# Math2111: Chapter 5: Recommended reading: The fundamental theorems

In this blog entry you can find lecture notes for Math2111, several variable calculus. See also the table of contents for this course. This blog entry printed to pdf is available here.

In this entry we show that all the fundamental theorems (fundamental theorem of calculus, fundamental theorem of line integrals, Green’s theorem, Stokes’ theorem and the divergence theorem) are based on the same principle. Further, we will see that those theorems are all the fundamental theorems in $\mathbb{R},$ $\mathbb{R}^2$ and $\mathbb{R}^3.$

All the fundamental theorems have the same basic structure, namely, they are of the form

$\displaystyle \int_\Omega \mathrm{d} \omega = \int_{\partial \Omega} \omega. \hspace{2cm} (1)$

At this stage it is not clear what is meant by this expression, but it will become more clear what we mean by this by considering several instances of this expression below.

Here, $\Omega \subset \mathbb{R}^n,$ for some $n = 1,2,$ or ${}3.$ We call $\mathbb{R}^n$ the ambient space. Further $\partial \Omega$ is the boundary of $\Omega.$ We require some restrictions on $\Omega,$ that is, we shall assume that $\Omega$ is compact and that the boundary $\partial \Omega$ is closed and connected and can be parameterised by a continuously differentiable function. Note that if the dimension of $\Omega$ is ${}k,$ then the dimension of $\partial \Omega$ is $k-1.$

Further, $\omega$ stands for the integrand and $\mathrm{d} \omega$ stands for some derivative of $\omega$ (to be more precise, $\mathrm{d}$ stands for either differentiation, gradient, divergence, scalar curl or curl; $\rhd$ the importance of these operators can for instance be seen from Helmholtz’s theorem).

We always assume that $\omega$ is continuously differentiable. Again, what is meant by $\omega$ and $\mathrm{d} \omega$ shall become clear by the examples considered below.

Note: We only consider the fundamental theorems of the form (1) only for line and surface integrals over vector fields.

We categorize all the theorems according to the dimension of the ambient space.

A. Ambient space $\mathbb{R}$

We start with the simplest case where the ambient space is $\mathbb{R}.$ This case requires some interpretation of the symbols introduced above.

In this case, $\Omega \subset \mathbb{R}$ (we shall assume that $\Omega=[a,b]$ is an interval). Then the fundamental theorem of calculus asserts that

$\displaystyle \int_a^b f^\prime(x)\,\mathrm{d} x = f(b)-f(a). \hspace{2cm} (2)$

Here, the dimension of $\Omega$ is ${}1,$ hence the dimension of $\partial \Omega$ is ${}0.$ Indeed, the boundary simply consists of the real numbers ${}a$ and ${}b,$ and this boundary has dimension ${}0.$ We define a ${}0$ dimensional subset consisting of two points to be closed and connected by definition (note that the points $a,b$ considered as a ${}1$ dimensional subset, i.e., $\{a,b\}=[a,a]\cup [b,b]$ are not connected if $a \neq b$).

Here $\omega=f$ and $\mathrm{d}\omega=f(x)\,\mathrm{d}x.$ We interpret $\int_{\{a,b\}} \omega$ as $f(b)-f(a).$ By interpreting the symbols in (1) in this way, we can say that (2) is of the form (1).

B. Ambient space $\mathbb{R}^2$

Here we have $\Omega \subset\mathbb{R}^2.$ Hence the dimension of $\Omega$ can be either ${}1$ or ${}2.$

1. Dimension of $\Omega$ is ${}1:$
2. In this case $\Omega$ is a curve and the boundary $\partial \Omega$ consists of two points. The integral over $\Omega$ must therefore be a line integral over some vector field. In fact, this vector field must be a derivative of some form, since it must fit into the form (1). Since the ambient space is $\mathbb{R}^2,$ we consider functions ${}f$ defined on some subset of $\mathbb{R}^2.$ The `derivatives’ of such a function are the gradient, divergence and scalar curl. Of these, only the gradient $\nabla f$ is a vector field defined on a subset of $\mathbb{R}^2.$ Hence the only possible form of the left-hand side of (1) is a line integral over the gradient of a scalar field ${}f.$ Therefore, similarly to the fundamental theorem of calculus, the right-hand side of (1) is just the difference of the function values at the boundary points. Therefore we obtain that (1) must be of the form

$\displaystyle \int_{\Omega} \nabla f\cdot \mathrm{d} \boldsymbol{s} = f(B)-f(A),$

where ${}A$ is the starting point and ${}B$ is the end point of the curve $\Omega.$

3. Dimension of $\Omega$ is ${}2:$
4. Here, $\Omega$ is a ${}2$ dimensional subset of ${}\mathbb{R}^2$ and $\partial \Omega$ is a ${}1$ dimensional subset, i.e. $\partial \Omega$ is a closed curve. Hence the right-hand side of (1) must be a line integral over a vector field $\boldsymbol{F}:\Omega\subset\mathbb{R}^2\to\mathbb{R}^2$ (and this is also the only possible choice). The left-hand side on the other hand is now a double integral over $\Omega$ of a scalar field. This scalar field must be some derivative of the vector field $\boldsymbol{F}.$ There are now two possibilities, namely $\mathrm{d} \omega$ can stand for the divergence of $\boldsymbol{F}$ or the scalar curl of $\boldsymbol{F}.$ In fact, both choices are possible. By this we obtain the two forms of Green’s theorem, namely

$\displaystyle \iint_\Omega \mbox{scalar curl } \boldsymbol{F} \,\mathrm{d} A = \int_{\partial \Omega} \boldsymbol{F}\cdot \widehat{\boldsymbol{T}} \,\mathrm{d} s,$

which is Green’s theorem in tangential form, and

$\displaystyle \iint_\Omega \mbox{div } \boldsymbol{F} \,\mathrm{d} A = \int_{\partial \Omega} \boldsymbol{F} \cdot \widehat{\boldsymbol{n}} \,\mathrm{d} s,$

which is Green’s theorem in normal form. Those two theorems are essentially the same since one can be obtained from the other by a simple substitution.

C. Ambient space $\mathbb{R}^3$

Here we have $\Omega\subset\mathbb{R}^3.$ Hence the dimension of $\Omega$ can be either $1,2$ or ${}3.$

1. Dimension of $\Omega$ is ${}1:$
2. In this case the dimension of $\Omega$ is ${}1$, that is $\Omega$ is a curve and the boundary $\partial \Omega$ consists again of the starting point ${}A$ of the curve and the end point ${}B$ of the curve. In this case, similarly to B.1 we obtain the fundamental theorem of line integrals

$\displaystyle \int_\Omega \nabla f \cdot \mathrm{d} \boldsymbol{s} = f(B)-f(A).$

3. Dimension of $\Omega$ is ${}2:$
4. Now the dimension of $\Omega$ is ${}2$ and hence the dimension of the boundary $\partial \Omega$ is ${}1.$ Thus $\Omega$ is a surface in $\mathbb{R}^3$ and the boundary $\partial \Omega$ is a curve. Hence, the left-hand side of (1) is a surface integral over some vector field which is a derivative and the right-hand side is a line integral over some vector field $\boldsymbol{F}.$ Hence in order to find over what kind of object we are integrating over, we have to ask what derivative of a vector field on $\mathbb{R}^3$ yields again a vector field on $\mathbb{R}^3$? Here, the choices for derivative are gradient, curl or divergence. The gradient and divergence of a vector field are not vector fields on $\mathbb{R}^3,$ hence the only possible choice is $\mbox{curl}.$ Thus $\mathrm{d} \omega$ in (1) must stand for the curl of $\boldsymbol{F}.$ Therefore, in this case (1) is of the form

$\displaystyle \iint_\Omega \mbox{curl }\boldsymbol{F} \cdot \mathrm{d} \boldsymbol{\mathcal{S}} = \int_{\partial \Omega} \boldsymbol{F} \cdot \mathrm{d} \boldsymbol{s}.$

This is Stokes’ theorem.

5. Dimension of $\Omega$ is ${}3:$
6. Here $\Omega$ is ${}3$ dimensional and $\partial \Omega$ is ${}2$ dimensional. Hence $\Omega$ is a ${}3$ dimensional region in space and the boundary $\partial \Omega$ is a surface. Thus the left-hand side of (1) is a triple integral over some region in space and the right-hand side is a surface integral. The surface integral must be over a vector field $\boldsymbol{F}:\Omega\subset \mathbb{R}^3\to\mathbb{R}^3$ and the triple integral must be over a scalar field. Further, this scalar field must be some derivative of $\boldsymbol{F}.$ The only possible choice is therefore that the derivative of $\boldsymbol{F}$ is the divergence, i.e. $\omega$ stands for the vector field $\boldsymbol{F}$ and $\mathrm{d}\omega$ must stand for $\mbox{div } \boldsymbol{F}.$ Thus we obtain

$\displaystyle \iiint_\Omega \mbox{div } \boldsymbol{F} \,\mathrm{d} V = \iint_{\partial \Omega} \boldsymbol{F} \cdot \mathrm{d} \boldsymbol{\mathcal{S}}.$

This is the divergence theorem.

Some further remarks

The above categorization shows that we have exhausted all possibilities, hence we have seen all the fundamental theorems in dimension up to ${}3.$ All of those theorems have the basic structure

$\displaystyle \int_\Omega \mathrm{d} \omega = \int_{\partial \Omega} \omega.$

Notice that this is also the only possible form. If the dimension of $\Omega$ is ${}k,$ then the dimension of $\partial \Omega$ is $k-1.$ Hence on the left-hand side we have one more integral than on the right-hand side. By the fundamental theorem of calculus, integration is undoing the effect of differentiation, i.e. we have $\int f^\prime = f.$ Hence, since we have one more integral on the left-hand side, we need to have some derivative on the left-hand side.

The opposite does not hold. For instance, consider the ambient space $\mathbb{R}^3$ and let $\Omega$ be a ${}3$ dimensional subset of $\mathbb{R}^3.$ Then $\partial \Omega$ is a surface. Now if we would have a derivative on the right-hand side instead of the left-hand side, we could get an expression of the form $\iiint_\Omega f \,\mathrm{d} V=\iint_{\partial \Omega} \nabla f\cdot \mathrm{d}\boldsymbol{\mathcal{S}}$ for some scalar field $f:\Omega\subset\mathbb{R}^3\to\mathbb{R}.$ But this does not hold. Consider for example $f(x,y,z)=1.$ Then $\iiint_\Omega f\,\mathrm{d} V = \mbox{Vol }(\Omega)$ and $\iint_{\partial \Omega} \nabla f \cdot \mathrm{d} \mathcal{\boldsymbol{S}}=0.$

There are also fundamental theorems for the ambient spaces $\boldsymbol{R}^n,$ where $n=4,5,6,\ldots .$ To obtain these, one needs to consider differential forms. How this works is explained in Chapter 5: Additional Material: Differential forms and the general Stokes’ theorem.