# 6.2 Line integrals  (Page 10/20)

 Page 10 / 20

In [link] , what if we had oriented the unit circle clockwise? We denote the unit circle oriented clockwise by $\text{−}C.$ Then

${\oint }_{\text{−}C}\text{F}·\text{T}ds=\text{−}{\oint }_{C}\text{F}·\text{T}ds=-2\pi .$

Notice that the circulation is negative in this case. The reason for this is that the orientation of the curve flows against the direction of F .

Calculate the circulation of $\text{F}\left(x,y\right)=⟨-\frac{y}{{x}^{2}+{y}^{2}},\frac{x}{{x}^{2}+{y}^{2}}⟩$ along a unit circle oriented counterclockwise.

$2\pi$

## Calculating work

Calculate the work done on a particle that traverses circle C of radius 2 centered at the origin, oriented counterclockwise, by field $\text{F}\left(x,y\right)=⟨-2,y⟩.$ Assume the particle starts its movement at $\left(1,0\right).$

The work done by F on the particle is the circulation of F along C : ${\oint }_{C}\text{F}·\text{T}ds.$ We use the parameterization $\text{r}\left(t\right)=⟨2\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t,2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t⟩,0\le t\le 2\pi$ for C . Then, ${r}^{\prime }\left(t\right)=⟨-2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t,2\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t⟩$ and $\text{F}\left(\text{r}\left(t\right)\right)=⟨-2,2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t⟩.$ Therefore, the circulation of F along C is

$\begin{array}{cc}\hfill {\oint }_{C}\text{F}·\text{T}ds& ={\int }_{0}^{2\pi }⟨-2,2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t⟩·⟨-2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t,2\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t⟩dt\hfill \\ & ={\int }_{0}^{2\pi }\left(4\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t+4\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t\right)dt\hfill \\ & ={\left[-4\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t+4\phantom{\rule{0.2em}{0ex}}{\text{sin}}^{2}t\right]}_{0}^{2\pi }\hfill \\ & =\left(-4\phantom{\rule{0.2em}{0ex}}\text{cos}\left(2\pi \right)+2\phantom{\rule{0.2em}{0ex}}{\text{sin}}^{2}\left(2\pi \right)\right)-\left(-4\phantom{\rule{0.2em}{0ex}}\text{cos}\left(0\right)+4\phantom{\rule{0.2em}{0ex}}{\text{sin}}^{2}\left(0\right)\right)\hfill \\ & =-4+4=0.\hfill \end{array}$

The force field does zero work on the particle.

Notice that the circulation of F along C is zero. Furthermore, notice that since F is the gradient of $f\left(x,y\right)=-2x+\frac{{y}^{2}}{2},$ F is conservative. We prove in a later section that under certain broad conditions, the circulation of a conservative vector field along a closed curve is zero.

Calculate the work done by field $\text{F}\left(x,y\right)=⟨2x,3y⟩$ on a particle that traverses the unit circle. Assume the particle begins its movement at $\left(-1,0\right).$

0

## Key concepts

• Line integrals generalize the notion of a single-variable integral to higher dimensions. The domain of integration in a single-variable integral is a line segment along the x -axis, but the domain of integration in a line integral is a curve in a plane or in space.
• If C is a curve, then the length of C is ${\int }_{C}ds.$
• There are two kinds of line integral: scalar line integrals and vector line integrals. Scalar line integrals can be used to calculate the mass of a wire; vector line integrals can be used to calculate the work done on a particle traveling through a field.
• Scalar line integrals can be calculated using [link] ; vector line integrals can be calculated using [link] .
• Two key concepts expressed in terms of line integrals are flux and circulation. Flux measures the rate that a field crosses a given line; circulation measures the tendency of a field to move in the same direction as a given closed curve.

## Key equations

• Calculating a scalar line integral
${\int }_{C}f\left(x,y,z\right)ds={\int }_{a}^{b}f\left(\text{r}\left(t\right)\right)\sqrt{{\left({x}^{\prime }\left(t\right)\right)}^{2}+{\left({y}^{\prime }\left(t\right)\right)}^{2}+{\left({z}^{\prime }\left(t\right)\right)}^{2}}dt$
• Calculating a vector line integral
${\int }_{C}\text{F}·ds={\int }_{C}\text{F}·\text{T}ds={\int }_{a}^{b}\text{F}\left(\text{r}\left(t\right)\right)·{{r}^{\prime }}^{}\left(t\right)dt$
or
${\int }_{C}Pdx+Qdy+Rdz={\int }_{a}^{b}\left(P\left(\text{r}\left(t\right)\right)\frac{dx}{dt}+Q\left(\text{r}\left(t\right)\right)\frac{dy}{dt}+R\left(\text{r}\left(t\right)\right)\frac{dz}{dt}\right)dt$
• Calculating flux
${\int }_{C}\text{F}·\frac{\text{n}\left(t\right)}{‖\text{n}\left(t\right)‖}\phantom{\rule{0.2em}{0ex}}ds={\int }_{a}^{b}\text{F}\left(\text{r}\left(t\right)\right)·\text{n}\left(t\right)dt$

True or False? Line integral ${\int }_{C}^{}f\left(x,y\right)ds$ is equal to a definite integral if C is a smooth curve defined on $\left[a,b\right]$ and if function $f$ is continuous on some region that contains curve C .

True

True or False? Vector functions ${\text{r}}_{1}=t\text{i}+{t}^{2}\text{j},$ $0\le t\le 1,$ and ${\text{r}}_{2}=\left(1-t\right)\text{i}+{\left(1-t\right)}^{2}\text{j},$ $0\le t\le 1,$ define the same oriented curve.

True or False? ${\int }_{\text{−}C}^{}\left(Pdx+Qdy\right)={\int }_{C}^{}\left(Pdx-Qdy\right)$

False

True or False? A piecewise smooth curve C consists of a finite number of smooth curves that are joined together end to end.

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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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