1
STOKES’S THEOREM
STOKES’S THEOREM
We return to Green’s theorem
ZZ "
Ω
#
I
∂Q
∂P
(x, y) −
(x, y) dxdy =
P (x, y) dx + Q(x, y) dy
∂x
∂y
C
Setting v = P i + Qj + Rk, we have
(∇ × v) · k =
Œ
Œ
Œ
Πi
Œ
Œ ∂
Œ ∂x
Œ
Œ
ΠP
j
Œ
Œ
k ŒŒ
∂
∂y
∂
∂z
Q
R
Œ
Œ·k
Œ
Œ
Œ
Œ
=
∂Q ∂P
−
.
∂x
∂y
Thus in terms of v, Green’s theorem can be written
ZZ
Ω
[(∇ × v) · k] dxdy =
I
C
v(r) · dr.
Since any plane can be coordinatized as the xy-plane, this result can be phrased as follows: Let S be a flat surface in space bounded by a Jordan curve C. If v is continuously
differentiable on S, then
ZZ
S
[(∇ × v) · n] dσ =
I
C
v(r) · dr
where n is a unit normal for S and the line integral is taken in the positive sense,
meaning, in the direction of the unit tangent T for which T × n points away from the
surface. (An observer marching along C in the attitude of n keeps the surface to his
or her left.)
Consider a polyhedral surface S bounded by a closed polygonal path C. The surface
S consists of a finite number of flat faces S1 , ..., SN with polygonal boundaries C1 , ..., CN
and unit normals n1 , ..., nN . We choose these unit normals in a consistent manner; that
is, they emanate from the same side of the surface. Now let n = n(x, y, z) be a vector
function of norm 1 which is n1 on S1 , n2 on S2 , n3 on S3 , etc. How n is defined on the
line segments that join the different faces is immaterial. Suppose now that v = v(x, y, z)
is a vector function continuously differentiable on an open set that contains S. Then
ZZ
S
[(∇ × v) · n] dσ =
N ZZ
X
i=1 S
i
[(∇ × v) · n] dσ =
N I
X
i=1 Ci
v(r) · dr,
March 17, 2001
2
STOKES’S THEOREM
the integral over Ci being taken in the positive sense with respect to ni . Now when we
add these line integrals, we find that all the line segments that make up the Ci but are
not part of C are traversed twice and in opposite directions. Thus these line segments
contribute nothing to the sum of the line integrals and we are left with the integral
around C. It follows that for a polyhedral surface S with boundary C
ZZ
S
[(∇ × v) · n] dσ =
I
C
v(r) · dr.
This result can be extended to smooth surfaces with smooth bounding curves by
approximating these configurations by polyhedral configurations of the type considered
and using a limit process. In an admittedly informal way we have arrived at Stokes’s
theorem.
THEOREM STOKES’S THEOREM
Let S be a smooth surface with a smooth bounding curve C. If v = v(x, y, z) is
continuously differentiable on an open set that contains S, then
ZZ
S
[(∇ × v) · n] dσ =
I
C
v(r) · dr
where n = n(x, y, z) is a unit normal that varies continuously on S and the line integral
is taken in the positive sense with respect to n.
Problem 1. Verify Stokes’s theorem for
v = −3y i + 3x j + z 4 k
taking S as the portion of the ellipsoid 2x2 + 2y 2 + z 2 = 1 that lies above the plane
√
z = 1/ 2.
Solution. A little algebra shows that S is the graph of
f (x, y) =
q
1 − 2(x2 + y 2 )
with (x, y) restricted to the disc Ω : x2 + y 2 ≤ 1/4. Now
∇×v =
Œ
Œ
Œ
Œ
Œ
Œ
Œ
Œ
Œ
Œ
i
∂
∂x
j
∂
∂y
−3y 3x
Œ
Œ
k ŒŒ
Œ
∂ Œ
∂z ŒŒ
Œ
z4 Œ
"
#
∂
∂
=
(3x) +
(3y) k = 6k.
∂x
∂y
March 17, 2001
3
STOKES’S THEOREM
Taking n as the upper unit normal, we have
ZZ
S
[(∇ × v) · n] dσ =
=
ZZ
=
S
(6k · n) dσ
(−(0)fx0 − (0)fy0 + 6) dxdy
Ω
ZZ
ZZ
6 dxdy = 6(area of Ω) =
Ω
3π
.
2
√
The bounding curve for C is the set of all (x, y, z) with x2 + y 2 = 1/4 and z = 1/ 2.
We can parametrize C by setting
1
r(u) = 1/2 cos u i + 1/2 sin u j + √ k, u ∈ [0, 2π].
2
Since n is the upper unit normal, this parametrization gives C in the positive sense.
Thus
I
C
v(r) · dr =
Z 2π
0
3
3
1
1
1
(− sin u i + cos u j + k) · (− sin u i + cos u j) du
2
2
4
2
2
=
Z 2π
0
3π
3
du =
.
4
2
This is the value we obtained for the surface integral.
Problem 2. Verify Stokes’s theorem for
v = z 2 i − 2x j + y 3 k
taking S as the upper half of the unit sphere x2 + y 2 + z 2 = 1.
Solution. We use the upper unit normal n = xi + yj + zk. Now
∇×v =
Therefore
ZZ
S
=
[(∇ × v) · n] dσ =
ZZ
S
Œ
Œ
Πi
Œ
Œ
Œ ∂
Œ
Œ ∂x
Π2
Πz
∂
∂y
−2x
ZZ
S
Œ
Œ
k ŒŒ
j
Œ
Œ
Œ
Œ
3 ŒŒ
y
∂
∂z
= 3y 2 i + 2z j − 2 k.
(3y 2 i + 2z j − 2 k) · (xi + yj + zk) dσ
(3xy 2 + 2yz − 2z) dσ =
ZZ
S
3xy 2 dσ +
ZZ
S
2yzdσ −
ZZ
2z dσ.
S
March 17, 2001
4
STOKES’S THEOREM
The first integral is zero because S is symmetric about the yz-plane and the integrand
is odd with respect to x. The second integral is zero because S is symmetric about the
xz-plane and the integrand is odd with respect to y. Thus
ZZ
S
[(∇ × v) · n] dσ = −
ZZ
1
2z dσ = −2z̄(area of S) = −2( )2π = −2π.
2
S
This is also the value of the integral along the bounding base circle taken in the positive
sense:
I
C
v(r) · dr =
I
I
2
C
z dx − 2x dy = −2
C
x dy = −2
Z 2π
0
cos 2 u du = −2π.
Earlier we saw that the curl of a gradient is zero. Using Stokes’s theorem we can
prove a partial converse.
If a vector field v = v(x, y, z) is continuously differentiable on an open convex set
U and ∇ × v = 0 on all of U , then v is the gradient of some scalar field φ defined on U.
Proof. Choose a point a in U and for each point x in U define
φ(x) =
Z x
a
v(r) · dr.
(This is the line integral from a to x taken along the line segment that joins these two
points. We know that this line segment lies in U because U is convex.)
Since U is open, x + h is in U for all khk sufficiently small. Assume then that khk
is sufficiently small for x + h to be in U . Since U is convex, the triangular region with
vertices at a, x, x + h lies in U . Since ∇ × v = 0 on U , we can conclude from Stokes’s
theorem that
Z x
a
Therefore
v(r) · dr +
Z x+h
x
Z x+h
x
v(r) · dr = −
=
Z x+h
a
v(r) · dr +
Z a
x+h
Z a
x+h
v(r) · dr −
v(r) · dr −
Z x
a
v(r) · dr = 0.
Z x
a
v(r) · dr
v(r) · dr.
By our definition of φ we have
φ(x + h) − φ(x) =
Z x+h
x
v(r) · dr.
We can parametrize the line segment from x to x + h by r(u) = x + uh with u ∈ [0, 1].
Therefore
φ(x + h) − φ(x) =
Z 1
0
[v(r(u)) · r0 (u)]du
March 17, 2001
5
STOKES’S THEOREM
=
Z 1
0
[v(r(u)) · h]du
= v(r(u0 )) · h for some u0 ∈ [0, 1]
= v(x + u0 h) · h = v(x) · h + [v(x + u0 h) − v(x)] · h.
That v = ∇φ follows from observing that [v(x + u0 h) − v(x)] · h is o(h): as h → 0,
kv(x + u0 h) − v(x)k khk
|[v(x + u0 h) − v(x)] · h|
≤
= kv(x + u0 h) − v(x)k → 0.
khk
khk
The Normal Component of ∇ × v as Circulation per Unit
Area; Irrotational Flow
Interpret v = v(x, y, z) as the velocity of a fluid flow. We stated that ∇ × v measures
the rotational tendency of the fluid. Now we can be more precise.
Take a point P within the flow and choose a unit vector n. Let D be the -disc
that is centered at P and is perpendicular to n. Let C  be the circular boundary of D
directed in the positive sense with respect to n. By Stokes’s theorem
ZZ
D
[(∇ × v) · n] dσ =
I
C
v(r) · dr.
The line integral on the right is called the circulation of v around C  . Thus we can say
that
(the average n-component of ∇ × v on D )× (the area of D ) = the circulation of v
around C  .
It follows that
the average n-component of ∇ × v on D =
the circulation of v aroundC 
.
the area of D
Taking the limit as  shrinks to 0, we see that
the n-component of ∇ × v at P = lim→0
the circulation of v aroundC 
.
the area of D
At each point P the component of ∇ × v in any direction n is the circulation of
v per unit area in the plane normal to n. If ∇ × v = 0 identically, the fluid has no
rotational tendency and the flow is called irrotational.
The Normal Component of ∇ × v as Circulation per Unit Area; Irrotational Flow March 17, 2001