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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
J. Robert Buchanan Stokes’ Theorem
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.
J. Robert Buchanan Stokes’ Theorem
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.
J. Robert Buchanan Stokes’ Theorem
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.
J. Robert Buchanan Stokes’ Theorem
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.
J. Robert Buchanan Stokes’ Theorem
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
J. Robert Buchanan Stokes’ Theorem
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〉.
J. Robert Buchanan Stokes’ Theorem
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
J. Robert Buchanan Stokes’ Theorem
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
J. Robert Buchanan Stokes’ Theorem
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
J. Robert Buchanan Stokes’ Theorem
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.
J. Robert Buchanan Stokes’ Theorem
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
J. Robert Buchanan Stokes’ Theorem
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
J. Robert Buchanan Stokes’ Theorem
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
J. Robert Buchanan Stokes’ Theorem
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
J. Robert Buchanan Stokes’ Theorem
Interpretation of the Curl (2 of 3)
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
J. Robert Buchanan Stokes’ Theorem
Interpretation of the Curl (2 of 3)
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
J. Robert Buchanan Stokes’ Theorem
Interpretation of the Curl (3 of 3)
[(∇× 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.
J. Robert Buchanan Stokes’ Theorem
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.
J. Robert Buchanan Stokes’ Theorem
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〉.
J. Robert Buchanan Stokes’ Theorem
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〉.
J. Robert Buchanan Stokes’ Theorem
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.
J. Robert Buchanan Stokes’ Theorem
Proof (1 of 2)
Suppose ∇× F = 0.According to Stokes’ Theorem∮
CF · dr =
∫∫S(∇× F) · n dS =
∫∫S
0 dS = 0.
J. Robert Buchanan Stokes’ Theorem
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.
J. Robert Buchanan Stokes’ Theorem
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.
J. Robert Buchanan Stokes’ Theorem
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〉
J. Robert Buchanan Stokes’ Theorem
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〉
J. Robert Buchanan Stokes’ Theorem
Homework
Read Section 14.8.Exercises: 1–33 odd
J. Robert Buchanan Stokes’ Theorem