STOKES'S THEOREM

We return to Green's theorem
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Setting $v=Pi+Qj+Rk,$ we have
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Thus in terms of $v$, Green's theorem can be written
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Since any plane can be coordinatized as the $xy$-plane, this result can be phrased as follows: Let $S$ be a flat surface in space bounded by a Jordan curve $C$. If $v$ is continuously differentiable on $S$, then
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where $n$ is a unit normal for $S$ and the line integral is taken in the positive sense, meaning, in the direction of the unit tangent $T$ for which $T\times n$ points away from the surface. (An observer marching along $C$ in the attitude of $n$ keeps the surface to his or her left.)

Consider a polyhedral surface $S$ bounded by a closed polygonal path $C $. The surface $S$ consists of a finite number of flat faces $S_1,...,S_N$ with polygonal boundaries $C_1,...,C_N$ and unit normals $n_1,...,n_N$. We choose these unit normals in a consistent manner; that is, they emanate from the same side of the surface. Now let $n=n(x,y,z)$ be a vector function of norm $1$ which is $n_1$ on $S_1$, $n_2$ on $S_2$, $n_3$ on $S_3$, etc. How $n $ is defined on the line segments that join the different faces is immaterial. Suppose now that $v=v(x,y,z)$ is a vector function continuously differentiable on an open set that contains $S$. Then
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the integral over $C_i$ being taken in the positive sense with respect to $n_i.$ Now when we add these line integrals, we find that all the line segments that make up the $C_i$ but are not part of $C$ 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 $C$. It follows that for a polyhedral surface $S$ with boundary $C$
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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 $S$ be a smooth surface with a smooth bounding curve $C$. If $v=v(x,y,z)$ is continuously differentiable on an open set that contains $S$, then
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where $n=n(x,y,z)$ is a unit normal that varies continuously on $S$ and the line integral is taken in the positive sense with respect to $n.$

Problem 1. Verify Stokes's theorem for
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taking $S$ as the portion of the ellipsoid $2x^2+2y^2+z^2=1$ that lies above the plane $z=1/\sqrt{2}.$

Solution. A little algebra shows that $S$ is the graph of
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with $(x,y)$ restricted to the disc MATH Now
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Taking $n$ as the upper unit normal, we have
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The bounding curve for $C$ is the set of all $(x,y,z)$ with $x^{2}+y^{2}=1/4$ and $z=1/\sqrt{2}$. We can parametrize $C$ by setting
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Since $n$ is the upper unit normal, this parametrization gives $C$ in the positive sense. Thus
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This is the value we obtained for the surface integral.

Problem 2. Verify Stokes's theorem for
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taking $S$ as the upper half of the unit sphere $x^2+y^2+z^2=1.$

Solution. We use the upper unit normal $n=xi+yj+zk$. Now
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Therefore
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The first integral is zero because $S$ is symmetric about the $yz$-plane and the integrand is odd with respect to $x$. The second integral is zero because $S$ is symmetric about the $xz$-plane and the integrand is odd with respect to $y$. Thus
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This is also the value of the integral along the bounding base circle taken in the positive sense:
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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 $v=v(x,y,z)$ is continuously differentiable on an open convex set $U$ and $\nabla \times v=0$ on all of $U$, then $v$ is the gradient of some scalar field $\QTR{group}{\phi }$ defined on $U.$

Proof. Choose a point $a$ in $U$ and for each point $x$ in $U$ define
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(This is the line integral from $a$ to $x$ taken along the line segment that joins these two points. We know that this line segment lies in $U$ because $U $ is convex.)

Since $U$ is open, $x+h$ is in $U$ for all $\left\| h\right\| $ sufficiently small. Assume then that $\left\| h\right\| $ is sufficiently small for $x+h$ to be in $U$. Since $U$ is convex, the triangular region with vertices at $a,x,x+h$ lies in $U$. Since $\nabla \times v=0$ on $U$, we can conclude from Stokes's theorem that
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Therefore
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By our definition of $\QTR{group}{\phi }$ we have
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We can parametrize the line segment from $x$ to $x+h$ by $r(u)=x+uh$ with $u\in [0,1]$. Therefore
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That $v=\nabla \phi $ follows from observing that MATH is $o(h)$: as $h\rightarrow 0$,
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The Normal Component of $\nabla \times v$ as Circulation per Unit Area; Irrotational Flow

Interpret $v=v(x,y,z)$ as the velocity of a fluid flow. We stated that $\nabla \times v$ measures the rotational tendency of the fluid. Now we can be more precise.

Take a point $P$ within the flow and choose a unit vector $n$. Let $D_\epsilon ${} be the $\epsilon $-disc that is centered at $P$ and is perpendicular to $n$. Let MATH be the circular boundary of $D_\epsilon $ directed in the positive sense with respect to $n$. By Stokes's theorem
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The line integral on the right is called the circulation of $v$ around MATH. Thus we can say that

(the average $n$-component of $\nabla \times v$ on MATH (the area of $D_\epsilon $) = the circulation of $v$ around MATH

It follows that

the average $n$-component of $\nabla \times v$ on MATH

Taking the limit as $\epsilon $ shrinks to $0$, we see that

the $n$-component of $\nabla \times v$ at P MATH MATH

At each point $P$ the component of $\nabla \times v$ in any direction $n $ is the circulation of $v$ per unit area in the plane normal to $n$. If $\nabla \times v=0$ 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
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when
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Theorem. Suppose that
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and suppose also that
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exists. Then
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exists, and
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PROOF. The hypothesis that MATH exists contains two implicit assumptions:

(1) there is an interval MATH such that $f^{\prime }(x)$ and $g^{\prime }(x)$ exist for all $x$ in MATH except, perhaps, for $x=\alpha $,

(2) in this interval MATH with, once again, the possible exception of $x=\alpha .$

On the other hand, $f$ and $g$ are not even assumed to be defined at $\alpha $. If we define MATH (changing the previous values of $f(\alpha )$ and $g(\alpha )$, if necessary), then $f$ and $g$ are continuous at $\alpha $. If MATH, then the Mean Value Theorem and the Cauchy Mean Value Theorem apply to $f$ and $g$ on the interval MATH (and a similar statement holds for MATH). First applying the Mean Value Theorem to $g$, we see that $g(x)\neq 0$, for if $g(x)=0$ there would be some $x_{1}$ in $(\alpha ,x)$ with MATH, contradicting (2). Now applying the Cauchy Mean Value Theorem to $f$ and $g$, we see that there is a number $b$. in $(\alpha ,x)$ such that
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or
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Now $b$ approaches $a$ as $x$ approaches $\alpha $, because $b$, is in $(\alpha ,x)$; since MATH exists, it follows that
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Taylor Polynomial

Theorem. Suppose that $f$ is a function for which
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all exists. Let
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and define
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Then
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Proof. Define
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and $g(x)=(x-a)^n$. We prove that
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Notice that
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Thus
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and
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We may therefore apply l'H\opital's Rule $n-1$ times to obtain
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Since $Q$ is a polynomial of degree $n-1$, its $(n-1)$st derivative is constant; in fact, MATH. Thus
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Therefore
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Definition. The polynomial $P_{n,a}(x)$ is called the Taylor polynomial of degree $n$ for $f$ at $a$.

Order of Equality

Two functions $f$ and $g$ are said to be equal up to order $n$ at $a$ if
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The above theorem says that the Taylor polynomial equals $f$ up to order $n$ near $a$.

Uniqueness

Theorem. Let $P$ and $Q$ be two polynomials in $(x-a)$, of degree $\leq n$, and suppose that $P$ and $Q$ are equal up to order $n$ neat $a$. Then $P=Q.$

Proof. Let $R=P-Q$. Then $R$ is a polynomial of degree $\leq n$. Write
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The given condition
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implies that
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For $j=0$, we have
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Thus $b_0=0$ and
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Therefore
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For $j=1$, we have
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Thus $b_1=0$ and
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Continuing this way we see that
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Corollary. Let $f$ be $n$-times differentiable at $a$, and suppose that $P$ is a polynomial in $(x-a)$ of degree $\leq n$, which equals $f$ up to order $n$ at $a$. Then $P\equiv P_{n,a}$, the Taylor polynomial of degree $n$ for $f $ near a.

Derivation of the Taylor Expansion from Integrations by Parts

Write
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Adding up these identities, we have
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Taylor's Theorem

Theorem. Suppose that $f^{\prime }$, MATH are defined on $[a,x]$ and $R_{n,a}(x)$ is defined by
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Then

(1) there exists some $t$ in $(a,x)$ such that
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(2) there exists some $t$ in $(a,x)$ such that
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Moreover, if $f^{(n+1)}$ is integrable on $[a,x]$ then

(3) MATH

Proof. Fix $a$ and $x$. Define
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for $t$ in $[a,x]$. Then
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Applying the Mean Value Theorem to $\ S$ on the interval $[a,x]:$there exists some $\ t$ in $(a,x)$ such that
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Note that
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Therefore
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This is called the Cauchy form of remainder.

Set $g(t)=(x-t)^{n+1}$. Now apply the Cauchy Mean value Theorem to $S$ and $g $ there exists some $t$ in $(a,x)$ such that
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Therefore
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or
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This is called the Lagrange form of the remainder.

If $f^{(n+1)}$ is integrable on $[a,x]$, it follows from the Fundamental Theorem of Calculus that
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Therefore
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This is called the integral form of the remainder.

Some Consequences


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From the equation
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we have
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for all $x>-1.$

From the equation
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we have
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