\ifdefined\inmaster\else\def\subonly{\jobname}\input{calc3}\fi \chapter{Line integrals} \label{cha:line-integrals} We introduced differential forms by saying that when you wrote \[ \itint x a b {f(x)} \] in one-variable calculus, you were actually integrating not the function $f$ but the differential form $f(x)\,\dx$. We will now make precise what it means to integrate a differential form, and generalize to the multivariable situation. \section{Line integrals} \label{sec:line-integrals} If we want to integrate a form in more than one dimension, we have a new problem: we need more information to specify the ``interval'' over which to integrate. The most appropriate sort of ``interval'' turns out to be a \textbf{parametrized curve} (or \emph{parametric curve}), a notion which you might have encountered in one-variable calculus. Recall that a parametrized curve $\curve$ in two dimensions consists of a \emph{parameter interval} $[a,b]$ and two functions $\curvex(t)$ and $\curvey(t)$ defined for all $t\in [a,b]$. The curve is then the set of points $\curvept(t) = (\curvex(t),\curvey(t))$ as $t$ ranges over some domain (we call $t$ the \emph{parameter}). Similarly, in three dimensions we have three functions $\curvex(t)$, $\curvey(t)$, and $\curvez(t)$ and we consider the points $\curvept(t) = (\curvex(t),\curvey(t),\curvez(t))$. We say that a parametrized curve is \textbf{continuous} or \textbf{differentiable} if the functions $\curvex$, $\curvey$, and (possibly) $\curvez$ are continuous or differentiable. \begin{stewart}Parametrized curves are discussed in sections 1.7 and 10.1 of the textbook.\end{stewart}% \begin{hugheshallett}Parametrized curves are discussed in sections 17.1--17.2 of the textbook.\end{hugheshallett}% \begin{rogawski}Parametrized curves are discussed in section 11.1 of the textbook.\end{rogawski}% Now suppose $\omega$ is a differential form in two dimensions, which we want to integrate over a parametrized curve $\curve$ defined on $[a,b]$. We regard $\omega(\ptx,\dpx)$ as telling us about the first-order changes in some quantity Q as $\ptx$ changes to $\ptx+\dpx$, whereas we are interested in the total change of Q ``along $\curve$''. We can approximate this total change by dividing $[a,b]$ into a large number $N$ of subintervals $[t_{i-1},t_i]$, and choosing a point $\rstag t i$ in each such subinterval. Then if we consider the part of $\curve$ between $t_{i-1}$ and $t_i$ to be approximately a straight line, we can approximate the first-order change in Q along that part by $\omega(\curvept(\rstag t i),\Delta \curvept_i)$, where $\Delta\curvept_i = \curvept(t_i)-\curvept(t_{i-1})$. Adding these up, we obtain the Riemann sum \[ \sum_{i=1}^N \omega(\curvept(\rstag t i),\Delta \curvept_i) \] and we define the \textbf{line integral} of $\omega$ over $\curve$ to be the limit of these sums, as the widths $\Delta t_i = t_i - t_{i-1}$ go to zero: \begin{equation} \lint{\curve} \omega = \lim_{\Delta t_i \to 0} \sum_{i=1}^N \omega(\curvept(\rstag t i),\Delta \curvept_i).\label{eq:line-integral} \end{equation} (It would be more appropriate to call this a \textit{curve integral}, since $\curve$ is a curve and not usually a straight line, but the term ``line integral'' is standard.) If the curve $\curve$ is differentiable, then we can write $\dcurvept = \curvept'(t)\,\dt$, or in components \[\dx = x'(t)\,\dt \qquad \dy = y'(t) \,\dt \qquad \dz = z'(t)\,\dt. \] (That is, $\curvept'(t)$ denotes the vector $(\curvex'(t),\curvey'(t))$ or $(\curvex'(t),\curvey'(t),\curvez'(t))$.) Thus, we might expect to be able to evaluate a line integral by substituting these values in $\omega$ to get an ordinary integral in terms of $t$. This is in fact possible, although it is not trivial to prove; it is a special case of the \emph{Change of Variables Theorem}. Here we have another situation like the chain rule, where the notation of differentials makes a deep fact appear obvious and easy-to-use. \begin{eg}\label{eg:lineint-cov} Suppose $x(t) = t^2$ and $y(t) = t^3$ define a curve in two dimensions with domain $[0,2]$, so that $\dx = 2t\,\dt$ and $\dy = 3t^2\,\dt$. Then we can evaluate a line integral such as \[ \lint{\curve} (x^2 y \,\dx - x \,\dy) \] by substituting these values for $x$, $y$, $\dx$, and $\dy$, obtaining the ordinary integral \[ \int_{t=0}^2 \Big( (t^2)^2(t^3)(2t\,\dt) - (t^2)(3t^2\,\dt)\Big) = \int_{t=0}^2 (2t^8 - 3t^4)\,\dt\] which we can evaluate using the ordinary fundamental theorem of calculus. \end{eg} Recall that differential forms are functions whose input is a point and a vector. Using this, we can formally state the Change of Variables Theorem for line integrals. \begin{thm}[Change of Variables] If $\curve$ is a differentiable curve with parameter in $[a,b]$, and $\omega$ is a differential form such that for each $\ptx$, the function $\omega(\ptx,\dpx)$ is Lipschitz continuous in $\dpx$, then \begin{equation} \lint{\curve} \omega = \lint{[a,b]} \omega\big(\curvept(t),\, \curvept'(t) \,\dt\big)\label{eq:lineint-chvar-2} \end{equation} \end{thm} Now suppose $\omega$ is a \emph{linear} differential form. Recall that that means it looks like $f(\ptx)\,\dx + g(\ptx)\,\dy + h(\ptx)\,\dz$. Then \begin{align*} \omega\big(\curvept(t),\, \curvept'(t) \,\dt\big) &= f(\curvept(t))\,\curvept'(t) \,\dt + g(\curvept(t))\,\curvept'(t) \,\dt + h(\curvept(t))\,\curvept'(t) \,\dt\\ &= \Big(f(\curvept(t))\,\curvept'(t) + g(\curvept(t))\,\curvept'(t) + h(\curvept(t))\,\curvept'(t)\Big) \,\dt\\ &= \omega\big(\curvept(t),\, \curvept'(t)\big)\,\dt \end{align*} and thus the right-hand side of \cref{eq:lineint-chvar-2} becomes \[ \itint t a b {\omega\big(\curvept(t),\, \curvept'(t)\big)}. \] This is an ordinary one-variable integral of the sort we already understand. Therefore, the line integral of any linear differential form over any differentiable curve can be reduced to a one-variable integral just as in \cref{eg:lineint-cov}, which we can at least hope to integrate using the fundamental theorem of calculus. More generally, if $\curve$ is only \emph{piecewise} differentiable, then we can break a line integral up into several integrals and apply this method on each piece. \begin{notextbook}Your textbook probably contains more details.\end{notextbook}% \begin{stewart}More details are in the textbook (section 13.2).\end{stewart}% \begin{hugheshallett}See the textbook (sections 18.1--18.2).\end{hugheshallett}% \begin{rogawski}See the textbook (section 16.2).\end{rogawski}% \begin{rmk}\label{rmk:lineint-orientation} Although we used a particular parametrization of the curve $\curve$ to define the line integral $\lint{\curve} \omega$ in \cref{eq:line-integral}, and to evaluate it as in \cref{eq:lineint-chvar-2}, in fact the value of the integral is independent of the parametrization. This makes sense if we look at the definition of the Riemann sums in \cref{eq:line-integral}, which depend only on the values of $\omega$ at the points $\curvept(\rstag t i)$ and the vectors $\Delta \curvept_i$; any other parametrization would induce a corresponding division into subintervals with the same points and vectors. It is necessary that any parametrization must traverse the curve in a specific direction, and \emph{continue} going in that direction: it cannot stop and turn around, ``re-covering'' part of the curve more than once. Moreover, the integral \emph{does} depend on the direction in which the parametrization traverses the curve; a choice of such a direction is called an \emph{orientation} of the curve $\curve$. Reversing the orientation would negate the vectors $\Delta \curvept_i$, and for a \emph{linear} differential form, this would negate the value of the integral. (There are some nonlinear forms whose line integrals are independent of orientation, however; we will see some in \cref{sec:arc-length}.) \end{rmk} Finally, recall that any vector field $\vc{v}$ gives rise to a linear differential form $\dotp{\vc{v}(\ptx)}{\dpx}$. We can then integrate this form over any curve, obtaining a line integral denoted \[ \lint{\curve} \dotp{\vc{v}(\ptx)}{\dpx} \] or any number of similar notations. \begin{notextbook}You will probably find this in your textbook.\end{notextbook}% \begin{stewart}The textbook writes $\lint{\curve} \dotp{\vc{v}}{d\pt{r}}$ (section 13.2).\end{stewart}% \begin{hugheshallett}The textbook writes $\lint{\curve} \dotp{\vc{v}}{d\pt{r}}$ (section 18.1).\end{hugheshallett}% \begin{rogawski}The textbook writes $\lint{\curve} \dotp{\vc{v}}{d\pt{s}}$ (section 16.2), calling it a \textbf{vector line integral}.\end{rogawski} This is the commonly used notation for line integrals in vector calculus, and it includes many cases of interest because many differential forms are linear. \section{Arc length} \label{sec:arc-length} The most important example of a \emph{nonlinear} line integral is one that you may already have encountered in one-variable calculus: the \emph{arc length} of a parametrized curve. To find this as an integral, we should choose $\omega$ to be a first-order approximation to the length of a bit of curve. If we consider the curve to be approximately a line, then this would be just the length of the vector $\dpx$. Thus, in two dimensions we should integrate the differential form \[ \sqrt{\dx^2+\dy^2}\] while in three dimensions we should integrate \[ \sqrt{\dx^2+\dy^2 + \dz^2}.\] It is traditional to denote these differential forms by $\mathrm{d}\ell$ or $\mathrm{d}s$, but this is potentially confusing because they are not the differential of anything. \begin{stewart}(The textbook uses $ds$.)\end{stewart}% \begin{rogawski}(The textbook uses $ds$.)\end{rogawski} % Hughes-Hallett doesn't mention arc length integrals. We will instead denote them by $\dl$ (note the stroke over the $\mathrm{d}$). That is, we define \[ \dl = \mgn{\dpx} = \begin{cases} \sqrt{\dx^2+\dy^2}\qquad\text{in two dimensions}\\[3pt] \sqrt{\dx^2+\dy^2+\dz^2}\qquad\text{in three dimensions}. \end{cases}\] The form $\dl$ is sometimes called the \textbf{length element} or the \textbf{line element}. Now the \textbf{arc length} of a curve $\curve$ is defined to be \[ \text{arc length of } \curve = \lint{\curve} \dl \] if this integral exists. More generally, if $f$ is a multivariable function, then the \textbf{line integral of $f$ with respect to arc length} over $\curve$ is defined to be \[ \lint{\curve} f(\ptx)\,\dl. \] \begin{rogawski}The textbook calls this a \textbf{scalar line integral}.\end{rogawski} Even though $\dl$ is not a linear differential form, it is Lipschitz continuous in $\dpx$. Thus, the Change of Variables Theorem applies to arc length integrals, and can reduce them to one-variable integrals. In fact, if $\dx = x'(t) \,\dt$ and $\dy = y'(t)\,\dt$, then we can write \begin{align*} \dl &= \sqrt{\dx^2+\dy^2}\\ &= \sqrt{(x'(t)\,\dt)^2 + (y'(t)\,\dt)^2}\\ &= \sqrt{(x'(t)^2 + y'(t)^2) \,\dt^2}\\ &= \sqrt{x'(t)^2 + y'(t)^2} \;|\dt|. \end{align*} and we know how to evaluate integrals of forms such as $f(t)\,|\dt|$: we use \cref{eq:onevar-absint}. In fact, since we nearly always parametrize curves over intervals $[a,b]$ with $a