) · n dS 6= 0

which contradicts our first assumption.


Inclusive Result

TheoremSuppose that F(x , y , z) has continuous first partial derivativesthroughout a simply connected region D, then the followingstatements are equivalent.

1 F is conservative in D, i.e. F = ∇f .

2

∫C

F · dr is independent of path in D.

3 F is irrotational in D, i.e. ∇× F = 0 in D.

4

∮C

F · dr = 0 for every simple closed curve in D.


Example

Let F(x , y , z) = y2i + (2xy + e3z)j + 3ye3zk and show that F isconservative.

We can accomplish this by showing that ∇× F = 0.

∇× F =

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

y2 2xy + e3z 3ye3z

∣∣∣∣∣∣ = 〈0,0,0〉


Example

Let F(x , y , z) = y2i + (2xy + e3z)j + 3ye3zk and show that F isconservative.

We can accomplish this by showing that ∇× F = 0.

∇× F =

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

y2 2xy + e3z 3ye3z

∣∣∣∣∣∣ = 〈0,0,0〉


Homework

Read Section 14.8.Exercises: 1–33 odd


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Stokes' Theorem - MATH 311, Calculus IIIbanach.millersville.edu/~bob/math311/Stokes/main.pdf · Stokes’ Theorem MATH 311, Calculus III J. Robert Buchanan Department of Mathematics