Stokes theorem and related theorems

LINE, SURFACE AND VOLUME INTEGRALS
11.9 Stokes’ theorem and related theorems
Stokes’ theorem is the ‘curl analogue’ of the divergence theorem and relates the
integral of the curl of a vector field over an open surface S to the line integral of
the vector field around the perimeter C bounding the surface.
Following the same lines as for the derivation of the divergence theorem, we
can divide the surface S into many small areas Si with boundaries Ci and unit
normals n̂i . Using (11.17), we have for each small area
0
(∇ × a) · n̂i Si ≈
a · dr.
Ci
Summing over i we find that on the RHS all parts of all interior boundaries
that are not part of C are included twice, being traversed in opposite directions
on each occasion and thus contributing nothing. Only contributions from line
elements that are also parts of C survive. If each Si is allowed to tend to zero
then we obtain Stokes’ theorem,
0
(∇ × a) · dS =
a · dr.
(11.23)
S
C
We note that Stokes’ theorem holds for both simply and multiply connected open
surfaces, provided that they are two-sided. Stokes’ theorem may also be extended
to tensor fields (see chapter 26).
Just as the divergence theorem (11.18) can be used to relate volume and surface
integrals for certain types of integrand,
Stokes’ theorem can be used in evaluating
/
surface integrals of the form S (∇ × a) · dS as line integrals or vice versa.
Given the vector field a = y i − x j + z k, verify Stokes’ theorem for the hemispherical
surface x2 + y 2 + z 2 = a2 , z ≥ 0.
Let us first evaluate the surface integral
(∇ × a) · dS
S
over the hemisphere. It is easily shown that ∇ × a = −2 k, and the surface element is
dS = a2 sin θ dθ dφ r̂ in spherical polar coordinates. Therefore
2π
π/2
(∇ × a) · dS =
dφ
dθ −2a2 sin θ r̂ · k
S
0
0
2π
= −2a2
π/2
dφ
0
2π
= −2a2
sin θ
0
π/2
dφ
0
z
a
dθ
sin θ cos θ dθ = −2πa2 .
0
We now evaluate the line integral around the perimeter curve C of the surface, which
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11.9 STOKES’ THEOREM AND RELATED THEOREMS
is the circle x2 + y 2 = a2 in the xy-plane. This is given by
0
0
a · dr = (y i − x j + z k) · (dx i + dy j + dz k)
C
0C
= (y dx − x dy).
C
Using plane polar coordinates, on C we have x = a cos φ, y = a sin φ so that dx =
−a sin φ dφ, dy = a cos φ dφ, and the line integral becomes
2π
2π
0
(y dx − x dy) = −a2
(sin2 φ + cos2 φ) dφ = −a2
dφ = −2πa2 .
C
0
0
Since the surface and line integrals have the same value, we have verified Stokes’ theorem
in this case. The two-dimensional version of Stokes’ theorem also yields Green’s theorem in
a plane. Consider the region R in the xy-plane shown in figure 11.11, in which a
vector field a is defined. Since a = ax i + ay j, we have ∇ × a = (∂ay /∂x − ∂ax /∂y) k,
and Stokes’ theorem becomes
0
∂ay
∂ax
−
dx dy = (ax dx + ay dy).
∂x
∂y
R
C
Letting P = ax and Q = ay we recover Green’s theorem in a plane, (11.4).
11.9.1 Related integral theorems
As for the divergence theorem, there exist two other integral theorems that are
closely related to Stokes’ theorem. If φ is a scalar field and b is a vector field,
and both φ and b satisfy our usual differentiability conditions on some two-sided
open surface S bounded by a closed perimeter curve C, then
0
dS × ∇φ =
φ dr,
(11.24)
S
0C
(dS × ∇) × b =
dr × b.
(11.25)
S
C
Use Stokes’ theorem to prove (11.24).
In Stokes’ theorem, (11.23), let a = φc, where c is a constant vector. We then have
0
[∇ × (φc)] · dS =
φc · dr.
(11.26)
S
C
Expanding out the integrand on the LHS we have
∇ × (φc) = ∇φ × c + φ∇ × c = ∇φ × c,
since c is constant, and the scalar triple product on the LHS of (11.26) can therefore be
written
[∇ × (φc)] · dS = (∇φ × c) · dS = c · (dS × ∇φ).
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LINE, SURFACE AND VOLUME INTEGRALS
Substituting this into (11.26) and taking c out of both integrals because it is constant, we
find
0
c · dS × ∇φ = c ·
φ dr.
S
C
Since c is an arbitrary constant vector we therefore obtain the stated result (11.24). Equation (11.25) may be proved in a similar way, by letting a = b × c in Stokes’
theorem, where c is again a constant vector. We also note that by setting b = r
in (11.25) we find
0
(dS × ∇) × r =
dr × r.
S
C
Expanding out the integrand on the LHS gives
(dS × ∇) × r = dS − dS(∇ · r) = dS − 3 dS = −2 dS.
Therefore, as we found in subsection 11.5.2, the vector area of an open surface S
is given by
0
1
r × dr.
S = dS =
2 C
S
11.9.2 Physical applications of Stokes’ theorem
Like the divergence theorem, Stokes’ theorem is useful in converting integral
equations into differential equations.
From Ampère’s law, derive Maxwell’s equation in the case where the currents are steady,
i.e. ∇ × B − µ0 J = 0.
Ampère’s rule for a distributed current with current density J is
0
B · dr = µ0 J · dS,
C
S
for any
circuit C bounding a surface S. Using Stokes’ theorem, the LHS can be transformed
into S (∇ × B) · dS; hence
(∇ × B − µ0 J) · dS = 0
S
for any surface S. This can only be so if ∇ × B − µ0 J = 0, which is the required relation.
Similarly, from Faraday’s law of electromagnetic induction we can derive Maxwell’s
equation ∇ × E = −∂B/∂t. In subsection 11.8.3 we discussed the flow of an incompressible fluid in the
presence of several sources and sinks. Let us now consider vortex flow in an
incompressible fluid with a velocity field
v=
1
êφ ,
ρ
in cylindrical polar coordinates ρ, φ, z. For this velocity field ∇ × v equals zero
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