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CLP-4 Vector Calculus

Joel Feldman, Andrew Rechnitzer, Elyse Yeager

Section 4.7 Optional — A Generalized Stokes’ Theorem

As we have seen, the fundamental theorem of calculus, the divergence theorem, Greens’ theorem and Stokes’ theorem share a number of common features. There is in fact a single framework which encompasses and generalizes all of them, and there is a single theorem of which they are all special cases. We now give a bare bones introduction to this framework and theorem. A proper treatment typically takes up a good part of a full course. Here is an outline of what we shall do:
  • First, we will define differential forms. To try and keep things as simple and concrete as possible, we’ll only define
     1 
    In general, a differential form is defined on a manifold, which is an abstract generalization of a multi-dimensional surface, like a sphere or a torus.
    differential forms on \(\bbbr^3\) — all of our functions will be defined on \(\bbbr^3\text{.}\) Very roughly speaking, a \(k\)-form is what you write after the integral sign of an integral over a \(k\) dimensional object. Here \(k\) is one of \(0\text{,}\) \(1\text{,}\) \(2\text{,}\) \(3\text{.}\) As a example, a \(1\)-form is an expression of the form \(F_1(x,y,z)\,\dee{x} + F_2(x,y,z)\,\dee{y} + F_3(x,y,z)\,\dee{z}\text{.}\) For \(k=0\text{,}\) think of a point as a zero dimensional object and think of evaluating a function at a point as “integrating the function over the point”.
  • Then we will define some operations on differential forms, so that we can add them, multiply them, differentiate them and, eventually, integrate them. The derivative of a \(k\)-form \(\om\) is a \((k+1)\)-form that is denoted \(\dee{\om}\text{.}\) It will turn out that
    • differentiating a \(0\)-form amounts to taking a gradient,
    • differentiating a \(1\)-form amounts to taking a curl, and
    • differentiating a \(2\)-form amounts to taking a divergence.
  • Finally we will get to the generalized Stokes’ theorem which says that, if \(\om\) is a \(k\)-form (with \(k=0,1,2\)) and \(D\) is a \((k+1)\)-dimensional domain of integration, then
    \begin{equation*} \int_D d\om=\int_{\partial D}\om \end{equation*}
    It will turn out that
    • when \(k=0\text{,}\) this is just the fundamental theorem of calculus and
    • when \(k=1\text{,}\) this is both Green’s theorem and our Stokes’ theorem, and
    • when \(k=2\text{,}\) this is the divergence theorem.
Now let’s get to work. For simplicity, we will assume throughout this section that all derivatives of all functions exist and are continuous. Our first task to define differential forms.
As we said above we will define a 1-form as an expression of the form \(F_1(x,y,z)\,\dee{x} + F_2(x,y,z)\,\dee{y} + F_3(x,y,z)\,\dee{z}\text{.}\) When you learned the definition of the integral the symbol “\(\dee{x}\)” was not given any mathematical meaning by itself. A meaning was given only to collections of symbols like the indefinite integral “\(\int f(x)\ \dee{x}\)” and the definite integral “\(\int_a^b f(x)\ \dee{x}\)”. Later in this section, we will give a meaning to \(\dee{x}\text{.}\) We will, in Definition 4.7.9, define a differentiation operator that we will call \(\dee{}\text{.}\) Then \(\dee{x}\) will be that differentiation operator applied to the function \(f(x)=x\text{.}\) However, until then we will have to treat \(\dee{x}\) and \(\dee{y}\) and \(\dee{z}\) just as symbols. Their sole role in \(F_1(x,y,z)\,\dee{x} + F_2(x,y,z)\,\dee{y} + F_3(x,y,z)\,\dee{z}\) is to allow us to distinguish
 2 
We could also define, for example, a \(1\)-form as an ordered list \(\big( F_1(x,y,z)\,,\, F_2(x,y,z)\,,\, F_3(x,y,z)\big)\) of three functions and just view \(F_1(x,y,z)\,\dee{x} + F_2(x,y,z)\,\dee{y} + F_3(x,y,z)\,\dee{z}\) as another notation for \(\big( F_1(x,y,z)\,,\, F_2(x,y,z)\,,\, F_3(x,y,z)\big)\text{.}\)
\(F_1(x,y,z)\text{,}\) \(F_2(x,y,z)\) and \(F_3(x,y,z)\text{.}\)
Similarly, we will define a 2-form as an expression of the form \(F_1(x,y,z)\,\dee{y}\wedge\dee{z} + F_2(x,y,z)\,\dee{z}\wedge\dee{x} + F_3(x,y,z)\,\dee{x}\wedge\dee{y}\text{.}\) Once again there is a symbol, namely “\(\wedge\)”, that we have not yet given a meaning to. We will, in Definition 4.7.3, define a product, called the wedge product, with \(\wedge\) as the multiplication symbol. Then \(\dee{x}\wedge\dee{y}\) will be the wedge product of \(\dee{x}\) and \(\dee{y}\text{.}\) Until then we will have to treat \(\dee{y}\wedge\dee{z}\text{,}\) \(\dee{z}\wedge\dee{x}\) and \(\dee{x}\wedge\dee{y}\) just as three more meaningless symbols.
Finally here is the definition.

Definition 4.7.1.

  1. A \(0\)-form is a function \(f(x,y,z)\text{.}\)
  2. A \(1\)-form is an expression of the form
    \begin{equation*} F_1(x,y,z)\,\dee{x} + F_2(x,y,z)\,\dee{y} + F_3(x,y,z)\,\dee{z} \end{equation*}
    with \(F_1(x,y,z)\text{,}\) \(F_2(x,y,z)\) and \(F_3(x,y,z)\) being functions of three variables.
  3. A \(2\)-form is an expression of the form
    \begin{equation*} F_1(x,y,z)\,\dee{y}\wedge\dee{z} + F_2(x,y,z)\,\dee{z}\wedge\dee{x} + F_3(x,y,z)\,\dee{x}\wedge\dee{y} \end{equation*}
    with \(F_1(x,y,z)\text{,}\) \(F_2(x,y,z)\) and \(F_3(x,y,z)\) being functions of three variables.
  4. A \(3\)-form is an expression of the form \(f(x,y,z)\,\dee{x}\wedge\dee{y}\wedge\dee{z}\text{,}\) with \(f(x,y,z)\) being a function of three variables.
At this stage (there’ll be more later), just think of “\(\dee{x}\)”, “\(\dee{y}\)”, “\(\dee{z}\)”, “\(\dee{x}\wedge\dee{y}\)”, and so on, as symbols. Do not yet attempt to attach any significance to them.
There are four operations involving differential forms — addition, multiplication (\(\wedge\)), differentiation (\(\dee{}\)) and integration. Here are their definitions. First, addition is defined, and works, just the way that you would expect it to.

Definition 4.7.2. Addition of differential forms.

  1. The sum of the \(0\)-forms \(f\) and \(g\) is the \(0\)-form \(f+g\text{.}\)
  2. The sum of two \(1\)-forms is the \(1\)-form
    \begin{align*} &\big[F_1\,\dee{x} + F_2\,\dee{y} + F_3\,\dee{z}\big]\\ +&\big[G_1\,\dee{x} + G_2\,\dee{y} + G_3\,\dee{z}\big]\\ =& (F_1+G_1)\,\dee{x} + (F_2+G_2)\,\dee{y} + (F_3+G_3)\,\dee{z} \end{align*}
  3. The sum of two \(2\)-forms is the \(2\)-form
    \begin{align*} &\big[F_1\,\dee{y}\wedge\dee{z} + F_2\,\dee{z}\wedge\dee{x} + F_3\,\dee{x}\wedge\dee{y}\big]\\ +&\big[G_1\,\dee{y}\wedge\dee{z} + G_2\,\dee{z}\wedge\dee{x} + G_3\,\dee{x}\wedge\dee{y}\big]\\ =&(F_1+G_1)\,\dee{y}\wedge\dee{z} + (F_2+G_2)\,\dee{z}\wedge\dee{x} + (F_3+G_3)\,\dee{x}\wedge\dee{y} \end{align*}
  4. The sum of two \(3\)-forms is the \(3\)-form
    \begin{gather*} f\,\dee{x}\wedge\dee{y}\wedge\dee{z} \ +\ g\,\dee{x}\wedge\dee{y}\wedge\dee{z} \ =\ \big(f+g\big)\,\dee{x}\wedge\dee{y}\wedge\dee{z} \end{gather*}
There is one wrinkle in multiplication. It is not commutative, meaning that \(\alpha\wedge\be\) need not be the same as \(\be\wedge\alpha\text{.}\) You have already seen some noncommutative products. If \(\va\) and \(\vb\) are two vectors in \(\bbbr^3\text{,}\) then \(\va\times\vb = -\vb\times \va\text{.}\) Also, if \(A\) and \(B\) are two \(n\times n\) matrices, the matrix product \(AB\) need not be the same as \(BA\text{.}\)

Definition 4.7.3. Multiplication of differential forms.

We now define a multiplication rule for differential forms. If \(\om\) is a \(k\)-form and \(\om'\) is a \(k'\)-form then the product will be a \((k+k')\)-form and will be denoted \(\om\wedge\om'\) (read “omega wedge omega prime”). It is determined by the following properties.
  1. If \(f\) is a function (i.e. a \(0\)-form), then
    \begin{align*} f\big[F_1\,\dee{x} + F_2\,\dee{y} + F_3\,\dee{z}\big] &= (fF_1)\,\dee{x} \!+\! (fF_2)\,\dee{y} \!+\! (fF_3)\,\dee{z}\\ f\big[F_1\,\dee{y}\wedge\dee{z} \!+\! F_2\,\dee{z}\wedge\dee{x} \!+\! F_3\,\dee{x}\wedge\dee{y}\big] &=(fF_1)\,\dee{y}\wedge\dee{z} + (fF_2)\,\dee{z}\wedge\dee{x}\\ &\hskip1.25in + (fF_3)\,\dee{x}\wedge\dee{y}\\ f\big[g\,\dee{x}\wedge\dee{y}\wedge\dee{z}\big] &= (fg)\,\dee{x}\wedge\dee{y}\wedge\dee{z} \end{align*}
    Traditionally, the \(\wedge\) is not written when multiplying a differential form by a function (i.e. a \(0\)-form).
  2. \(\om\wedge\om'\) is linear in \(\om\) and in \(\om'\text{.}\) This means that if \(\om = f_1\om_1+f_2\om_2\text{,}\) where \(f_1\text{,}\)\(f_2\) are functions and \(\om_1,\om_2\) are forms, then
    \begin{equation*} \big(f_1\om_1+f_2\om_2\big)\wedge \om'= f_1 (\om_1\wedge\om') +f_2 (\om_2\wedge\om') \end{equation*}
    Similarly,
    \begin{equation*} \om\wedge\big(f_1\om'_1+f_2\om'_2\big)= f_1 (\om\wedge\om'_1) +f_2 (\om\wedge\om'_2) \end{equation*}
  3. If \(\om\) is a \(k\)-form and \(\om'\) is a \(k'\)-form then
    \begin{equation*} \om\wedge\om'=(-1)^{kk'}\om'\wedge\om \end{equation*}
    That is, if at least one of \(k\) and \(k'\) is even, then
    \begin{equation*} \om\wedge\om'=\om'\wedge\om \end{equation*}
    (so that the wedge product is commutative) and if \(k\) and \(k'\) are both odd then
    \begin{equation*} \om\wedge\om'=-\om'\wedge\om \end{equation*}
    (so that the wedge product is anticommutative). In particular, if \(\om\) is a \(d\)-form with \(d\) odd
    \begin{equation*} \om\wedge\om = 0 \end{equation*}
  4. The wedge product is associative. This means that
    \begin{equation*} (\om\wedge\om')\wedge\om''=\om\wedge\big(\om'\wedge\om''\big) \end{equation*}
So the wedge product obeys most of the usual multiplication rules, with the one big exception that if \(\om\) is \(k\)-form and \(\om'\) is a \(k'\)-form with \(k\) and \(k'\) both odd then \(\om\wedge\om'=-\om'\wedge\om\text{.}\)
The best way to get a handle on the wedge product is to work through some examples, like these.

Example 4.7.4.

Let \(\om = F_1\,\dee{x} + F_2\,\dee{y} + F_3\,\dee{z}\) and \(\om' = G_1\,\dee{x} + G_2\,\dee{y} + G_3\,\dee{z}\) be any two \(1\)-forms. Their product is
\begin{align*} \om\wedge\om' &=\big[F_1\,\dee{x} + F_2\,\dee{y} + F_3\,\dee{z}\big] \wedge \big[G_1\,\dee{x} + G_2\,\dee{y} + G_3\,\dee{z}\big]\\ &= \ \big(F_1\,\dee{x}\big)\wedge\big(G_1\,\dee{x}\big) +\big(F_1\,\dee{x}\big)\wedge\big(G_2\,\dee{y}\big) +\big(F_1\,\dee{x}\big)\wedge\big(G_3\,\dee{z}\big)\\ &\hskip0.1in +\big(F_2\,\dee{y}\big)\wedge\big(G_1\,\dee{x}\big) +\big(F_2\,\dee{y}\big)\wedge\big(G_2\,\dee{y}\big) +\big(F_2\,\dee{y}\big)\wedge\big(G_3\,\dee{z}\big)\\ &\hskip0.1in+ \big(F_3\,\dee{z}\big)\wedge\big(G_1\,\dee{x}\big) +\big(F_3\,\dee{z}\big)\wedge\big(G_2\,\dee{y}\big) +\big(F_3\,\dee{z}\big)\wedge\big(G_3\,\dee{z}\big)\\ &\hskip0.5in\text{(by linearity, i.e. by part (b) of the last Definition)}\\ &= \ F_1G_1\,\dee{x}\wedge\,\dee{x} + F_1G_2\,\dee{x}\wedge\,\dee{y} + F_1G_3\,\dee{x}\wedge\,\dee{z}\\ &\hskip0.1in +F_2G_1\,\dee{y}\wedge\,\dee{x} + F_2G_2\,\dee{y}\wedge\,\dee{y} + F_2G_3\,\dee{y}\wedge\,\dee{z}\\ &\hskip0.1in+ F_3G_1\,\dee{z}\wedge\,\dee{x} + F_3G_2\,\dee{z}\wedge\,\dee{y} + F_3G_3\,\dee{z}\wedge\,\dee{z}\\ &= \big(F_1G_2-F_2G_1)\,\dee{x}\wedge\dee{y} +\big(F_3G_1-F_1G_3)\,\dee{z}\wedge\dee{x}\\ &\hskip0.1in +\big(F_2G_3-F_3G_2)\,\dee{y}\wedge\dee{z} \end{align*}
because
\begin{equation*} \dee{x}\wedge\,\dee{x}=\dee{y}\wedge\,\dee{y}=\dee{z}\wedge\,\dee{z}=0 \end{equation*}
and
\begin{equation*} \dee{x}\wedge\,\dee{y}=-\dee{y}\wedge\,\dee{x}\qquad \dee{x}\wedge\,\dee{z}=-\dee{z}\wedge\,\dee{x}\qquad \dee{z}\wedge\,\dee{y}=-\dee{y}\wedge\,\dee{z} \end{equation*}
Note that, looking at the last example, if we view \(\vF=(F_1,F_2,F_3)\) and \(\vG=(G_1,G_2,G_3)\) as vectors, we can write the product simply as
where we are using \((\vF\times\vG)_\ell\) to denote the \(\ell^{\rm th}\) component of the cross product \(\vF\times\vG\text{.}\) In the special case that \(F_3=G_3=0\text{,}\) we have
We can now see why in the Definition 4.7.1.c of \(2\)-forms
  • there were no \(\dee{x}\wedge\dee{x}\) or \(\dee{y}\wedge\dee{y}\) or \(\dee{z}\wedge\dee{z}\) terms — they are all zero and
  • there were no \(\dee{y}\wedge\dee{x}\) or \(\dee{z}\wedge\dee{y}\) or \(\dee{x}\wedge\dee{z}\) terms — they can all be rewritten using \(\dee{x}\wedge\dee{y}\text{,}\) \(\dee{y}\wedge\dee{z}\) and \(\dee{z}\wedge\dee{x}\) terms (or vice versa).
The reason that we chose to write the Definition 4.7.1.c as
\begin{equation*} F_1\,\dee{y}\wedge\dee{z} + F_2\,\dee{z}\wedge\dee{x} + F_3\,\dee{x}\wedge\dee{y} \end{equation*}
as opposed to in the form, for example,
\begin{equation*} f_1\,\dee{x}\wedge\dee{y} + f_2\,\dee{x}\wedge\dee{z} + f_3\,\dee{y}\wedge\dee{z} \end{equation*}
was to make formulae like 4.7.5 work. The easy way to remember
\begin{equation*} F_1\,\dee{y}\wedge\dee{z} + F_2\,\dee{z}\wedge\dee{x} + F_3\,\dee{x}\wedge\dee{y} \end{equation*}
is to rename (in your head) \(x,y,z\) to \(x_1,x_2,x_3\text{.}\) Then the subscripts in the three terms of
\begin{equation*} F_1\,\dee{x_2}\wedge\dee{x_3} + F_2\,\dee{x_3}\wedge\dee{x_1} + F_3\,\dee{x_1}\wedge\dee{x_2} \end{equation*}
are just \(1,2,3\) and \(2,3,1\) and \(3,1,2\) — the three cyclic permutations of \(1,2,3\text{.}\)

Example 4.7.7.

The product of the (general) \(1\)-form \(\om = F_1\,\dee{x} + F_2\,\dee{y} + F_3\,\dee{z}\) and the (general) \(2\)-form \(\om'=\big[G_1\,\dee{y}\wedge\dee{z} + G_2\,\dee{z}\wedge\dee{x} + G_3\,\dee{x}\wedge\dee{y}\big]\) (again note the numbering of the coefficients in the \(2\)-form) is
\begin{align*} \om\wedge\om' &=\big[F_1\,\dee{x} + F_2\,\dee{y} + F_3\,\dee{z}\big] \wedge \big[G_1\,\dee{y}\wedge\dee{z} + G_2\,\dee{z}\wedge\dee{x} + G_3\,\dee{x}\wedge\dee{y}\big]\\ &= \ F_1G_1\,\dee{x}\wedge\dee{y}\wedge\dee{z} + F_2G_2\,\dee{y}\wedge\dee{z}\wedge\dee{x} + F_3G_3\,\dee{z}\wedge\dee{x}\wedge\dee{y}\\ & = \big(F_1G_1+F_2G_2+F_3G_3)\,\dee{x}\wedge\dee{y}\wedge\dee{z} \end{align*}
Here we have used that, for \(1\)-forms, \(\alpha\wedge\beta=-\beta\wedge\alpha\text{,}\) so that
\begin{align*} \dee{y}\wedge\dee{z}\wedge\dee{x} &=-\dee{y}\wedge\dee{x}\wedge\dee{z}=\dee{x}\wedge\dee{y}\wedge\dee{z}\\ \dee{z}\wedge\dee{x}\wedge\dee{y} &=-\dee{x}\wedge\dee{z}\wedge\dee{y} =\dee{x}\wedge\dee{y}\wedge\dee{z} \end{align*}
We have also used that any wedge product of three \(\dee{\{x\text{ or }y\text{ or }z\}}\)’s with at least two of the coordinates being the same is zero. For example
\begin{equation*} \dee{x}\wedge\dee{z}\wedge\dee{x} = - \dee{x}\wedge\dee{x}\wedge\dee{z} =0 \end{equation*}
So
\begin{align*} &\big[F_1\,\dee{x} + F_2\,\dee{y} + F_3\,\dee{z}\big] \wedge \big[G_1\,\dee{y}\wedge\dee{z} + G_2\,\dee{z}\wedge\dee{x} + G_3\,\dee{x}\wedge\dee{y}\big]\\ &\hskip4in = \vF\cdot\vG\,\dee{x}\wedge\dee{y}\wedge\dee{z} \end{align*}

Example 4.7.8.

Combining Examples 4.7.4 and 4.7.7, we have the wedge product of any three (general) \(1\)-forms \(F_1\,\dee{x} + F_2\,\dee{y} + F_3\,\dee{z}\) and \(G_1\,\dee{x} + G_2\,\dee{y} + G_3\,\dee{z}\) and \(H_1\,\dee{x} + H_2\,\dee{y} + H_3\,\dee{z}\) is
\begin{align*} &\big[F_1\,\dee{x} \!+\! F_2\,\dee{y} \!+\! F_3\,\dee{z}\big] \wedge \big[G_1\,\dee{x} \!+\! G_2\,\dee{y} \!+\! G_3\,\dee{z}\big] \wedge \big[H_1\,\dee{x} + H_2\,\dee{y} + H_3\,\dee{z}\big]\\ &\hskip0.1in=\big[F_1\,\dee{x} + F_2\,\dee{y} + F_3\,\dee{z}\big] \wedge\\ &\hskip1in \big[(\vG\times\vH)_1\, \dee{y}\wedge\dee{z} +(\vG\times\vH)_2\, \dee{z}\wedge\dee{x} +(\vG\times\vH)_3\, \dee{x}\wedge\dee{y}\big]\\ &\hskip0.1in =\big\{F_1(\vG\times\vH)_1 + F_2(\vG\times\vH)_2 + F_3(\vG\times\vH)_3\big\} \dee{x}\wedge\dee{y}\wedge\dee{z}\\ &\hskip0.1in =\big\{F_1(G_2H_3\!-\!G_3H_2) \!+\! F_2(G_3H_1\!-\!G_1H_3) \!+\! F_3(G_1H_2\!-\!G_2H_1\big\} \dee{x}\wedge\dee{y}\wedge\dee{z} \end{align*}
This can be expressed cleanly in terms of determinants. Recalling the rule for expanding a determinant along its top row
\begin{align*} &\big[F_1\,\dee{x} \!+\! F_2\,\dee{y} \!+\! F_3\,\dee{z}\big] \wedge \big[G_1\,\dee{x} \!+\! G_2\,\dee{y} \!+\! G_3\,\dee{z}\big] \wedge \big[H_1\,\dee{x} \!+\! H_2\,\dee{y} \!+\! H_3\,\dee{z}\big]\\ &\hskip3in =\det\left[\begin{matrix} F_1 & F_2 & F_3 \\ G_1 & G_2 & G_3 \\ H_1 & H_2 & H_3 \end{matrix}\right] \dee{x}\wedge\dee{y}\wedge\dee{z} \end{align*}
Our next operation is a differential operator which unifies and generalizes gradient, curl and divergence.

Definition 4.7.9. Differentiation of differential forms.

If \(\om\) is a \(k\)-form, then \(\dee{\om}\) is a \(k+1\)-form, with \(\dee{}\) being the unique
 3 
That \(\dee{}\) is unique just means that the action of \(\dee{}\) on any differential form is completely determined by the four rules (a), (b), (c), (d). We will see in Example 4.7.10.c,d,e, that this is indeed the case.
such operator that obeys
  1. \(\dee{}\) is linear. That is, if \(\om_1,\om_2\) are \(k\)-forms and \(a_1,a_2\in\bbbr\text{,}\) then
    \begin{equation*} \dee{\big(a_1\om_1+a_2\om_2\big)} =a_1\dee{\om_1}+a_2\dee{\om_2} \end{equation*}
  2. \(\dee{}\) obeys a “graded product rule”. Precisely, if \(\om^{(k)}\) is a \(k\)-form and \(\om^{(\ell)}\) is an \(\ell\)-form, then
    \begin{equation*} \dee{\big(\om^{(k)}\wedge\om^{(\ell)}\big)} =\big(d\om^{(k)}\big)\wedge\om^{(\ell)} +(-1)^k\om^{(k)} \wedge \big(\dee{\om^{(\ell)}}\big) \end{equation*}
  3. If \(f(x,y,z)\) is a \(0\)-form, then
    \begin{align*} \dee{f} &=\frac{\partial f}{\partial x}(x,y,z)\ \dee{x} +\frac{\partial f}{\partial y}(x,y,z)\ \dee{y} +\frac{\partial f}{\partial z}(x,y,z)\ \dee{z}\\ &=\vnabla f(x,y,z)\cdot\dee{\vr} \qquad\text{where } \dee{\vr} = \dee{x}\,\hi + \dee{y}\,\hj + \dee{z}\,\hk \end{align*}
  4. For any differential form \(\om\text{,}\)
    \begin{equation*} \dee{\big(\dee{\om}\big)}=0 \end{equation*}

Example 4.7.10.

  1. If \(f(x,y,z) = x\text{,}\) then
    \begin{equation*} \dee{f} =\frac{\partial x}{\partial x}(x,y,z)\ \dee{x} +\frac{\partial x}{\partial y}(x,y,z)\ \dee{y} +\frac{\partial x}{\partial z}(x,y,z)\ \dee{z} =\dee{x} \end{equation*}
    That is, \(\dee{x}\) really is the operator \(\dee{}\) applied to the function \(x\text{.}\) Similarly, \(\dee{y}\) really is the operator \(\dee{}\) applied to the function \(y\) and \(\dee{z}\) really is the operator \(\dee{}\) applied to the function \(z\text{.}\)
  2. For any \(k\)-form \(\om\)
    \begin{align*} \dee{\big[\om\wedge\dee{x}\big]} &=\dee{\om}\wedge\dee{x} + (-1)^k\om\wedge\dee{\big(\dee{x}\big)}\\ &=\dee{\om}\wedge\dee{x} \end{align*}
    Similarly
    \begin{equation*} \dee{\big[\om\wedge\dee{y}\big]}=\dee{\om}\wedge\dee{y}\qquad \dee{\big[\om\wedge\dee{z}\big]}=\dee{\om}\wedge\dee{z} \end{equation*}
  3. For any \(1\)-form
    \begin{align*} &\dee{}\big[F_1\dee{x} + F_2\dee{y} + F_3\dee{z}\big] =\dee{F_1}\wedge\dee{x} + \dee{F_2}\wedge\dee{y} + \dee{F_3}\wedge\dee{z}\\ &\hskip0.5in=\Big(\frac{\partial F_1}{\partial x}\ \dee{x} +\frac{\partial F_1}{\partial y}\ \dee{y} +\frac{\partial F_1}{\partial z}\ \dee{z}\Big)\wedge\dee{x}\\ &\hskip2in +\Big(\frac{\partial F_2}{\partial x}\ \dee{x} +\frac{\partial F_2}{\partial y}\ \dee{y} +\frac{\partial F_2}{\partial z}\ \dee{z}\Big)\wedge\dee{y}\\ &\hskip2in +\Big(\frac{\partial F_3}{\partial x}\ \dee{x} +\frac{\partial F_3}{\partial y}\ \dee{y} +\frac{\partial F_3}{\partial z}\ \dee{z}\Big)\wedge\dee{z}\\ &\hskip0.5in= \Big(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}\Big) \ \dee{y}\wedge\dee{z} +\Big(\frac{\partial F_1}{\partial z}-\frac{\partial F_3}{\partial x}\Big) \ \dee{z}\wedge\dee{x}\\ &\hskip2in +\Big(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\Big) \ \dee{x}\wedge\dee{y}\\ &\hskip0.5in= (\vnabla\times\vF)_1\,\dee{y}\wedge\dee{z} +(\vnabla\times\vF)_2\,\dee{z}\wedge\dee{x} +(\vnabla\times\vF)_3\,\dee{x}\wedge\dee{y} \end{align*}
  4. For any \(2\)-form
    \begin{align*} &\dee{}\big[F_1\,\dee{y}\wedge\dee{z} + F_2\,\dee{z}\wedge\dee{x} + F_3\,\dee{x}\wedge\dee{y}\big]\\ &\hskip0.5in=\dee{F_1}\wedge\dee{y}\wedge\dee{z} + \dee{F_2}\wedge\dee{z}\wedge\dee{x} + \dee{F_3}\wedge\dee{x}\wedge\dee{y}\\ &\hskip0.5in=\Big(\frac{\partial F_1}{\partial x}\ \dee{x} +\frac{\partial F_1}{\partial y}\ \dee{y} +\frac{\partial F_1}{\partial z}\ \dee{z}\Big)\wedge\dee{y}\wedge\dee{z}\\ &\hskip2in +\Big(\frac{\partial F_2}{\partial x}\ \dee{x} +\frac{\partial F_2}{\partial y}\ \dee{y} +\frac{\partial F_2}{\partial z}\ \dee{z}\Big)\wedge\dee{z}\wedge\dee{x}\\ &\hskip2in +\Big(\frac{\partial F_3}{\partial x}\ \dee{x} +\frac{\partial F_3}{\partial y}\ \dee{y} +\frac{\partial F_3}{\partial z}\ \dee{z}\Big)\wedge\dee{x}\wedge\dee{y}\\ &\hskip0.5in= \Big(\frac{\partial F_1}{\partial x} +\frac{\partial F_2}{\partial y} +\frac{\partial F_3}{\partial z}\Big) \ \dee{x}\wedge\dee{y}\wedge\dee{z}\\ &\hskip0.5in= \vnabla\cdot\vF\ \dee{x}\wedge\dee{y}\wedge\dee{z} \end{align*}
  5. For any \(3\)-form
    \begin{align*} \dee{}\big[f\,\dee{x}\wedge\dee{y}\wedge\dee{z}\big] &=\Big(\frac{\partial f}{\partial x}\ \dee{x} +\frac{\partial f}{\partial y}\ \dee{y} +\frac{\partial f}{\partial z}\ \dee{z}\Big) \wedge\dee{x}\wedge\dee{y}\wedge\dee{z}\\ &=0 \end{align*}

Example 4.7.11.

In Definition 4.7.9.c, we defined, for any function \(f(x,y,z)\) of three variables
\begin{align*} \dee{f} &=\frac{\partial f}{\partial x}(x,y,z)\ \dee{x} +\frac{\partial f}{\partial y}(x,y,z)\ \dee{y} +\frac{\partial f}{\partial z}(x,y,z)\ \dee{z} \end{align*}
The analogous formulae
 4 
Indeed, you can view \(f(t)\) as a function of three variables that happens to be independent of two of the three variables. Similarly you can view \(f(u,v)\) as a function of three variables that happens to be independent of one of the three variables.
for functions of one or two variables also apply.
\begin{align*} \dee{f(t)} & = \diff{f}{t}(t)\,\dee{t}\\ \dee{f(u,v)} &=\frac{\partial f}{\partial u}(u,v)\ \dee{u} +\frac{\partial f}{\partial v}(u,v)\ \dee{v} \end{align*}
  1. Let \(F_1(x,y,z)\,\dee{x} + F_2(x,y,z)\,\dee{y} + F_3(x,y,z)\,\dee{z}\) be a \(1\)-form. Suppose that we substitute \(x=x(t)\text{,}\) \(y=y(t)\) and \(z=z(t)\text{,}\) so that we are restricting our \(1\)-form to a parametrized curve. Then, writing \(\vr(t) = \big(x(t),y(t),z(t)\big)\text{,}\)
    \begin{align*} &F_1\big(x(t),y(t),z(t)\big)\,\dee{x(t)} + F_2\big(x(t),y(t),z(t)\big)\,\dee{y(t)}\\ &\hskip2in + F_3\big(x(t),y(t),z(t)\big)\,\dee{z(t)}\\ &\hskip0.5in=F_1\big(\vr(t)\big)\diff{x}{t}(t)\,\dee{t} + F_2\big(\vr(t)\big)\diff{y}{t}(t)\,\dee{t} + F_3\big(\vr(t)\big)\diff{z}{t}(t)\,\dee{t}\\ &\hskip0.5in= \vF\big(\vr(t)\big)\cdot\diff{\vr}{t}(t)\,\dee{t} \end{align*}
  2. Let \(F_1(x,y,z)\,\dee{y}\wedge\dee{z} + F_2(x,y,z)\,\dee{z}\wedge\dee{x} + F_3(x,y,z)\,\dee{x}\wedge\dee{y}\) be a \(2\)-form. Suppose that we substitute \(x=x(u,v)\text{,}\) \(y=y(u,v)\) and \(z=z(u,v)\text{,}\) so that we are restricting our \(2\)-form to a parametrized surface. Then, writing \(\vr(u,v) = \big(x(u,v),y(u,v),z(u,v)\big)\text{,}\)
    \begin{align*} &F_1\big(x(u,v),y(u,v),z(u,v)\big)\,\dee{y(u,v)}\wedge\dee{z(u,v)}\\ &\hskip1in + F_2\big(x(u,v),y(u,v),z(u,v)\big)\,\dee{z(u,v)}\wedge\dee{x(u,v)}\\ &\hskip1in + F_3\big(x(u,v),y(u,v),z(u,v)\big)\,\dee{x(u,v)}\wedge\dee{y(u,v)}\\ &=F_1\big(\vr(u,v)\big)\, \Big(\frac{\partial y}{\partial u}\dee{u} +\frac{\partial y}{\partial v}\dee{v}\Big)\wedge \Big(\frac{\partial z}{\partial u}\dee{u} +\frac{\partial z}{\partial v}\dee{v}\Big)\\ &\hskip1in + F_2\big(\vr(u,v)\big)\, \Big(\frac{\partial z}{\partial u}\dee{u} +\frac{\partial z}{\partial v}\dee{v}\Big)\wedge \Big(\frac{\partial x}{\partial u}\dee{u} +\frac{\partial x}{\partial v}\dee{v}\Big)\\ &\hskip1in + F_3\big(\vr(u,v)\big)\, \Big(\frac{\partial x}{\partial u}\dee{u} +\frac{\partial x}{\partial v}\dee{v}\Big)\wedge \Big(\frac{\partial y}{\partial u}\dee{u} +\frac{\partial y}{\partial v}\dee{v}\Big)\\ &=\Big[F_1\big(\vr(u,v)\big)\, \Big(\frac{\partial y}{\partial u}\frac{\partial z}{\partial v} -\frac{\partial y}{\partial v}\frac{\partial z}{\partial u}\Big) +F_2\big(\vr(u,v)\big)\, \Big(\frac{\partial z}{\partial u}\frac{\partial x}{\partial v} -\frac{\partial z}{\partial v}\frac{\partial x}{\partial u}\Big)\\ &\hskip1in +F_3\big(\vr(u,v)\big)\, \Big(\frac{\partial x}{\partial u}\frac{\partial y}{\partial v} -\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}\Big) \Big] \dee{u}\wedge\dee{v}\\ &=\Big[\vF\big(\vr(u,v)\big)\cdot \frac{\partial\vr}{\partial u}(u,v) \times \frac{\partial\vr}{\partial v}(u,v)\Big] \dee{u}\wedge\dee{v} \end{align*}
Let us summarize what we have seen in the Example 4.7.10.
Our final operation is integration of differential forms.

Definition 4.7.13. Integration of differential forms.

  1. Let \(f(x,y,z)\) be a \(0\)-form and \(P=(x_0,y_0,z_0)\in\bbbr^3\) be a point. Then
    \begin{equation*} \int_{P} f = f\big(x_0,y_0,z_0\big) \end{equation*}
    More generally if, for each \(1\le i\le \ell\text{,}\) \(P_i=(x_i,y_i,z_i)\in\bbbr^3\) is a point and \(n_i\) is an integer, then
    \begin{equation*} \int_{\Sigma_{i=1}^\ell n_iP_i} f = \sum_{i=1}^\ell n_i f\big(x_i,y_i,z_i\big) \end{equation*}
  2. Let \(\om = \vF(\vr)\cdot\dee{\vr} = F_1(x,y,z)\,\dee{x} + F_2(x,y,z)\,\dee{y} + F_3(x,y,z)\,\dee{z} \) be a \(1\)-form. Let \(\cC\) be a curve that is parametrized by \(\vr(t) = \big(x(t)\,,\,y(t)\,,\,z(t)\big)\text{,}\) \(a\le t\le b\text{.}\) Then, motivated by Example 4.7.11.a above,
    \begin{equation*} \int_{\cC}\om = \int_a^b \vF\big(\vr(t)\big)\cdot \diff{\vr}{t}(t)\ \dee{t} =\int_{\cC} \vF\cdot\dee{\vr} \end{equation*}
  3. Let \(\om = F_1(x,y,z)\,\dee{y}\wedge\dee{z} + F_2(x,y,z)\,\dee{z}\wedge\dee{x} + F_3(x,y,z)\,\dee{x}\wedge\dee{y}\) be a \(2\)-form. Let \(S\) be an oriented surface that is parametrized by \(\vr(u,v) = \big(x(u,v)\,,\,y(u,v)\,,\,z(u,v)\big)\text{,}\) with \((u,v)\) running over a region \(R\) in the \(uv\)-plane. Assume that \(\vr(u,v)\) is orientation preserving in the sense that \(\hn\,\dee{S} = +\frac{\partial \vr}{\partial u} \times \frac{\partial \vr}{\partial v}\,\dee{u}\,\dee{v}\text{.}\)Then, motivated by Example 4.7.11.b above,
    \begin{align*} \int_{S}\om &= \dblInt_R \Big[\vF\big(\vr(u,v)\big)\cdot \frac{\partial\vr}{\partial u}(u,v) \times \frac{\partial\vr}{\partial v}(u,v)\Big] \dee{u}\wedge\dee{v} = \dblInt_S \vF\cdot \hn\,\dee{S} \end{align*}
  4. Let \(\om = f(x,y,z)\,\dee{x}\wedge\dee{y}\wedge\dee{z}\) be a \(3\)-form. Let \(V\) be a solid in \(\bbbr^3\text{.}\) Then
    \begin{align*} \int_{V}\om &= \tripInt_V f(x,y,z)\,\dee{x}\dee{y}\dee{z} \end{align*}
Finally, after all of these definitions, we have a very compact theorem that simultaneously covers the fundamental theorem of calculus, Green’s theorem. Stokes’ theorem and the divergence theorem. Had we given all of our definitions in \(n\) dimensions, rather than just three dimensions, it would cover a lot more. This general theorem is also called Stokes’ theorem.
To see the connection between the general Stokes’ theorem 4.7.14 and the Stokes’ and divergence theorems of the earlier part of this chapter, here are the \(k=1\) and \(k=2\) cases of Theorem 4.7.14 again.
  • Let \(\om = F_1 \dee{x} + F_2 \dee{y} + F_3 \dee{z}\) be a \(1\)-form and let \(S\) be a piecewise smooth oriented surface as in (our original) Stokes’ theorem 4.4.1. Then, by Lemma 4.7.12.b,
    \begin{equation*} d\om = (\vnabla\times\vF)_1\,\dee{y}\wedge\dee{z} +(\vnabla\times\vF)_2\,\dee{z}\wedge\dee{x} +(\vnabla\times\vF)_3\,\dee{x}\wedge\dee{y} \end{equation*}
    So, by parts (c) (but with \(\vF\) replaced by \(\vnabla\times\vF\)) and (b) of Definition 4.7.13, the conclusion \(\int_D d\om=\int_{\partial D}\om\) of (the general) Stokes’ theorem 4.7.14 is
    \begin{gather*} \dblInt_S \vnabla\times\vF\cdot\hn\,\dee{S} =\int_S d\om=\int_{\partial S}\om =\int_{\partial S}\vF\cdot\dee{\vr} \end{gather*}
    which is the conclusion of (our original) Stokes’ theorem 4.4.1.
  • \(\om = F_1(x,y,z)\,\dee{y}\wedge\dee{z} + F_2(x,y,z)\,\dee{z}\wedge\dee{x} + F_3(x,y,z)\,\dee{x}\wedge\dee{y}\) be a \(2\)-form and let \(V\) be a solid as in the divergence theorem 4.2.2. Then, by Lemma 4.7.12.c,
    \begin{equation*} d\om = \vnabla\cdot\vF\,\dee{x}\wedge\dee{y}\wedge\dee{z} \end{equation*}
    So, by parts (d) (with \(f =\vnabla\cdot\vF\)) and (c) of Definition 4.7.13, the conclusion \(\int_D d\om=\int_{\partial D}\om\) of (the general) Stokes’ theorem 4.7.14 is
    \begin{gather*} \tripInt_V \vnabla\cdot\vF\,\dee{x}\dee{y}\dee{z} =\int_V d\om=\int_{\partial V}\om =\dblInt_{\partial V}\vF\cdot\hn\,\dee{S} \end{gather*}
    which is the conclusion of the divergence theorem 4.2.2.
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