# 6.2 Line integrals  (Page 9/20)

 Page 9 / 20

We now give a formula for calculating the flux across a curve. This formula is analogous to the formula used to calculate a vector line integral (see [link] ).

## Calculating flux across a curve

Let F be a vector field and let C be a smooth curve with parameterization $\text{r}\left(t\right)=⟨x\left(t\right),y\left(t\right)⟩,a\le t\le b.$ Let $\text{n}\left(t\right)=⟨{y}^{\prime }\left(t\right),\text{−}{x}^{\prime }\left(t\right)⟩.$ The flux of F across C is

${\int }_{C}\text{F}·\text{N}ds={\int }_{a}^{b}\text{F}\left(\text{r}\left(t\right)\right)·\text{n}\left(t\right)dt$

## Proof

The proof of [link] is similar to the proof of [link] . Before deriving the formula, note that $‖\text{n}\left(t\right)‖=‖⟨y\prime \left(t\right),\text{−}x\prime \left(t\right)⟩‖=\sqrt{{\left(y\prime \left(t\right)\right)}^{2}+{\left(x\prime \left(t\right)\right)}^{2}}=‖{r}^{\prime }\left(t\right)‖.$ Therefore,

$\begin{array}{cc}\hfill {\int }_{C}\text{F}·\text{N}ds& ={\int }_{C}\text{F}·\frac{\text{n}\left(t\right)}{‖\text{n}\left(t\right)‖}ds\hfill \\ & ={\int }_{a}^{b}\text{F}·\frac{\text{n}\left(t\right)}{‖\text{n}\left(t\right)‖}‖{r}^{\prime }\left(t\right)‖dt\hfill \\ & ={\int }_{a}^{b}\text{F}\left(\text{r}\left(t\right)\right)·\text{n}\left(t\right)dt.\hfill \end{array}$

## Flux across a curve

Calculate the flux of $\text{F}=⟨2x,2y⟩$ across a unit circle oriented counterclockwise ( [link] ).

To compute the flux, we first need a parameterization of the unit circle. We can use the standard parameterization $\text{r}\left(t\right)=⟨\text{cos}\phantom{\rule{0.2em}{0ex}}t,\text{sin}\phantom{\rule{0.2em}{0ex}}t⟩,$ $0\le t\le 2\pi .$ The normal vector to a unit circle is $⟨\text{cos}\phantom{\rule{0.2em}{0ex}}t,\text{sin}\phantom{\rule{0.2em}{0ex}}t⟩.$ Therefore, the flux is

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

Calculate the flux of $\text{F}=⟨x+y,2y⟩$ across the line segment from $\left(0,0\right)$ to $\left(2,3\right),$ where the curve is oriented from left to right.

3/2

Let $\text{F}\left(x,y\right)=⟨P\left(x,y\right),Q\left(x,y\right)⟩$ be a two-dimensional vector field. Recall that integral ${\int }_{C}\text{F}·\text{T}ds$ is sometimes written as ${\int }_{C}Pdx+Qdy.$ Analogously, flux ${\int }_{C}\text{F}·\text{N}ds$ is sometimes written in the notation ${\int }_{C}\text{−}Qdx+Pdy,$ because the unit normal vector N is perpendicular to the unit tangent T . Rotating the vector $d\text{r}=⟨dx,dy⟩$ by 90° results in vector $⟨dy,\text{−}dx⟩.$ Therefore, the line integral in [link] can be written as ${\int }_{C}-2ydx+2xdy.$

Now that we have defined flux, we can turn our attention to circulation. The line integral of vector field F along an oriented closed curve is called the circulation    of F along C . Circulation line integrals have their own notation: ${\oint }_{C}\text{F}·\text{T}ds.$ The circle on the integral symbol denotes that C is “circular” in that it has no endpoints. [link] shows a calculation of circulation.

To see where the term circulation comes from and what it measures, let v represent the velocity field of a fluid and let C be an oriented closed curve. At a particular point P , the closer the direction of v ( P ) is to the direction of T ( P ), the larger the value of the dot product $\text{v}\left(P\right)·\text{T}\left(P\right).$ The maximum value of $\text{v}\left(P\right)·\text{T}\left(P\right)$ occurs when the two vectors are pointing in the exact same direction; the minimum value of $\text{v}\left(P\right)·\text{T}\left(P\right)$ occurs when the two vectors are pointing in opposite directions. Thus, the value of the circulation ${\oint }_{C}\text{v}·\text{T}ds$ measures the tendency of the fluid to move in the direction of C .

## Calculating circulation

Let $\text{F}=⟨-y,x⟩$ be the vector field from [link] and let C represent the unit circle oriented counterclockwise. Calculate the circulation of F along C .

We use the standard parameterization of the unit circle: $\text{r}\left(t\right)=⟨\text{cos}\phantom{\rule{0.2em}{0ex}}t,\text{sin}\phantom{\rule{0.2em}{0ex}}t⟩,0\le t\le 2\pi .$ Then, $\text{F}\left(\text{r}\left(t\right)\right)=⟨\text{−}\text{sin}\phantom{\rule{0.2em}{0ex}}t,\text{cos}\phantom{\rule{0.2em}{0ex}}t⟩$ and ${r}^{\prime }\left(t\right)=⟨\text{−}\text{sin}\phantom{\rule{0.2em}{0ex}}t,\text{cos}\phantom{\rule{0.2em}{0ex}}t⟩.$ Therefore, the circulation of F along C is

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

Notice that the circulation is positive. The reason for this is that the orientation of C “flows” with the direction of F . At any point along the circle, the tangent vector and the vector from F form an angle of less than 90°, and therefore the corresponding dot product is positive.

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