1. Fundamental Theorem for Line Integrals
Theorem 1.1. LetD be a region inR3andf :D→Rbe a smooth function. Suppose thatC is a smooth parametrized curve parametrized by the functionr: [a, b]→R3.Denoter(a) byAandr(b) byB.Then
Z
C
∇f·dr=f(B)−f(A).
Proof. Letg(t) =f(r(t)) for t∈[a, b].By chain rule,
g0(t) =fx(r(t))x0(t) +fy(r(t))y0(t) +fz(r(t))z0(t) =∇f(r(t))·r0(t), fort∈[a, b],wherer(t) =x(t)i+y(t)j+z(t)k.By the fundamental Theorem of calculus,
Z
C
∇f(r)dr= Z b
a
∇f(r(t))·r0(t)dt= Z b
a
g0(t)dt=g(b)−g(a).
Notice thatg(b) =f(r(b)) =f(B) and g(a) =f(r(a)) =f(A). Therefore we proved our assertion.
Corollary 1.1. Letf :D→Rbe as above. SupposeC1andC2 are two parametrized curves with the same initial and terminal points. Then
Z
C1
∇f ·dr= Z
C2
∇f ·dr.
Proof. Let A and B be the initial point and the terminal point of C1 and C2 respectively. By fundamental Theorem of calculus for line integrals,
Z
C1
∇f·dr=f(B)−f(A) = Z
C2
∇f·dr.
A parametrized curveCis called a loop if the terminal point equals to its initial point.
Corollary 1.2. Letf :D→Rbe as above. Suppose thatCis a closed loop. Then Z
C
∇f·dr= 0.
Proof. This follows directly from the fundamental Theorem of calculus.
Definition 1.1. Let F be a vector field on a region D. We say that F is conservative if for any A, B∈D the integralR
CF·dris independent of the choice ofC with initial pointAand terminal pointB.
One sees that if F = ∇f for some smooth function ∇f : D → R, then F is conservative by Corollary 1.1.
Definition 1.2. A vector field Fon a regionDin R3has an antiderivative or a potential function if there exists a smooth function f : D → R such that F = ∇f. If such f exists, we call f an antiderivative or a potential function ofF.
Given a vector fieldFon a regionD,when doesFposses an antiderivative? We need the following theorem.
Theorem 1.2. Letf :D→Rbe a smooth function. Then
∇ × ∇f =0.
1
2
Proof. By definition,∇f =fxi+fyj+fzk.Then
∇ ×(∇f) =
i j k
∂x ∂y ∂z fx fy fz
= (fzy−fyz)i+ (fxz−fzx)j+ (fyx−fxy)k.
Sincef is smooth,fzy =fyz andfxz =fzx andfyx=fxy.We find that∇ ×(∇f) = 0.
If F= ∇f, then ∇ ×F= 0by Theorem 1.2. In other words, if F has an antiderivative, then
∇ ×F=0.In fact, the converse is also true whenD is a simply connected domain:
Theorem 1.3. LetD be a simply connected domain. ThenFhas an antiderivative if and only if
∇ ×F=0.
Proof. We have seen one direction. We will prove the other direction after Green’s Theorem and Stoke’s Theorem.
Theorem 1.4. LetD be any region inR3 andFbe a vector field on D.ThenFis conservative if and only ifFhas an antiderivative.
Proof. We have seen that if F has an antiderivative, then F is conservative. Let us prove the converse. Let us fixed a pointA∈D.For (x, y, z)∈D,we define
Z (x,y,z)
A
F·dr= Z
C
F·dr,
whereC is a curve from Ato (x, y, z). SinceFis conservative, the left hand side is independent of choice ofC.This allows us to define a function f :D→Rby
f(x, y, z) =
Z (x,y,z)
A
F·dr.
Let us show that∇f =F.Assume thatF=P(x, y, z)i+Q(x, y, z)j+R(x, y, z)k.At first, we prove thatfx=P.
Let us choose a curve C1 from A to (x, y, z) and let C2 be the line segment from (x, y, z) to (x+h, y, z) defined by the parametrizationr2(t) = (x+th, y, z) for 0≤t≤1.Then
f(x+h, y, z) = Z
C1+C2
F·dr= Z
C1
F·dr+ Z
C2
F·dr=f(x, y, z) + Z
C2
F·dr.
This implies that
f(x+h, y, z)−f(x, y, z) = Z
C2
F·dr= Z 1
0
P(x+th, y, z)hdt.
We see that
f(x+h, y, z)−f(x, y, z)
h =
Z 1
0
P(x+th, y, z)dt.
By takingh→0,we obtain that
fx(x, y, z) = lim
h→0
f(x+h, y, z)−f(x, y, z) h
= lim
h→0
Z 1
0
P(x+th, y, z)dt
= Z 1
0
P(x, y, z)dt=P(x, y, z) Similarly, we can show thatfy=Qand fz=R.
