Definition 18.4.2A vector field F defined in a three dimensional region is said to beconservative^{3}if for every piecewise smooth closed curve C, it follows∫_{C}F⋅dR = 0.
Definition 18.4.3Let
(x,p1,⋅⋅⋅,pn,y)
be an ordered list of points in ℝ^{p}. Let
p (x,p1,⋅⋅⋅,pn,y)
denote the piecewise smooth curve consisting of a straight line segment from x to p_{1}and thenthe straight line segment from p_{1}to p_{2}
⋅⋅⋅
and finally the straight line segment from p_{n}toy.This is called a polygonal curve. An open set in ℝ^{p}, U, is said to be a region if it hasthe property that for any two pointsx,y∈ U, there exists a polygonal curve joining the twopoints.
Conservative vector fields are important because of the following theorem, sometimes called the
fundamental theorem for line integrals.
Theorem 18.4.4Let U be a region in ℝ^{p}and let F : U → ℝ^{p}be a continuous vector field. Then F isconservative if and only if there exists a scalar valued function of p variables ϕ such that F = ∇ϕ.Furthermore, if C is an oriented curve which goes from x to y in U, then
∫
F ⋅dR = ϕ(y)− ϕ (x ). (18.5)
C
(18.5)
Thus the line integral is path independent in this case. This function ϕ is called ascalar potential forF.
Proof: To save space and fussing over things which are unimportant, denote by p
(x0,x )
a polygonal
curve from x_{0} to x. Thus the orientation is such that it
goes from x_{0} to x. The curve p
(x,x0)
denotes the same set of points but in the opposite order. Suppose
first F is conservative. Fix x_{0}∈ U and let
∫
ϕ(x) ≡ F ⋅dR.
p(x0,x)
This is well defined because if q
(x0,x)
is another polygonal curve joining x_{0} to x, Then the curve obtained
by following p
(x0,x)
from x_{0} to x and then from x to x_{0} along q
(x,x0)
is a closed piecewise smooth curve
and so by assumption, the line integral along this closed curve equals 0. However, this integral is
just
∫ ∫ ∫ ∫
F ⋅dR+ F ⋅dR = F⋅dR− F⋅dR
p(x0,x) q(x,x0) p(x0,x) q(x0,x)
centered at x is contained in U. Therefore, the line
segment from x to x + te_{i} is also contained in U and so one can take p
(x,x+ tei)
(s)
= x + s
(tei)
for
s ∈
[0,1]
. Therefore, the above difference quotient reduces to
∫ ∫
1 1 1
t 0 F (x+ s (tei)) ⋅teids = 0 Fi(x + s(tei))ds
= Fi(x +st(tei))
by the mean value theorem for integrals. Here s_{t} is some number between 0 and 1. By continuity of F, this
converges to F_{i}
(x)
as t → 0. Therefore, ∇ϕ = F as claimed.
Conversely, if ∇ϕ = F, then if R :
[a,b]
→ ℝ^{p} is any C^{1} curve joining x to y,
∫ b ′ ∫ b ′
a F (R (t))⋅R (t)dt = a ∇ ϕ(R (t))⋅R (t) dt
∫ b
= -d(ϕ (R (t)))dt
a dt
= ϕ (R (b))− ϕ(R (a))
= ϕ (y) − ϕ (x)
and this verifies (18.5) in the case where the curve joining the two points is smooth. The general case
follows immediately from this by using this result on each of the pieces of the piecewise smooth curve.
For example if the curve goes from x to p and then from p to y, the above would imply the
integral over the curve from x to p is ϕ
(p )
− ϕ
(x)
while from p to y the integral would yield
ϕ
(y )
− ϕ
(p)
. Adding these gives ϕ
(y )
− ϕ
(x)
. The formula (18.5) implies the line integral over any
closed curve equals zero because the starting and ending points of such a curve are the same.
■
Example 18.4.5Let F
(x,y,z)
=
(cosx− yz sin(xz),cos(xz ),− yxsin(xz))
. Let C be a piecewisesmooth curve which goes from
(π,1,1)
to
(π )
2,3,2
. Find∫_{C}F ⋅dR.
The specifics of the curve are not given so the problem is nonsense unless the vector field is
conservative. Therefore, it is reasonable to look for the function ϕ satisfying ∇ϕ = F. Such a function
satisfies
ϕx = cosx− y(sin xz)z
and so, assuming ϕ exists,
ϕ(x,y,z) = sinx + ycos(xz)+ ψ(y,z).
I have to add in the most general thing possible, ψ
(y,z)
to ensure possible solutions are not being thrown
out. It wouldn’t be good at this point to only add in a constant since the answer could involve a function of
either or both of the other variables. Now from what was just obtained,
ϕy = cos(xz)+ ψy = cosxz
and so it is possible to take ψ_{y} = 0. Consequently, ϕ, if it exists is of the form
ϕ (x,y,z) = sinx + ycos(xz)+ ψ(z).
Now differentiating this with respect to z gives
ϕz = − yx sin(xz)+ ψz = − yx sin(xz)
and this shows ψ does not depend on z either. Therefore, it suffices to take ψ = 0 and
ϕ(x,y,z) = sin (x)+ ycos(xz).
Therefore, the desired line integral equals
(π)
sin 2 + 3cos(π)− (sin(π)+ cos(π)) = − 1.
The above process for finding ϕ will not lead you astray in the case where there does not exist a scalar
potential. As an example, consider the following.
Example 18.4.6Let F
(x,y,z)
=
( 2 )
x,y x,z
. Find a scalar potential for F if it exists.
If ϕ exists, then ϕ_{x} = x and so ϕ =
x22
+ ψ
(y,z)
. Then ϕ_{y} = ψ_{y}
(y,z)
= xy^{2} but this is impossible
because the left side depends only on y and z while the right side depends also on x. Therefore, this vector
field is not conservative and there does not exist a scalar potential.
Definition 18.4.7A set of points in three dimensional space V is simply connected if everypiecewise smooth closed curve C is the edge of a surface S which is contained entirely within V insuch a way that Stokes theorem holds for the surface S and its edge, C.
PICT
This is like a sock. The surface is the sock and the curve C goes around the opening of the
sock.
As an application of Stoke’s theorem, here is a useful theorem which gives a way to check whether a
vector field is conservative.
Theorem 18.4.8For a three dimensional simply connected open set V and F a C^{1}vector fielddefined in V , F is conservative if ∇×F = 0in V .
Proof:If ∇×F = 0 then taking an arbitrary closed curve C, and letting S be a surface bounded by C
which is contained in V , Stoke’s theorem implies
∫ ∫
0 = ∇ × F⋅ndA = F ⋅dR.
S C
Thus F is conservative. ■
Example 18.4.9Determine whether the vector field
( 3 ( ( 2 2)) ( ( 2 2)) )
4x + 2 cos x + z x,1,2 cos x + z z
is conservative.
Since this vector field is defined on all of ℝ^{3}, it only remains to take its curl and see if it is the zero
vector.
|| ||
|| i j k ||
|| ( ∂x( )) ∂y ( ( ∂z )) ||.
|4x3 + 2 cos x2 +z2 x 1 2 cos x2 + z2 z |
This is obviously equal to zero. Therefore, the given vector field is conservative. Can you find a
potential function for it? Let ϕ be the potential function. Then ϕ_{z} = 2
(cos(x2 + z2))
z and so
ϕ
(x,y,z)
= sin
(x2 + z2)
+ g
(x,y)
. Now taking the derivative of ϕ with respect to y, you see g_{y} = 1 so
g
(x,y)
= y + h
(x)
. Hence ϕ
(x,y,z)
= y + g
(x)
+ sin
(x2 + z2)
. Taking the derivative with respect to x,
you get 4x^{3} + 2