20. Stokes’ Theorem I. March 6, 2013
20.1. Formulation
This is a 3D version of Green’s Theorem. Consider an oriented surface Σ with an orientation given
by a normal vector field n. Assume its boundary is a simple closed curve C which is traversed in the
following direction: if you walk around C in this direction with your head pointing in the direction
of n, then the surface Σ will always be on your left. We say the orientations of C and Σ agree. Take
a 3D vector field F. Then
Z
ZZ
F · dr =
curl F · dS
C
Σ
The proof of this theorem is not very difficult, but quite computational and tedious, and not quite
illuminating and interesting. It is done by reduction to Green’s Theorem. One can read it in the
textbook.
20.2. Relation to Green’s Theorem
Green’s Theorem is a particular case of Stokes’ Theorem. Consider a closed simple parametric curve
C in R2 traversed counterclockwise. Let D be the region inside this curve. Take a 2D vector field
F(x, y) = P (x, y)i + Q(x, y)j. Calculate its curl:
i
j
k
∂
∂Q
∂P
∂
∂
curl F = ∂x
−
k.
∂y
∂z =
∂x
∂y
P (x, y) Q(x, y) 0 The domain D can be viewed as a parametric surface Σ with the following parametrization: r(u, v) =
ui + vj + 0k, (u, v) ∈ D. The normal vector is n = k, since D lies on the xy-plane. Therefore,
ZZ ZZ
Z
ZZ ∂Q ∂R
∂Q ∂R
curl F · dS =
F · dr =
−
−
k · ndS =
dA.
∂x
∂y
∂x
∂y
Σ
Σ
C
D
20.3. Example 1
R
Evaluate C F · dr, where F(x, y, z) = xj, and C is the intersection of y + z = 2 and x2 + y 2 = 1. We
traverse C counterclockwise when viewed from above. We have: P = 0, Q = x, R = 0, and
i
j
k
∂ ∂ ∂
curl F = ∂x ∂y ∂z = k.
0 x 0
Therefore, for every surface Σ with boundary C, we have:
Z
ZZ
F · dr =
k · dS.
C
Σ
Although there are many surfaces with boundary C, the simplest one is the piece Σ of plane y +z = 2
inside the cylinder x2 + y 2 = 1. It can be parametrized as r(x, y) =< x, y, 2 − y >, where (x, y) ∈
D := {x2 + y 2 ≤ 1}. We have: ru =< 1, 0, 0 >, and rv =< 0, 1, −1 >, so ru × rv =< 0, 1, 1 >, and
k · (ru × rv ) = 1. Thus,
ZZ
ZZ
ZZ
k · dS =
k · (ru × rv )dA =
dA = Area(D) = π
Σ
D
D
1
20.4. Example 2
Consider C to be the following triangular curve: from the origin O to A = (0, 3, 0), then to B =
(1, 0, 1), then back to the origin. Let
F(x, y, z) = (2x − 3z)i + (y + 7x)j + (5y − z)k.
Then we have:
i
j
k
∂
∂
∂
= 5i − 3j + 7k.
curl F = ∂x
∂y
∂z
2x − 3y y + 7x 5y − z The planar surface Σ enclosed by C is a triangle with the normal√vector OA
√ × OB =< 3, 0, −3 >.
(Remember from Math 126!) The unit normal vector is n =< 1/ 2, 0, −1/ 2 >. This vector gives
the right orientation. This triangle is right, because√the angle O equals π/2: OA · OB =< 0, 3, 0 >
· < 1, 0, 1 >= 0. Its area is (1/2)|OA||OB| = (1/2)3 2. Then we have:
Z
ZZ
ZZ
ZZ
√
√
F · dr =
curl F · dS =
< 5, −3, 7 > ·n dS =
(− 2) dS = − 2 Area(Σ) = −3
C
Σ
Σ
Σ
This problem is taken from Professor Arms’ Practice Final from Dr Loveless’ archive. One might as
well express this surface as a graph z = g(x, y) of some function (in fact, just z = x!), and use the
standard formula. But here we used a clever trick to avoid these calculations.
20.5. Remarks
1. Just like Green’s Theorem, Stokes’ Theorem can be extended for surfaces with holes and for
curves which are not simple (i.e. they are self-intersecting).
2. If S is a oriented surface which is closed, i.e. does not have any boundary C, then
ZZ
curl F · dS = 0.
S
Examples of such surfaces: a sphere, a cube, and, more generally, any boundary of a solid body
(provided this body is bounded and does not have holes).
2