# Math2111: Chapter 3: Recommended reading: Line integrals and orientation

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.

We discuss in what sense the scalar line integrals do not depend on the orientation of the curve.

Let us start with integrals $\int_a^b f(t)\,\mathrm{d} t$ where $f$ is a Riemann integrable function. Let $a \textless b$ and let $P_n=\{x_0,\ldots, x_n\}$ be such that $a=x_0 \textless x_1 \textless x_2 \textless \cdots \textless x_{n-1} \textless x_n =b,$ let $x_{i-1} \le t_i \le x_{i}$ for $1\le i \le n$ and $\delta(P_n)=\max_{1\le i\le n} x_{i+1}-x_i.$ We can now form the Riemann sum

$\displaystyle \mathcal{S}(P_n) = \sum_{i=1}^n f(t_i) (x_i-x_{i-1}).$

By considering the limit $\lim \mathcal{S}(P_n)$ where $n\to\infty$ such that $\delta(P_n) \to 0,$ we obtain the Riemann integral $\int_a^b f(t)\,\mathrm{d} t.$

Notice that we now have only defined the integral for the case when $a \textless b.$ This definition does not apply to integrals of the form $\int_\alpha^\beta f(t)\,\mathrm{d} t$ where $\alpha \textgreater \beta.$ Hence we are still free to choose what we mean by $\int_\alpha^\beta f(t)\,\mathrm{d} t$ for $\alpha \textgreater \beta.$ It is customary, however, to define

$\displaystyle \int_\alpha^\beta f(t)\,\mathrm{d} t := -\int_\beta^\alpha f(t)\,\mathrm{d} t \quad \mbox{for } \alpha \textgreater \beta. \qquad\qquad (1)$

This corresponds to reversing the partition ${}P$ in the sense that we now have $\beta=x_n\textless x_{n-1}\textless \cdots \textless x_1\textless x_0=\alpha,$ that is, we now have $x_{i+1}\textless x_i$ instead of $x_i\textless x_{i+1}.$

Let now $a \textless b.$ If $f(t)\ge 0$ for all $t\in [a,b],$ then $\mathcal{S}(P_n)\ge 0$ for all $P_n$ and hence it follows that $\int_a^b f(t) \,\mathrm{d} t \ge 0.$ Let now $g(t)=f(a+b-t),$ that is, we reverse the orientation in the sense that $g(a)=f(b)$ and $g(b)=f(a).$ Then we still have $g(t)=f(a+b-t) \ge 0.$ What happens now to $\int_a^b g(t)\,\mathrm{d} t?$ Using the substitution $u = a+b-t$ we obtain

$\displaystyle \int_a^b g(t)\,\mathrm{d} t= -\int_b^a g(a+b-t)\,\mathrm{d} t=\int_a^b f(t)\,\mathrm{d} t \ge 0$

since $g(a+b-t)=f(t)$. Thus, in this sense, integrating the function ${}f$ from ${}a$ to ${}b$ yields the same result as when integrating ${}f$ from ${}b$ to ${}a.$

Scalar line integrals

A similar result holds for scalar line integrals. Assume we have a curve $\mathcal{C}$ with end points ${}P$ and ${}Q$ and a continuous function ${}f$ defined on the curve $\mathcal{C}.$ Let $\boldsymbol{c}:[a,b]\to\mathbb{R}^3,$ where $a\textless b,$ be a continuously differentiable parameterisation of $\mathcal{C}$ such that $\boldsymbol{c}(a)=P$ and $\boldsymbol{c}(b)=Q.$ Then, if we define $\boldsymbol{r}:[a,b]\to\mathbb{R}^3$ by $\boldsymbol{r}(t)=\boldsymbol{c}(a+b-t),$ then $\boldsymbol{r}(a)=Q$ and $\boldsymbol{r}(b)=P.$

Hence, as ${}t$ goes from ${}a$ to ${}b,$ the parameterisation $\boldsymbol{c}$ traverses the curve $\mathcal{C}$ from ${}P$ to ${}Q,$ whereas the parameterisation $\boldsymbol{r}$ traverses the curve in the opposite direction from ${}Q$ to ${}P.$ Then, by using the substitution $u=a+b-t,$ we obtain

$\displaystyle \begin{array}{rcl} \int_a^b f(\boldsymbol{c}(t)) \|\boldsymbol{c}^\prime(t)\|\,\mathrm{d} t &=& -\int_b^a f(\boldsymbol{c}(a+b-u)) \|\boldsymbol{c}^\prime(a+b-u)\| \,\mathrm{d} u \\&&\\ &=& \int_a^b f(\boldsymbol{r}(u)) \|\boldsymbol{r}^\prime(u)\|\,\mathrm{d} u,\end{array}$

since $\boldsymbol{c}(a+b-u)=\boldsymbol{r}(u).$ In this sense the scalar line integral is independent of the orientation of the curve $\mathcal{C}.$ For instance, this implies that if $a \textless b$ and $f\ge 0$ for all points on the curve, then

$\displaystyle \int_{a}^b f(\boldsymbol{c}(t)) \|\boldsymbol{c}^\prime(t)\|\,\mathrm{d} t \ge 0,$

regardless in which direction the parameterisation $\boldsymbol{c}$ traverses the curve $\mathcal{C}.$ Hence, when computing scalar line integrals, we do not need to specify the direction in which the parameterisation traces out the curve $\mathcal{C}.$

However, we have

$\displaystyle \int_b^a f(\boldsymbol{c}(t)) \|\boldsymbol{c}^\prime(t)\|\,\mathrm{d} t =- \int_a^b f(\boldsymbol{c}(t))\|\boldsymbol{c}(t)\|\,\mathrm{d}t.$

In this case the orientation of the curve is also reversed, but the minus sign comes from the convention (1) and not from changing the orientation of the curve, in the sense that if we left out the minus sign in (1), we would also get $\displaystyle \int_b^a f(\boldsymbol{c}(t)) \|\boldsymbol{c}^\prime(t)\|\,\mathrm{d} t = \int_a^b f(\boldsymbol{c}(t))\|\boldsymbol{c}(t)\|\,\mathrm{d}t.$

Note that this implies that the scalar line integral needs to be set up such that we have $\int_a^b \cdots$ where $a \textless b.$

Line integrals of vector fields

We now consider line integrals of vector fields. In this case the situation is different. We use the same notation as in the part on scalar line integrals.

Let $\boldsymbol{F}$ be a vector field defined on the curve $\mathcal{C}.$ We have $\boldsymbol{r}^\prime(t)= \frac{\mathrm{d}}{\mathrm{d} t} \boldsymbol{c}(a+b-t) = -\boldsymbol{c}^\prime(a+b-t).$ Hence, if we now go through the same steps as above we obtain

$\displaystyle \int_a^b \boldsymbol{F}(\boldsymbol{c}(t)) \cdot \boldsymbol{c}^\prime(t)\,\mathrm{d} t= - \int_a^b \boldsymbol{F}(\boldsymbol{r}(t))\cdot \boldsymbol{r}^\prime(t)\,\mathrm{d} t.$

Hence, changing the orientation of the curve for line integrals over vector fields also changes the sign in the result.

Both of these results are also clear from the interpretation of scalar line integrals as the area of a fence and the interpretation of line integrals of vector fields as the work done by a force.