13.2 Stoke’s Theorem
We will consider surfaces S which are of the form
where R has values in ℝ3 and R is a one to one C1 map such that Ru × Rv≠0. For such a
surface, we define the integral on the surface simply as
How does this compare with the earlier general notion of integration on a manifold? In
particular, what is
? In the earlier notation with charts and such, we would denote
. Thus we would be using a partition of unity and J
this the same thing as
? The short answer is yes. Here is why.
as claimed. As to this definition of area on the surface
being well defined in the sense that the same answer would be returned with another
parameterization, this follows from the change of variables formula just as in the case of
general manifolds except that it is easier here because there is no partition of unity to fuss
Recall from the definition of the cross product that the cross product is perpendicular to
the two vectors. Now the map u → R
v → R
are both smooth
curves in the surface and
are vectors tangent to these curves. Since Ru × Rv
is perpendicular to both of these vectors, we say that it is normal to the surface
With this definition, here is a useful identity.
Lemma 13.2.1 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
Proof: To begin with, we use the short labor saving notation. Let
Then by the chain rule, the above equals
Now here is the long way. Letting x,y,z denote the components of R
denote the components of F
, and letting a subscripted variable denote the partial derivative
with respect to that variable, the left side of 13.2
At this point add in and subtract off certain terms. Then the above equals
Stoke’s theorem comes through the use of these identities and Green’s theorem.
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
Here the u and v axes are in the same relation as the x and y axes. That is, the following
picture holds. That is, the positive x and u axes both point to the right and the positive y
and v axes point up.
Theorem 13.2.2 (Stoke’s Theorem) Let U be any region in ℝ2 for which the
conclusion of Green’s theorem holds. Let R ∈ C2
be a one to one function.
where α is the parametrization for the oriented closed curve making up the boundary of U
such that the conclusion of Green’s theorem holds. Say α
has t ∈
. Let S denote the
Then for F a C1 vector field defined near S,
Proof: Then from the definition of the line integral and change of variables, for ∂S the
oriented curve having orientation coming from γ,
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
. Now by Lemma 13.2.1
, this equals
What is the geometric significance? You note that from properties of the cross product,
is normal to the surface because it is normal to the direction vectors
. Also a short computation shows that
Now consider the increment of surface area. It is the same thing.
Therefore, the result of Stoke’s theorem says the following.
and n is a unit normal to the surface equal to
at the point on the surface R
Note that every generalization in the conditions for Green’s theorem leads to a
corresponding generalization in the conditions for Stoke’s theorem. Also observe that the
theorem itself only depends on the first order derivatives of R. Therefore, the same theorem
should hold assuming only that R is C1 rather than C2. This is in fact the case. You can use
a mollifier argument to establish this.