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Reversing orientation

Find the value of integral C F · d r , where C is the semicircle parameterized by r ( t ) = cos t + π , sin t , 0 t π and F = y , x .

Notice that this is the same problem as [link] , except the orientation of the curve has been traversed. In this example, the parameterization starts at r ( 0 ) = π , 0 and ends at r ( π ) = 0 , 0 . By [link] ,

C F · d r = 0 π sin t , cos t + π · sin t + π , cos t d t = 0 π sin t , cos t · sin t , cos t d t = 0 π ( sin 2 t cos 2 t ) d t = 0 π −1 d t = π .

Notice that this is the negative of the answer in [link] . It makes sense that this answer is negative because the orientation of the curve goes against the “flow” of the vector field.

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Let C be an oriented curve and let − C denote the same curve but with the orientation reversed. Then, the previous two examples illustrate the following fact:

C F · d r = C F · d r .

That is, reversing the orientation of a curve changes the sign of a line integral.

Let F = x i + y j be a vector field and let C be the curve with parameterization t , t 2 for 0 t 2 . Which is greater: C F · T d s or C F · T d s ?

C F · T d s

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Another standard notation for integral C F · d r is C P d x + Q d y + R d z . In this notation, P , Q , and R are functions, and we think of d r as vector d x , d y , d z . To justify this convention, recall that d r = T d s = r ( t ) d t = d x d t , d y d t , d z d t d t . Therefore,

F · d r = P , Q , R · d x , d y , d z = P d x + Q d y + R d z .

If d r = d x , d y , d z , then d r d t = d x d t , d y d t , d z d t , which implies that d r d t = d x d t , d y d t , d z d t d t . Therefore

C F · d r = C P d x + Q d y + R d z = ( P ( r ( t ) ) d x d t + Q ( r ( t ) ) d y d t + R ( r ( t ) ) d z d t ) d t .

Finding the value of an integral of the form C P d x + Q d y + R d z

Find the value of integral C z d x + x d y + y d z , where C is the curve parameterized by r ( t ) = t 2 , t , t , 1 t 4 .

As with our previous examples, to compute this line integral we should perform a change of variables to write everything in terms of t . In this case, [link] allows us to make this change:

C z d x + x d y + y d z = 1 4 ( t ( 2 t ) + t 2 ( 1 2 t ) + t ) d t = 1 4 ( 2 t 2 + t 3 / 2 2 + t ) d t = [ 2 t 3 3 + t 5 / 2 5 + 2 t 3 / 2 3 ] t = 1 t = 4 = 793 15 .
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Find the value of C 4 x d x + z d y + 4 y 2 d z , where C is the curve parameterized by r ( t ) = 4 cos ( 2 t ) , 2 sin ( 2 t ) , 3 , 0 t π 4 .

−26

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We have learned how to integrate smooth oriented curves. Now, suppose that C is an oriented curve that is not smooth, but can be written as the union of finitely many smooth curves. In this case, we say that C is a piecewise smooth curve    . To be precise, curve C is piecewise smooth if C can be written as a union of n smooth curves C 1 , C 2 ,… , C n such that the endpoint of C i is the starting point of C i + 1 ( [link] ). When curves C i satisfy the condition that the endpoint of C i is the starting point of C i + 1 , we write their union as C 1 + C 2 + + C n .

Three curves: C_1, C_2, and C_3. One of the endpoints of C_2 is also an endpoint of C_1, and the other endpoint of C_2 is also an end point of C_3. C_1’s and C_3’s other endpoints are not connect to any other curve. C_1 and C_3 appear to be nearly straight lines while C_2 is an increasing concave down curve. There are three arrowheads on each curve segment all pointing in the same direction: C_1 to C_2, C_2 to C_3, and C_3 to its other endpoint.
The union of C 1 , C 2 , C 3 is a piecewise smooth curve.

The next theorem summarizes several key properties of vector line integrals.

Properties of vector line integrals

Let F and G be continuous vector fields with domains that include the oriented smooth curve C . Then

  1. C ( F + G ) · d r = C F · d r + C G · d r
  2. C k F · d r = k C F · d r , where k is a constant
  3. C F · d r = C F · d r
  4. Suppose instead that C is a piecewise smooth curve in the domains of F and G , where C = C 1 + C 2 + + C n and C 1 , C 2 ,… , C n are smooth curves such that the endpoint of C i is the starting point of C i + 1 . Then
    C F · d s = C 1 F · d s + C 2 F · d s + + C n F · d s .

Notice the similarities between these items and the properties of single-variable integrals. Properties i. and ii. say that line integrals are linear, which is true of single-variable integrals as well. Property iii. says that reversing the orientation of a curve changes the sign of the integral. If we think of the integral as computing the work done on a particle traveling along C , then this makes sense. If the particle moves backward rather than forward, then the value of the work done has the opposite sign. This is analogous to the equation a b f ( x ) d x = a b f ( x ) d x . Finally, if [ a 1 , a 2 ] , [ a 2 , a 3 ] ,… , [ a n 1 , a n ] are intervals, then

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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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