Definition 28.4.2A vector field F defined in a three dimensional region issaid to beconservative^{3}if for every piecewise smooth closed curve C, it follows∫_{C}F⋅dR = 0.
Definition 28.4.3Let
(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}andthen the straight line segment from p_{1}to p_{2}
⋅⋅⋅
and finally the straight line segment fromp_{n}toy.This is called a polygonal curve. An open set in ℝ^{p}, U, is said to be a regionif it has the property that for any two pointsx,y∈ U, there exists a polygonal curvejoining the two points.
Conservative vector fields are important because of the following theorem, sometimes
called the fundamental theorem for line integrals.
Theorem 28.4.4Let U be a region in ℝ^{p}and let F : U → ℝ^{p}be a continuousvector field. Then F is conservative if and only if there exists a scalar valued function of pvariables ϕ such that F = ∇ϕ. Furthermore, if C is an oriented curve which goes from xto y in U, then
∫
F ⋅dR = ϕ(y)− ϕ (x ). (28.5)
C
(28.5)
Thus the line integral is path independent in this case. This function ϕ is called ascalarpotential for F.
Proof: To save space and fussing over things which are unimportant, denote by
p
(x ,x )
0
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,x )
0
denotes the same set of points but in the opposite
order. Suppose first F is conservative. Fix x_{0}∈ U and let
∫
ϕ(x) ≡ p(x0,x)F ⋅dR.
This is well defined because if q
(x ,x)
0
is another polygonal curve joining x_{0} to x, Then
the curve obtained by following p
(x ,x)
0
from x_{0} to x and then from x to x_{0} along
q
(x,x )
0
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)
which shows
∫ ∫
F⋅dR = F⋅dR
p(x0,x) q(x0,x)
and that ϕ is well defined. For small t,
∫ ∫
ϕ (x + tei)− ϕ (x) p(x0,x+tei)F⋅dR − p(x0,x)F ⋅dR
--------t------- = --------------t--------------
∫ F ⋅dR+ ∫ F ⋅dR − ∫ F ⋅dR
= -p(x0,x)--------p(x,x+tei)---------p(x0,x)------.
t
Since U is open, for small t, the ball of radius
|t|
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
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,
∫ ∫
bF (R (t))⋅R ′(t)dt = b∇ ϕ(R (t))⋅R′(t)dt
a a
∫ b d
= dt (ϕ (R (t)))dt
a
= ϕ (R (b))− ϕ(R (a))
= ϕ (y )− ϕ(x)
and this verifies 28.5 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 28.5 implies the line integral over any closed curve
equals zero because the starting and ending points of such a curve are the same.
■
Example 28.4.5Let F
(x,y,z)
=
(cosx− yz sin(xz),cos(xz),− yxsin(xz))
. LetC be a piecewise smooth 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 (sinxz)z
and so, assuming ϕ exists,
ϕ(x,y,z) = sin x+ 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
ϕ = − yx sin(xz)+ ψ = − yx sin(xz)
z z
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
sin (π-)+ 3cos(π)− (sin (π) +cos(π)) = − 1.
2
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 28.4.6Let F
(x,y,z)
=
( 2 )
x,y x,z
. Find a scalar potential for F if itexists.
If ϕ exists, then ϕ_{x} = x and so ϕ =
x2
2
+ 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 28.4.7A set of points in three dimensional space V is simplyconnected if every piecewise smooth closed curve C is the edge of a surface S which iscontained entirely within V in such a way that Stokes theorem holds for the surfaceS 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 28.4.8For a three dimensional simply connected open set V andF a C^{1}vector field defined 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 28.4.9Determine 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