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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] ).
Let F be a vector field and let C be a smooth curve with parameterization $\text{r}(t)=\u27e8x(t),y(t)\u27e9,a\le t\le b.$ Let $\text{n}(t)=\u27e8{y}^{\prime}(t),\text{\u2212}{x}^{\prime}(t)\u27e9.$ The flux of F across C is
The proof of [link] is similar to the proof of [link] . Before deriving the formula, note that $\Vert \text{n}(t)\Vert =\Vert \u27e8y\prime (t),\text{\u2212}x\prime (t)\u27e9\Vert =\sqrt{{\left(y\prime (t)\right)}^{2}+{\left(x\prime (t)\right)}^{2}}=\Vert {r}^{\prime}(t)\Vert .$ Therefore,
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Calculate the flux of $\text{F}=\u27e82x,2y\u27e9$ 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 $\text{r}(t)=\u27e8\text{cos}\phantom{\rule{0.2em}{0ex}}t,\text{sin}\phantom{\rule{0.2em}{0ex}}t\u27e9,$ $0\le t\le 2\pi .$ The normal vector to a unit circle is $\u27e8\text{cos}\phantom{\rule{0.2em}{0ex}}t,\text{sin}\phantom{\rule{0.2em}{0ex}}t\u27e9.$ Therefore, the flux is
Calculate the flux of $\text{F}=\u27e8x+y,2y\u27e9$ across the line segment from $(0,0)$ to $(2,3),$ where the curve is oriented from left to right.
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Let $\text{F}(x,y)=\u27e8P(x,y),Q(x,y)\u27e9$ be a two-dimensional vector field. Recall that integral ${\int}_{C}\text{F}\xb7\text{T}}ds$ is sometimes written as ${\int}_{C}Pdx+Qdy}.$ Analogously, flux ${\int}_{C}\text{F}\xb7\text{N}ds$ is sometimes written in the notation ${\int}_{C}\text{\u2212}Qdx+Pdy,$ because the unit normal vector N is perpendicular to the unit tangent T . Rotating the vector $d\text{r}=\u27e8dx,dy\u27e9$ by 90° results in vector $\u27e8dy,\text{\u2212}dx\u27e9.$ Therefore, the line integral in [link] can be written as ${\int}_{C}\mathrm{-2}ydx+2xdy}.$
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: ${\oint}_{C}\text{F}\xb7\text{T}ds}.$ 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 $\text{v}(P)\xb7\text{T}(P).$ The maximum value of $\text{v}(P)\xb7\text{T}(P)$ occurs when the two vectors are pointing in the exact same direction; the minimum value of $\text{v}(P)\xb7\text{T}(P)$ occurs when the two vectors are pointing in opposite directions. Thus, the value of the circulation ${\oint}_{C}\text{v}\xb7\text{T}ds$ measures the tendency of the fluid to move in the direction of C .
Let $\text{F}=\u27e8-y,x\u27e9$ 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: $\text{r}(t)=\u27e8\text{cos}\phantom{\rule{0.2em}{0ex}}t,\text{sin}\phantom{\rule{0.2em}{0ex}}t\u27e9,0\le t\le 2\pi .$ Then, $\text{F}\left(\text{r}(t)\right)=\u27e8\text{\u2212}\text{sin}\phantom{\rule{0.2em}{0ex}}t,\text{cos}\phantom{\rule{0.2em}{0ex}}t\u27e9$ and ${r}^{\prime}(t)=\u27e8\text{\u2212}\text{sin}\phantom{\rule{0.2em}{0ex}}t,\text{cos}\phantom{\rule{0.2em}{0ex}}t\u27e9.$ Therefore, the circulation of F along C is
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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