Part 4: Integral around a Closed Curve

Moreover, since ò-CF·dr = -òCF·dr, theorem 2 implies that
ó
õ 		A

B 
 F·dr = - ó
õ 		B

A 
 F·dr
(6)
for any conservative vector field F. Moreover, if C is a closed curve, then it begins and ends at a point A, and (6) implies that
ó
õ 		A

A 
 F·dr = 0
(7)
If we now let denote integration around a closed curve, then (7) implies the following:       

Theorem 3: A vector field F( x,y) is conservative on a simply connected solid S if and only if


C 
 F·dr = 0
for any closed curve C contained in S.

       

Because theorems 2 and 3 both involve the work integral, these theorems are important throughout mathematics and science.

       

EXAMPLE 7    How much work is performed in moving a box with a mass of 1 slug from the origin to the point (0,10  feet,5  feet) through the gravitational force field F( x,y,z ) = á 0,0,-32 ñ along each of the curves C1, C2, and C3 shown below?

Solution: First, let's use theorem 3 to show that F is a conservative field. If C is a closed curve in R3 parametrized by r( t) = á x( t) ,y( t),z( t) ñ , t in [ a,b] , then



C 
 F·dr = ó
õ 		b

a 
 F·  dr
dt
 dt = ó
õ 		b

a 
 á 0,0,-32 ñ ·
 dx
dt
 ,  dy
dt
 ,  dz
dt
 dt
Since C is a closed curve, r( a) = r(b) , which implies that z( a) = z( b) . Thus,


C 
 F·dr = -32 ó
õ 		b

a 
  dz
dt
  dt = -32[ z( b) -z( a) ] = 0
Thus, CF·dr = 0 for any closed curve C, which by theorem 3 implies that F is conservative.
        Indeed, it is easy to show that F has a potential of
U( x,y,z) = -32z+k
As a result, the work performed in moving an object from (0,0,0) to ( 0,10,5) is U( x,y,z) = -32z+k, so that using theorem 1 we also have
W = ó
õ 		( 0,10,5)

( 0,0,0)  
 F·dr = -32z+k   (0,10,5)

( 0,0,0)

  = -160   slug  ft2
sec2
Moreover, path independence implies that the work is W = -160 slug-ft2/sec2 along any path from ( 0,0,0) to (0,10,5) , so that specifically, the work along each of the three curves C1, C2, and C3 is -160 slug-ft2/sec2.