Stokes’ Theorem: Background
ƒ Similar to Green's Theorem: a line integral equals a surface integral.
[
]
[
]
Pdx
+
Qdy
=
Q
−
P
dxdy
x
y
∫C
∫∫R
R
C
ƒ The line integral is still the work around a curve:
r r
∫ F • dr ,
C
r
F = Piˆ + Qˆj
ƒ The surface integral in Green's Theorem is
⎡ ∂Q ∂P ⎤
∫∫R ⎢⎣ ∂x − ∂y ⎥⎦ dxdy
ƒ R is flat (in the xy plane). Its normal direction is k. (Qx - Py) is the
k component of the curl.
ƒ Green's Theorem uses only this component because the normal
direction is always k.
ƒ For Stokes' Theorem on a curved S, we need all three components
of curl F.
1
Stokes’ Theorem (Cont.)
ƒ Stokes’ Theorem: a line integral equals a surface integral in 3D.
n and C follow RHR
r r
r
∫ F • dr = ∫∫ (∇ × F ) • nˆdS
C
S
ƒ The total flux of the curl of a
vector field, F, through a surface
equals to the line integral of the
vector field, F, around the edge.
ƒ The right side adds up small spins in the surface. The left side is
the total circulation (or work) around C.
ƒ If curl F = 0, then the line integral is zero.
This applies above all to conservative fields.
2
Stokes’ Theorem: conservative field
ƒ A conservative field F has no curl and does no work over a
closed path:
r
r
r
∫ F • dr = 0
C
& ∇×F = 0
ƒ Since ∇ x (∇f) = 0, a gradient field has no curl.
ƒ Therefore conservative fields are gradient fields:
r
r
r
if ∇ × F = 0 then F can be expressed as F = ∇f
ƒ Stokes' Theorem is used in electricity and magnetism to yield
Faraday's Law.
r r
Faraday' s Law (integral form) : ∫ E.dr = − ∂φ ∂t
C
r
r
Faraday' s Law (different ial form) : ∇ × E = − ∂B ∂t
3
r r
∫ F • dr
• Example (P9-14.9): Evaluate the integral
C
using Stokes’ theorem, where
r
F = y 3 iˆ − x 3 ˆj + z 3 kˆ ,
Step 2
C : CCW , trace of x 2 + y 2 = 1
in the plane x + y + z = 1
Step 1
r r
r
∫ F • dr = ∫∫ (∇ × F ) • nˆdS
C
S
3
r
2
2 ˆ
∇ × F = −3( x + y )k , nˆ =< 1,1,1 > / 3
4
dS = 1 + ∂z
5
(
I =……
)
2
⎛
⎞
∂
z
+⎜
dA = 3 dA
⎟
∂x
⎝ ∂y ⎠
2
Complete the solution
4
• Example (P9-14.3): Verify Stokes’ theorem
r
F = ziˆ + xˆj + ykˆ,
S : portion of 2 x + y + 2 z = 6 in the first octant
r r
r
ˆ
F
•
d
r
=
(
∇
×
F
)
•
n
dS
∫
∫∫
C
S
(
dS = 1 + ∂z
)
+ ⎛⎜ ∂z ⎞⎟ dA
∂x
⎝ ∂y ⎠
2
2
1.
2.
nˆ = ∇g / | ∇g |
3.
Sketch the surface
4.
Calculate LHS and RHS, and show they are equal
5
Example (P9-14.13):r Evaluate the integral
r
(∇ × F ) • nˆ dS
∫∫x ˆ
S
ˆ
ˆ
F = 6 yzi + 5 xj + yze k ,
2
S : upward , z = 0.25 x + y , 0 ≤ z ≤ 4
2
1
∫∫
S
2
r
r r
(∇ × F ) • nˆ dS = ∫ F • dr
C
C
2
3
4
6