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A simple diagram of an increasing concave down curve C in vector field F, with no coordinate plane. Towards the top of the curve, the normal n is drawn perpendicular to the curve C. Another arrow F is drawn sharing n’s endpoint. This flux points up and to the right at about a 90-degree angle to n. The arrows in the vector field to the left of n are drawn pointing straight up. The arrows after n point in the same direction as the flux.
The flux of vector field F across curve C is computed by an integral similar to a vector line integral.

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 r ( t ) = x ( t ) , y ( t ) , a t b . Let n ( t ) = y ( t ) , x ( t ) . The flux of F across C is

C F · N d s = a b F ( r ( t ) ) · n ( t ) d t


The proof of [link] is similar to the proof of [link] . Before deriving the formula, note that n ( t ) = y ( t ) , x ( t ) = ( y ( t ) ) 2 + ( x ( t ) ) 2 = r ( t ) . Therefore,

C F · N d s = C F · n ( t ) n ( t ) d s = a b F · n ( t ) n ( t ) r ( t ) d t = a b F ( r ( t ) ) · n ( t ) d t .

Flux across a curve

Calculate the flux of F = 2 x , 2 y across a unit circle oriented counterclockwise ( [link] ).

A unit circle in a vector field in two dimensions. The arrows point away from the origin in a radial pattern. Shorter vectors are near the origin, and longer ones are further away. A unit circle is drawn around the origin to fit the pattern, and arrowheads are drawn on the circle in a counterclockwise manner.
A unit circle in vector field F = 2 x , 2 y .

To compute the flux, we first need a parameterization of the unit circle. We can use the standard parameterization r ( t ) = cos t , sin t , 0 t 2 π . The normal vector to a unit circle is cos t , sin t . Therefore, the flux is

C F · N d s = 0 2 π 2 cos t , 2 sin t · cos t , sin t d t = 0 2 π ( 2 cos 2 t + 2 sin 2 t ) d t = 2 0 2 π ( cos 2 t + sin 2 t ) d t = 2 0 2 π d t = 4 π .
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Calculate the flux of F = x + y , 2 y across the line segment from ( 0 , 0 ) to ( 2 , 3 ) , where the curve is oriented from left to right.


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Let F ( x , y ) = P ( x , y ) , Q ( x , y ) be a two-dimensional vector field. Recall that integral C F · T d s is sometimes written as C P d x + Q d y . Analogously, flux C F · N d s is sometimes written in the notation C Q d x + P d y , because the unit normal vector N is perpendicular to the unit tangent T . Rotating the vector d r = d x , d y by 90° results in vector d y , d x . Therefore, the line integral in [link] can be written as C −2 y d x + 2 x d y .

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: C F · T d s . 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 v ( P ) · T ( P ) . The maximum value of v ( P ) · T ( P ) occurs when the two vectors are pointing in the exact same direction; the minimum value of v ( P ) · T ( P ) occurs when the two vectors are pointing in opposite directions. Thus, the value of the circulation C v · T d s measures the tendency of the fluid to move in the direction of C .

Calculating circulation

Let 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: r ( t ) = cos t , sin t , 0 t 2 π . Then, F ( r ( t ) ) = sin t , cos t and r ( t ) = sin t , cos t . Therefore, the circulation of F along C is

C F · T d s = 0 2 π sin t , cos t · sin t , cos t d t = 0 2 π ( sin 2 t + cos 2 t ) d t = 0 2 π d t = 2 π .

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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Questions & Answers

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Practice Key Terms 8

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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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