We return to Green's theorem

Setting
we have

Thus in terms of
,
Green's theorem can be written

Since any plane can be coordinatized as the
-plane,
this result can be phrased as follows: Let
be a flat surface in space bounded by a Jordan curve
.
If
is continuously differentiable on
,
then

where
is a unit normal for
and the line integral is taken in the *positive sense*, meaning, in the
direction of the unit tangent
for which
points away from the surface. (An observer marching along
in the attitude of
keeps the surface to his or her left.)

Consider a *polyhedral surface*
bounded by a closed polygonal path
.
The surface
consists of a finite number of flat faces
with polygonal boundaries
and unit normals
.
We choose these unit normals in a consistent manner; that is, they emanate
from the same side of the surface. Now let
be a vector function of norm
which is
on
,
on
,
on
,
etc. How
is defined on the line segments that join the different faces is immaterial.
Suppose now that
is a vector function continuously differentiable on an open set that contains
.
Then

the integral over
being taken in the positive sense with respect to
Now when we add these line integrals, we find that all the line segments that
make up the
but are not part of
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
.
It follows that for a polyhedral surface
with boundary

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
be a smooth surface with a smooth bounding curve
.
If
is continuously differentiable on an open set that contains
,
then

where
is a unit normal that varies continuously on
and the line integral is taken in the positive sense with respect to

**Problem 1.** Verify Stokes's theorem for

taking
as the portion of the ellipsoid
that lies above the plane

Solution. A little algebra shows that
is the graph of

with
restricted to the disc
Now

Taking
as the upper unit normal, we have

The bounding curve for
is the set of all
with
and
.
We can parametrize
by setting

Since
is the upper unit normal, this parametrization gives
in the positive sense. Thus

This is the value we obtained for the surface integral.

**Problem 2.** Verify Stokes's theorem for

taking
as the upper half of the unit sphere

*Solution.* We use the upper unit normal
.
Now

Therefore

The first integral is zero because
is symmetric about the
-plane
and the integrand is odd with respect to
.
The second integral is zero because
is symmetric about the
-plane
and the integrand is odd with respect to
.
Thus

This is also the value of the integral along the bounding base circle taken in
the positive sense:

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 is continuously differentiable on an open convex set and on all of , then is the gradient of some scalar field defined on

**Proof. **Choose a point
in
and for each point
in
define

(This is the line integral from
to
taken along the line segment that joins these two points. We know that this
line segment lies in
because
is convex.)

Since
is open,
is in
for all
sufficiently small. Assume then that
is sufficiently small for
to be in
.
Since
is convex, the triangular region with vertices at
lies in
.
Since
on
,
we can conclude from Stokes's theorem that

Therefore

By our definition of
we have

We can parametrize the line segment from
to
by
with
.
Therefore

That
follows from observing that
is
:
as
,

Interpret as the velocity of a fluid flow. We stated that measures the rotational tendency of the fluid. Now we can be more precise.

Take a point
within the flow and choose a unit vector
.
Let
{}
be the
-disc
that is centered at
and is perpendicular to
.
Let
be the circular boundary of
directed in the positive sense with respect to
.
By Stokes's theorem

The line integral on the right is called the *circulation* of
around
.
Thus we can say that

(the average -component of on (the area of ) = the circulation of around

It follows that

the average -component of on

Taking the limit as shrinks to , we see that

the -component of at P

*At each point
the component of
in any direction
is the circulation of
per unit area in the plane normal to
.
If
identically, the fluid has no rotational tendency and the flow is called
irrotational.*

The Cauchy Mean Value Theorem is the basic tool needed to prove a theorem
which facilitates evaluation of limits of the form

when

Theorem. Suppose that

and suppose also that

exists. Then

exists, and

PROOF. The hypothesis that exists contains two implicit assumptions:

(1) there is an interval such that and exist for all in except, perhaps, for ,

(2) in this interval with, once again, the possible exception of

On the other hand,
and
are not even assumed to be defined at
.
If we define
(changing the previous values of
and
,
if necessary), then
and
are continuous at
.
If
,
then the Mean Value Theorem and the Cauchy Mean Value Theorem apply to
and
on the interval
(and a similar statement holds for
).
First applying the Mean Value Theorem to
,
we see that
,
for if
there would be some
in
with
,
contradicting (2). Now applying the Cauchy Mean Value Theorem to
and
,
we see that there is a number
.
in
such that

or

Now
approaches
as
approaches
,
because
,
is in
;
since
exists, it follows that

Theorem. Suppose that
is a function for which

all exists. Let

and define

Then

Proof. Define

and
.
We prove that

Notice that

Thus

and

We may therefore apply l'H\opital's Rule
times to obtain

Since
is a polynomial of degree
,
its
st
derivative is constant; in fact,
.
Thus

Therefore

Definition. The polynomial
is called the **Taylor polynomial** of degree
for
at
.

Two functions
and
are said to be **equal up to order**
at
if

The above theorem says that the Taylor polynomial equals
up to order
near
.

Theorem. Let and be two polynomials in , of degree , and suppose that and are equal up to order neat . Then

Proof. Let
.
Then
is a polynomial of degree
.
Write

The given condition

implies that

For
,
we have

Thus
and

Therefore

For
,
we have

Thus
and

Continuing this way we see that

Corollary. Let be -times differentiable at , and suppose that is a polynomial in of degree , which equals up to order at . Then , the Taylor polynomial of degree for near a.

Write

Adding up these identities, we have

Theorem. Suppose that
,
are defined on
and
is defined by

Then

(1) there exists some
in
such that

(2) there exists some
in
such that

Moreover, if
is integrable on
then

(3)

Proof. Fix
and
.
Define

for
in
.
Then

Applying the Mean Value Theorem to
on the interval
there
exists some
in
such that

Note that

Therefore

This is called the **Cauchy form of remainder**.

Set
.
Now apply the Cauchy Mean value Theorem to
and
there exists some
in
such that

Therefore

or

This is called the **Lagrange form of the remainder.**

If
is integrable on
,
it follows from the Fundamental Theorem of Calculus that

Therefore

This is called the **integral form of the remainder.**

From the equation

we have

for all

From the equation

we have