# 6.2 Line integrals  (Page 6/20)

 Page 6 / 20

## Reversing orientation

Find the value of integral ${\int }_{C}\text{F}·d\text{r},$ where $C$ is the semicircle parameterized by $\text{r}\left(t\right)=⟨\text{cos}\phantom{\rule{0.2em}{0ex}}t+\pi ,\text{sin}\phantom{\rule{0.2em}{0ex}}t⟩,0\le t\le \pi$ and $\text{F}=⟨\text{−}y,x⟩.$

Notice that this is the same problem as [link] , except the orientation of the curve has been traversed. In this example, the parameterization starts at $\text{r}\left(0\right)=⟨\pi ,0⟩$ and ends at $\text{r}\left(\pi \right)=⟨0,0⟩.$ By [link] ,

$\begin{array}{cc}\hfill {\int }_{C}\text{F}·d\text{r}& ={\int }_{0}^{\pi }⟨\text{−}\text{sin}\phantom{\rule{0.2em}{0ex}}t,\text{cos}\phantom{\rule{0.2em}{0ex}}t+\pi ⟩·⟨\text{−}\text{sin}\phantom{\rule{0.2em}{0ex}}t+\pi ,\text{cos}\phantom{\rule{0.2em}{0ex}}t⟩dt\hfill \\ & ={\int }_{0}^{\pi }⟨\text{−}\text{sin}\phantom{\rule{0.2em}{0ex}}t,\text{−}\text{cos}\phantom{\rule{0.2em}{0ex}}t⟩·⟨\text{sin}\phantom{\rule{0.2em}{0ex}}t,\text{cos}\phantom{\rule{0.2em}{0ex}}t⟩dt\hfill \\ & ={\int }_{0}^{\pi }\left(\text{−}{\text{sin}}^{2}t-{\text{cos}}^{2}t\right)dt\hfill \\ & ={\int }_{0}^{\pi }-1dt\hfill \\ & =\text{−}\pi .\hfill \end{array}$

Notice that this is the negative of the answer in [link] . It makes sense that this answer is negative because the orientation of the curve goes against the “flow” of the vector field.

Let C be an oriented curve and let − C denote the same curve but with the orientation reversed. Then, the previous two examples illustrate the following fact:

${\int }_{C}\text{F}·d\text{r}=\text{−}{\int }_{C}\text{F}·d\text{r}.$

That is, reversing the orientation of a curve changes the sign of a line integral.

Let $\text{F}=x\text{i}+y\text{j}$ be a vector field and let C be the curve with parameterization $⟨t,{t}^{2}⟩$ for $0\le t\le 2.$ Which is greater: ${\int }_{C}\text{F}·\text{T}ds$ or ${\int }_{\text{−}C}\text{F}·\text{T}ds?$

${\int }_{C}\text{F}·\text{T}ds$

Another standard notation for integral ${\int }_{C}\text{F}·d\text{r}$ is ${\int }_{C}Pdx+Qdy+Rdz.$ In this notation, P , Q , and R are functions, and we think of d r as vector $⟨dx,dy,dz⟩.$ To justify this convention, recall that $d\text{r}=\text{T}ds={r}^{\prime }\left(t\right)dt=⟨\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}⟩dt.$ Therefore,

$\text{F}·d\text{r}=⟨P,Q,R⟩·⟨dx,dy,dz⟩=Pdx+Qdy+Rdz.$

If $d\text{r}=⟨dx,dy,dz⟩,$ then $\frac{d\text{r}}{dt}=⟨\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}⟩,$ which implies that $\frac{d\text{r}}{dt}=⟨\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}⟩dt.$ Therefore

$\begin{array}{cc}\hfill {\int }_{C}\text{F}·d\text{r}& ={\int }_{C}Pdx+Qdy+Rdz\hfill \\ & =\left(P\left(\text{r}\left(t\right)\right)\frac{dx}{dt}+Q\left(\text{r}\left(t\right)\right)\frac{dy}{dt}+R\left(\text{r}\left(t\right)\right)\frac{dz}{dt}\right)dt.\hfill \end{array}$

## Finding the value of an integral of the form ${\int }_{C}Pdx+Qdy+Rdz$

Find the value of integral ${\int }_{C}zdx+xdy+ydz,$ where C is the curve parameterized by $\text{r}\left(t\right)=⟨{t}^{2},\sqrt{t},t⟩,1\le t\le 4.$

As with our previous examples, to compute this line integral we should perform a change of variables to write everything in terms of t . In this case, [link] allows us to make this change:

$\begin{array}{cc}\hfill {\int }_{C}zdx+xdy+ydz& ={\int }_{1}^{4}\left(t\left(2t\right)+{t}^{2}\left(\frac{1}{2\sqrt{t}}\right)+\sqrt{t}\right)dt\hfill \\ & ={\int }_{1}^{4}\left(2{t}^{2}+\frac{{t}^{3\text{/}2}}{2}+\sqrt{t}\right)dt\hfill \\ & ={\left[\frac{2{t}^{3}}{3}+\frac{{t}^{5\text{/}2}}{5}+\frac{2{t}^{3\text{/}2}}{3}\right]}_{t=1}^{t=4}\hfill \\ & =\frac{793}{15}.\hfill \end{array}$

Find the value of ${\int }_{C}4xdx+zdy+4{y}^{2}dz,$ where $C$ is the curve parameterized by $\text{r}\left(t\right)=⟨4\phantom{\rule{0.2em}{0ex}}\text{cos}\left(2t\right),2\phantom{\rule{0.2em}{0ex}}\text{sin}\left(2t\right),3⟩,0\le t\le \frac{\pi }{4}.$

$-26$

We have learned how to integrate smooth oriented curves. Now, suppose that C is an oriented curve that is not smooth, but can be written as the union of finitely many smooth curves. In this case, we say that C is a piecewise smooth curve    . To be precise, curve C is piecewise smooth if C can be written as a union of n smooth curves ${C}_{1},{C}_{2}\text{,…},{C}_{n}$ such that the endpoint of ${C}_{i}$ is the starting point of ${C}_{i+1}$ ( [link] ). When curves ${C}_{i}$ satisfy the condition that the endpoint of ${C}_{i}$ is the starting point of ${C}_{i+1},$ we write their union as ${C}_{1}+{C}_{2}+\cdots +{C}_{n}.$ The union of C 1 , C 2 , C 3 is a piecewise smooth curve.

The next theorem summarizes several key properties of vector line integrals.

## Properties of vector line integrals

Let F and G be continuous vector fields with domains that include the oriented smooth curve C . Then

1. ${\int }_{C}\left(\text{F}+\text{G}\right)·d\text{r}={\int }_{C}\text{F}·d\text{r}+{\int }_{C}\text{G}·d\text{r}$
2. ${\int }_{C}k\text{F}·d\text{r}=k{\int }_{C}\text{F}·d\text{r},$ where k is a constant
3. ${\int }_{C}\text{F}·d\text{r}={\int }_{\text{−}C}\text{F}·d\text{r}$
4. Suppose instead that C is a piecewise smooth curve in the domains of F and G , where $C={C}_{1}+{C}_{2}+\cdots +{C}_{n}$ and ${C}_{1},{C}_{2}\text{,…},{C}_{n}$ are smooth curves such that the endpoint of ${C}_{i}$ is the starting point of ${C}_{i+1}.$ Then
${\int }_{C}\text{F}·d\text{s}={\int }_{{C}_{1}}\text{F}·d\text{s}+{\int }_{{C}_{2}}\text{F}·d\text{s}+\cdots +{\int }_{{C}_{n}}\text{F}·d\text{s}.$

Notice the similarities between these items and the properties of single-variable integrals. Properties i. and ii. say that line integrals are linear, which is true of single-variable integrals as well. Property iii. says that reversing the orientation of a curve changes the sign of the integral. If we think of the integral as computing the work done on a particle traveling along C , then this makes sense. If the particle moves backward rather than forward, then the value of the work done has the opposite sign. This is analogous to the equation ${\int }_{a}^{b}f\left(x\right)dx=\text{−}{\int }_{a}^{b}f\left(x\right)dx.$ Finally, if $\left[{a}_{1},{a}_{2}\right],\left[{a}_{2},{a}_{3}\right]\text{,…},\left[{a}_{n-1},{a}_{n}\right]$ are intervals, then

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