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In [link] , what if we had oriented the unit circle clockwise? We denote the unit circle oriented clockwise by C . Then

C F · T d s = C F · T d s = −2 π .

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 F ( x , y ) = y x 2 + y 2 , x x 2 + y 2 along a unit circle oriented counterclockwise.

2 π

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Calculating work

Calculate the work done on a particle that traverses circle C of radius 2 centered at the origin, oriented counterclockwise, by field F ( x , y ) = −2 , y . Assume the particle starts its movement at ( 1 , 0 ) .

The work done by F on the particle is the circulation of F along C : C F · T d s . We use the parameterization r ( t ) = 2 cos t , 2 sin t , 0 t 2 π for C . Then, r ( t ) = −2 sin t , 2 cos t and F ( r ( t ) ) = −2 , 2 sin t . Therefore, the circulation of F along C is

C F · T d s = 0 2 π −2 , 2 sin t · −2 sin t , 2 cos t d t = 0 2 π ( 4 sin t + 4 sin t cos t ) d t = [ −4 cos t + 4 sin 2 t ] 0 2 π = ( −4 cos ( 2 π ) + 2 sin 2 ( 2 π ) ) ( −4 cos ( 0 ) + 4 sin 2 ( 0 ) ) = −4 + 4 = 0.

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 ( x , y ) = −2 x + 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.

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Calculate the work done by field F ( x , y ) = 2 x , 3 y on a particle that traverses the unit circle. Assume the particle begins its movement at ( −1 , 0 ) .

0

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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 C d s .
  • 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
    C f ( x , y , z ) d s = a b f ( r ( t ) ) ( x ( t ) ) 2 + ( y ( t ) ) 2 + ( z ( t ) ) 2 d t
  • Calculating a vector line integral
    C F · d s = C F · T d s = a b F ( r ( t ) ) · r ( t ) d t
    or
    C P d x + Q d y + R d z = a b ( P ( r ( t ) ) d x d t + Q ( r ( t ) ) d y d t + R ( r ( t ) ) d z d t ) d t
  • Calculating flux
    C F · n ( t ) n ( t ) d s = a b F ( r ( t ) ) · n ( t ) d t

True or False? Line integral C f ( x , y ) d s is equal to a definite integral if C is a smooth curve defined on [ a , b ] and if function f is continuous on some region that contains curve C .

True

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True or False? Vector functions r 1 = t i + t 2 j , 0 t 1 , and r 2 = ( 1 t ) i + ( 1 t ) 2 j , 0 t 1 , define the same oriented curve.

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True or False? C ( P d x + Q d y ) = C ( P d x Q d y )

False

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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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