# STOKES'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 ,

## The Normal Component of as Circulation per Unit Area; Irrotational Flow

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.

# L'Hôpital's Rule

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

# Taylor Polynomial

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 .

# Order of Equality

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 .

# Uniqueness

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.

# Derivation of the Taylor Expansion from Integrations by Parts

Write

Adding up these identities, we have

# Taylor's Theorem

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.

# Some Consequences

From the equation

we have

for all

From the equation

we have

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