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6.3 Conservative vector fields  (Page 3/16)

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Proof

By [link] ,

∫ C ∇ f • d r = ∫ a b ∇ f ( r ( t ) ) • r ′ ( t ) d t .

By the chain rule,

d d t ( f ( r ( t ) ) = ∇ f ( r ( t ) ) • r ′ ( t ) .

Therefore, by the Fundamental Theorem of Calculus,

∫ C ∇ f • d r = ∫ a b ∇ f ( r ( t ) ) • r ′ ( t ) d t = ∫ a b d d t ( f ( r ( t ) ) d t = [ f ( r ( t ) ) ] t = a t = b = f ( r ( b ) ) − f ( r ( a ) ) .

□

We know that if F is a conservative vector field, there are potential functions f such that ∇ f = F . Therefore ∫ C F · d r = ∫ C ∇ f · d r = f ( r ( b ) ) − f ( r ( a ) ) . In other words, just as with the Fundamental Theorem of Calculus, computing the line integral ∫ C F · d r , where F is conservative, is a two-step process: (1) find a potential function (“antiderivative”) f for F and (2) compute the value of f at the endpoints of C and calculate their difference f ( r ( b ) ) − f ( r ( a ) ) . Keep in mind, however, there is one major difference between the Fundamental Theorem of Calculus and the Fundamental Theorem for Line Integrals. A function of one variable that is continuous must have an antiderivative. However, a vector field, even if it is continuous, does not need to have a potential function.

Applying the fundamental theorem

Calculate integral ∫ C F • d r , where F ( x , y , z ) = ⟨ 2 x ln y , x 2 y + z 2 , 2 y z ⟩ and C is a curve with parameterization r ( t ) = ⟨ t 2 , t , t ⟩ , 1 ≤ t ≤ e

  1. without using the Fundamental Theorem of Line Integrals and
  2. using the Fundamental Theorem of Line Integrals.
  1. First, let’s calculate the integral without the Fundamental Theorem for Line Integrals and instead use [link] :
    ∫ C F • d r = ∫ 1 e F ( r ( t ) ) • r ′ ( t ) d t = ∫ 1 e ⟨ 2 t 2 ln t , t 4 t + t 2 , 2 t 2 ⟩ • ⟨ 2 t , 1 , 1 ⟩ d t = ∫ 1 e ( 4 t 3 ln t + t 3 + 3 t 2 ) d t = ∫ 1 e 4 t 3 ln t d t + ∫ 1 e ( t 3 + 3 t 2 ) d t = ∫ 1 e 4 t 3 ln t d t + [ t 4 4 + t 3 ] 1 e = r ∫ 1 e t 3 ln t d t + e 4 4 + e 3 − 5 4 .

    Integral ∫ 1 e t 3 ln t d t requires integration by parts. Let u = ln t and d v = t 3 . Then u = ln t , d v = t 3
    and
    d u = 1 t d t , v = t 4 4 .

    Therefore,
    ∫ 1 e t 3 ln t d t = [ t 4 4 ln t ] 1 e − 1 4 ∫ 1 e t 3 d t = e 4 4 − 1 r ( e 4 4 − 1 4 ) .

    Thus,
    ∫ C F • d r = 4 ∫ 1 e t 3 ln t d t + e 4 4 + e 3 − 5 4 = 4 ( e 4 4 − 1 4 ( e 4 4 − 1 4 ) ) + e 4 4 + e 3 − 5 4 = e 4 − e 4 4 + 1 4 + e 4 4 + e 3 − 5 4 = e 4 + e 3 − 1.
  2. Given that f ( x , y , z ) = x 2 ln y + y z 2 is a potential function for F , let’s use the Fundamental Theorem for Line Integrals to calculate the integral. Note that
    ∫ C F • d r = ∫ C ∇ f • d r = f ( r ( e ) ) − f ( r ( 1 ) ) = f ( e 2 , e , e ) − f ( 1 , 1 , 1 ) = e 4 + e 3 − 1.

    This calculation is much more straightforward than the calculation we did in (a). As long as we have a potential function, calculating a line integral using the Fundamental Theorem for Line Integrals is much easier than calculating without the theorem.
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[link] illustrates a nice feature of the Fundamental Theorem of Line Integrals: it allows us to calculate more easily many vector line integrals. As long as we have a potential function, calculating the line integral is only a matter of evaluating the potential function at the endpoints and subtracting.

Given that f ( x , y ) = ( x − 1 ) 2 y + ( y + 1 ) 2 x is a potential function for F = ⟨ 2 x y − 2 y + ( y + 1 ) 2 , ( x − 1 ) 2 + 2 y x + 2 x ⟩ , calculate integral ∫ C F · d r , where C is the lower half of the unit circle oriented counterclockwise.

A vector field in two dimensions. The arrows near the origin are the shortest, and the arrows in the upper right and lower left corners of quadrants 1 and 3 are the shortest. The arrows go up and to the left in quadrants 1 and 3. In quadrant 2, the arrows stretch up and to the right for values greater than x=-1. The closer the arrows are to y=1, the more horizontal they become. For values less than x=-1, the arrows point up and form a curve to the left. The closer the arrows are to y=1, the more horizontal they become. Above y=1, it looks like the arrows are shifting from vertical, going down to horizontal. In quadrant 4, the arrows go up and to the right fairly regularly, but they tend to be curving to the right the larger the x value becomes. For y values less than -1, the arrows shift from pointing up to pointing down, following x=1. The lower half of the unit circle with center at the origin is drawn in quadrants 3 and 4.

2

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The Fundamental Theorem for Line Integrals has two important consequences. The first consequence is that if F is conservative and C is a closed curve, then the circulation of F along C is zero—that is, ∫ C F · d r = 0 . To see why this is true, let f be a potential function for F . Since C is a closed curve, the terminal point r (b) of C is the same as the initial point r (a) of C —that is, r ( a ) = r ( b ) . Therefore, by the Fundamental Theorem for Line Integrals,
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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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