Course: Accelerated Engineering Calculus II Instructor: Michael Medvinsky 6.3 The Fundamental Theorem for Line Integrals (13.3)
Recall: Fundamental Theorem of Calculus ∫ F ' ( x ) dx = F ( b ) − F ( a ) . b
a
Def: A vector field F is called a conservative vector field if there is exist a potential, a function f, s.t. F = ∇f . 
Thm: Let C be a smooth curve given by the vector function r ( t ) , a ≤ t ≤ b . Let F be a continuous conservative vector field, and f is a differentiable function (of 2/3 variables) that satisfy the 


equation F = ∇f . Then 
∫ F ⋅ dr = ∫ ∇f ⋅ dr = f ( r (b )) − f ( r ( a )) C
C
⎛ df dx df dy df dz ⎞






Proof: ∫ ∇f ⋅ dr = ∫ ∇f ( r ( t )) ⋅ r ' ( t ) dt = ∫ ⎜
+
+
dt = ∫ dtd f ( r ( t )) dt = f ( r ( b )) − f ( r ( a )) ⎟
⎝ dx dt dy dt dz dt ⎠
C
a
a
a
b
b
b
Examples: 
∫ F ⋅ dr = f ( x2 , y2 ) − f ( x1, y1 ) and ∫ F ⋅ dr = f ( x2 , y2 , z2 ) − f ( x1, y1, z1 ) C
C
Def: Let vector field F be continuous on domain D. If 
∫ F ⋅ dr = ∫ F ⋅ dr for any two curves, C1,C2 , C1
C2
with the same initial and end points, then the line integral 
∫ F ⋅ dr is called independent of path. C
Corollary: A line integral of conservative vector field is independent of path\curve. Def: A curve C is called closed if its terminal points coincide. Thm: 
∫ F ⋅ dr is independent of path in D if and only if ∫ F ⋅ dr = 0 for any closed curve C in D. C
C
Proof: Let 
∫ F ⋅ dr be independent of path in D and C is an arbitrary closed curve in D. We can C
choose any two points A and B such that C1 is a curve from A to B and C2 is a curve from B to A, so that C = C1 ∪ C2 . Then: 
∫ F ⋅ dr = ∫ F ⋅ dr + ∫ F ⋅ dr = ∫ F ⋅ dr −
C
C1
C2
C1
∫ F ⋅ dr = 0 . −C2
Conversely, let 
∫ F ⋅ dr = 0 on any closed curve C in D. Then one shows independency of path by C
taking arbitrary pathes C1 and C2 from A to B in D and defines a closed curve by C = −C1 ∪ C2 , thus 0 = 
∫ F ⋅ dr =
C
∫ F ⋅ dr + ∫ F ⋅ dr = − ∫ F ⋅ dr + ∫ F ⋅ dr = 0 and therefore ∫ F ⋅ dr = ∫ F ⋅ dr . −C1
C2
C1
C2
C1
C2
Thm: Suppose F is a vector field that is continuous (all components are continuous) on an open connected region D. If 
∫ F ⋅ dr be independent of path in D, then F is conservative vector field on C
D, that is there is exists function f such that F = ∇f . Proof: Let ( a,b ) ∈D be an arbitrary fixed point. Define f ( x, y ) =
( x,y )
∫
( a,b )
F ⋅ dr . Since 
∫ F ⋅ dr is C
independent of path, we can construct a path C = C1 ∪ C2 from ( a,b ) to ( x, y ) that serve our 79 Course: Accelerated Engineering Calculus II Instructor: Michael Medvinsky needs. We choose a point x1 , such that ( x1 , y ) ∈D and define an arbitrary path C1 from ( a,b ) to ( x1, y ) and C2 from ( x1, y ) to ( x, y ) . d
d
d ( x1 ,y)
d ( x,y)
d ( x,y)
f ( x, y ) =
F ⋅ dr =
F ⋅ dr + ∫ F ⋅ dr = 0 + ∫ F ⋅ dr ∫
∫
( a,b )
dx
dx C
dx 
dx ( x1 ,y)
 dx ( x1 ,y)
independent of x
Now let F = P,Q and note that C2 is a line with constant y and therefore dy=0. ( x,y )
( x,y )
x
d
d
d
fx ( x, y ) =
F ⋅ dr =
P dx + Q dy =
P ( t, y ) dt = P ( x, y ) ∫
∫
dx ( x1 ,y)
dx ( x1 ,y)
dx x∫1
Similarly one shows that fy ( x, y ) = Q ( x, y ) by choosing a point y1 , such that ( x, y1 ) ∈D and define an arbitrary path C1 from ( a,b ) to ( x, y1 ) and C2 from ( x, y1 ) to ( x, y ) . Using he ( x,y )
d 1
arguments that ∫ F ⋅ dr = 0 and that C2 is a line with constant x so dx=0. dy ( a,b
)
The following theorems gives a way to determine whether or not a vector field is conservative. Thm: If F = P,Q is a conservative vector field where P,Q have continuous first order partial ∂P ∂Q
=
derivatives on domain D then in D. ∂y ∂x
Proof: Since F is conservative there is differentiable function f such that F = ∇f and therefore
∂P ∂Q
fxy =
=
= fyx . ∂y ∂x
Defs: 1) A simply connected curve is a curve that doesn’t intersect itself between endpoints. 2) A simple closed curve is a curve with 



r ( a ) = r ( b ) but r ( t1 ) ≠ r ( t 2 ) for any a < t1 < t 2 < b . 3) A simply connected region: is a region D in which every simple closed curve encloses only points from D. In other words D consist of one piece and has no hole. Thm: Let F = P,Q be a vector field on an open simply connected region D. If P,Q have ∂P ∂Q
=
continuous first order partial derivatives on domain D and , then F is conservative. ∂y ∂x
Ex 2. If F ( x, y ) = ( x sin y, y sin x ) is conservative, find the potential:
F isn’t conservative, since
∂P ∂
∂Q ∂
=
x sin y = x cos y ≠
=
y sin x = y cos x
∂y ∂y
∂x ∂x
Ex 3. If F ( x, y ) = ( x + y, x − y ) is conservative, find the potential: F is conservative since
∂P ∂
∂Q ∂
=
x + y) = 1 =
= ( x − y ) ⇒ P = f x ,Q = f y , therefore
(
∂y ∂y
∂x ∂x
f ( x, y ) = ∫ f x ( x, y ) dx = ∫ x + y dx =
x2
+ yx + g ( y ) where g ( y ) plays the role of constant of
2
integration. Q = f y = x + g ' ( y ) = x − y ⇒ g ' ( y ) = − y ⇒ g ( y ) = −
80 y2
x2
y2
⇒ f ( x, y ) = + yx −
2
2
2