12.2 Stoke’s Theorem
When you have a region for which Green’s theorem holds, you can quickly get a version of Stoke’s theorem.
This approach follows Davis and Snyder . Thus the difficult questions concerning Stoke’s
theorem on a two dimensional surface in ℝ3 can be reduced completely to Green’s theorem in
the plane. I am assuming you have seen some discussion of Stoke’s theorem in multivariable
Recall from calculus the following definition of the curl of a vector field. Recall also how the cross
product of two vectors defined as
is perpendicular to both vectors u,v and how u,v,u × v forms a right handed system. In this formula, it is
assumed that the vectors i,j,k pointing respectively in the direction of the positive x axis, y axis and z axis
also form a right handed system.
Also recall from calculus that when you have U a bounded region in ℝ2, you can define a parametric
surface S as R
R : U → ℝ3 has continuous partial derivatives and satisfies
Then when you write ∫
SfdS, you mean
Stoke’s theorem is really just a version of Green’s theorem for a surface in three dimensions rather than a
surface in the plane. It relates a line integral around the boundary of the surface to a surface integral like
what was just described.
Definition 12.2.1 Let
be a C1 vector field defined on an open set, V in ℝ3. Then
This is also called curl
and written as indicated, ∇× F. In symbols, the ith component of ∇× F
where εijk is the permutation symbol and summation is over repeated indices. This symbol is defined
In words, when you switch two subscripts, the result is −1 times what you had. Note that the above implies
that if two indices are equal, then the symbol equals 0. A fundamental identity is available.
it follows that
Note that δijxj = xi where it is always understood that summation over repeated indices is taking
The following lemma gives the fundamental identity which will be used in the proof of Stoke’s
Lemma 12.2.2 Let R : U → V ⊆ ℝ3 where U is an open subset of ℝ2 and V is an open subset of ℝ3.
Suppose R is C2 and let F be a C1 vector field defined in V.
Proof: Start with the left side and let xi = Ri
for short, then using
which proves 12.4
The proof of Stoke’s theorem given next follows . First, it is convenient to give a definition.
Definition 12.2.3 A vector valued function, R : U ⊆ ℝm → ℝn is said to be in Ck
is the restriction to U of a vector valued function which is defined on ℝm and is Ck. That is, this
function has continuous partial derivatives up to order k.
Let U be a region in ℝ2 for which Green’s theorem holds. Thus Green’s theorem says that if
P,Q are in C1
and the oriented boundary ∂U
is taken in the counter clockwise direction,
Here the u and v axes are in the same relation as the x and y axes. That is, the following picture
holds. The positive x and u axes both point to the right and the positive y and v axes point
Theorem 12.2.4 (Stoke’s Theorem) Let U be any region in ℝ2 for which the conclusion of Green’s
theorem holds and let R ∈ C2
be a one to one function satisfying
0 for all
∈ U and let S denote the surface, where the orientation on ∂S is consistent with the counter clockwise orientation on ∂U. (U
is on the left as you walk around ∂U as described above. For reasonably simple regions this
notion of counter clockwise motion is obvious.) Then for F a C1 vector field defined near S,
where n is the normal to S defined by
and dS is the increment of surface area defined as
for dm2 indicating 2 dimensional Lebesgue measure discussed earlier.
Proof: Letting C be an oriented part of ∂U having parametrization,
for t ∈
denote the oriented part of
corresponding to C,
Since this holds for each such piece of ∂U,
By the assumption that the conclusion of Green’s theorem holds for U, this equals
the last step holding by equality of mixed partial derivatives, a result of the assumption that R
by Lemma 12.2.2
, this equals
Note that there is no mention made in the final result that R is C2. Therefore, it is not
surprising that versions of this theorem are valid in which this assumption is not present. It
is possible to obtain extremely general versions of Stoke’s theorem if you use the Lebesgue