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Math 241 - Calculus III
Spring 2012, section CL
§ 16.5. Conservative vector fields in R
3
In these notes, we discuss conservative vector fields in 3 dimensions, and highlight the similar-
ities and differences with the 2-dimensional case. Compare with the notes on § 16.3.
1 Conservative vector fields
Let us recall the basics on conservative vector fields.
Definition 1.1. Let
F : D → R
n be a vector field with domain D ⊆ R
n
. The vector field
F is said to be conservative if it is the gradient of a function. In other words, there is a
differentiable function f : D → R satisfying
F = ∇f. Such a function f is called a potential
function for
F.
Example 1.2.
F (x, y, z) = (y
2
z
3
, 2 xyz
3
, 3 xy
2
z
2
) is conservative, since it is
F = ∇f for the
function f (x, y, z) = xy
2 z
3 .
Example 1.3.
F (x, y, z) = (3x
2
z, z
2
, x
3
+2yz) is conservative, since it is
F = ∇f for the function
f (x, y, z) = x
3 z + yz
2 .
The fundamental theorem of line integrals makes integrating conservative vector fields along
curves very easy. The following proposition explains in more detail what is nice about conser-
vative vector fields.
Proposition 1.4. The following properties of a vector field
F are equivalent.
F is conservative.
C
F · d~r is path-independent, meaning that it only depends on the endpoints of the curve
C.
C
F · d~r = 0 around any closed curve C.
Example 1.5. Find the line integral
C
F · d~r of the vector field
F (x, y, z) = (3x
2 z, z
2 , x
3
along the curve C parametrized by
~r(t) =
ln t
ln 2
, t
3
2 , t cos(πt)
, 1 ≤ t ≤ 4.
Solution. We know that
F is conservative, with potential function f (x, y) = x
3 z + yz
2
. The
endpoints of C are
~r(1) = (0, 1 , −1)
~r(4) = (
ln 4
ln 2
The fundamental theorem of line integrals yields
C
F · d~r =
C
∇f · d~r
= f (~r(4)) − f (~r(1))
= f (2, 8 , 0) − f (0, 1 , −1)
2 Necessary conditions
To know if a vector field
F is conservative, the first thing to check is the following criterion.
Proposition 2.1. Let D ⊆ R
3
be an open subset and let
F : D → R
3
be a continuously
differentiable vector field with domain D. If
F is conservative, then it satisfies curl
F =
Explicitly,
F = (F
1
, F
2
, F
3
) satisfies the three conditions
1
F
2
2
F
1
1
F
3
3
F
1
2
F
3
3
F
2
everywhere on D.
Proof. Assume there is a differentiable function f : D → R satisfying
F = ∇f on D. Because
f is twice continuously differentiable (meaning it has all second partial derivatives and they are
all continuous), Clairaut’s theorem applies, meaning the mixed partial derivatives agree. Since
the first partial derivatives of f are (fx, fy, fz ) = (∂ 1 f, ∂ 2 f, ∂ 3 f ) = (F 1 , F 2 , F 3 ), we obtain
1
2
f = ∂ 2
1
f
1
F
2
2
F
1
1
3
f = ∂ 3
1
f
1
F
3
3
F
1
2
3
f = ∂ 3
2
f
2
F
3
3
F
2
These conditions are equivalent to curl
F =
0, because of the formula:
curl
F =
i
j
k
F
1
F
2
F
3
i(∂ 2
F
3
3
F
2
j(∂ 1
F
3
3
F
1
k(∂ 1
F
2
2
F
1
= (∂ 2 F 3 − ∂ 3 F 2 , ∂ 3 F 1 − ∂ 1 F 3 , ∂ 1 F 2 − ∂ 2 F 1 ).
Solution. First we compute
curl
F =
i
j
k
∂x ∂y ∂z
3 x
2 y
2 z + 5y
3 2 x
3 yz + 15xy
2 − 7 z x
3 y
2 − 7 y + 4z
3
i
2 x
3
y − 7 − (2x
3
y − 7)
j(3x
2
y
2
− 3 x
2
y
2
) +
k
6 x
2
yz + 15y
2
− (6x
2
yz + 15y
2
)
Moreover,
F is defined (and smooth) on all of R
3 , hence it is conservative. Let us find a
potential function f (x, y, z) for
F. We want
fx = F 1 = 3x
2
y
2
z + 5y
3
f y
= F
2
= 2x
3
yz + 15xy
2
− 7 z
f z
= F
3
= x
3 y
2 − 7 y + 4z
3 .
Using the first equation, we obtain
f =
F
1
dx
3 x
2
y
2
z + 5y
3
dx
= x
3
y
2
z + 5xy
3
whose derivative with respect to y is
2 x
3
yz + 15xy
2
(y, z).
Using the second equation, we equate this with F 2 :
2 x
3
yz + 15xy
2
(y, z) = 2x
3
yz + 15xy
2
− 7 z
⇒ gy(y, z) = − 7 z
⇒ g(y, z) =
− 7 z dy
= − 7 yz + h(z).
Plugging this back into the expression for f , we obtain
f = x
3
y
2
z + 5xy
3
− 7 yz + h(z)
whose derivative with respect to z is
x
3
y
2
− 7 y + h
′
(z).
Using the third equation, we equate this with F 3 :
x
3
y
2
− 7 y + h
′
(z) = x
3
y
2
− 7 y + 4z
3
⇒ h
′
(z) = 4z
3
⇒ h(z) =
4 z
3
dz
= z
4
Choosing the constant c = 0, we obtain h(z) = z
4 and thus the potential function
f (x, y, z) = x
3
y
2
z + 5xy
3
− 7 yz + z
4
.
4 Simply connected domains
Asking for
F to be defined (and continuously differentiable) on all of R
3 is somewhat restrictive.
That condition can be loosened.
Definition 4.1. A subset D of R
n is called simply connected if it is path-connected moreover,
and every loop in D can be contracted to a point.
Example 4.2. R
3 itself is simply connected.
Example 4.3. The first octant {(x, y, z) ∈ R
3
| x, y, z > 0 } is simply connected.
Example 4.4. The open ball {(x, y, z) ∈ R
3 | x
2
2
2 < 1 } is simply connected.
Example 4.5. The closed ball {(x, y, z) ∈ R
3 | x
2
2
2 ≤ 1 } is simply connected.
Example 4.6. The (surface of the) sphere {(x, y, z) ∈ R
3 | x
2
2
2 = 1} is simply connected.
See the nice picture here:
http://en.wikipedia.org/wiki/Simply-connected#Informal_discussion.
Example 4.7. Interesting fact: The punctured space
R
3 \ {(0, 0 , 0)} = {(x, y, z) ∈ R
3 | (x, y, z) 6 = (0, 0 , 0)}
is simply connected. This might seem confusing, since the punctured plane R
2 \ {(0, 0)} is
not simply connected. That is because in 3 dimensions, there is enough room to move a loop
around the puncture and then contract it to a point. Therefore the informal idea that “simply
connected means no holes” is not really accurate.
Example 4.8. The “thick sphere” {(x, y, z) ∈ R
3 | 1 < x
2
2
2 < 4 } between radii 1 and 2
is simply connected.
Example 4.9. R
3
with a line removed, for example
D = R
3
\ {z-axis} = {(x, y, z) ∈ R
3
| x
2
2
6 = 0}
is not simply connected. Indeed, any curve in D going once around the z-axis cannot be
contracted to a point.
Example 4.10. The “thick cylinder” {(x, y, z) ∈ R
3
| 1 ≤ x
2
2
≤ 4 } between radii 1 and 2 is
not simply connected, for the same reason.
Example 4.11. The solid torus {((3 + u cos α) cos θ, (3 + u cos α) sin θ, u sin α) | α, θ ∈ R, 0 ≤
u ≤ 1 } is not simply connected. A curve going once around the “hole” in the middle (e.g. u, α
constant, θ goes from 0 to 2π) cannot be contracted to a point.
Example 4.12. The (surface of the) torus {((3 + cos α) cos θ, (3 + cos α) sin θ, sin α) | α, θ ∈ R} is
not simply connected. A curve going once around the “hole” in the middle (e.g. α constant, θ
goes from 0 to 2π) cannot be contracted to a point. Also, a curve going once around the “tire”
(e.g. θ constant, α goes from 0 to 2π) cannot be contracted to a point.
With that notion, we obtain the following improvement on proposition 3.1.
Let us find a potential f for
F. We want
fx = F 1 =
x
ρ
2
f y
= F
2
y
ρ
2
f z
= F
3
z
ρ
2
Using the first equation, we obtain
f =
F 1 dx
x
x
2
2
2
dx Take u = x
2
2
2
, du = 2xdx
u
du
ln u + g(y, z)
ln(x
2
2
2
) + g(y, z)
whose derivative with respect to y is
y
x
2
2
2
(y, z).
Using the second equation, we equate this with F 2
y
x
2
2
2
(y, z) =
y
x
2
2
2
⇒ g y
(y, z) = 0
⇒ g(y, z) = 0 + h(z).
Plugging this back into the expression for f , we obtain
f =
ln(x
2
2
2
) + h(z)
whose derivative with respect to z is
z
x
2
2
2
′
(z).
Using the third equation, we equate this with F 3
z
x
2
2
2
′
(z) =
z
x
2
2
2
⇒ h
′
(z) = 0
⇒ h(z) = c.
Choosing the constant c = 0, we obtain h(z) = 0 and thus the potential function
f (x, y, z) =
ln(x
2
2
2
).
5 Summary
Suppose we are given a vector field
F in 3 dimensions and we want to know if
F is conservative.
- Compute curl
F. If curl
F 6 =
0, then
F is certainly not conservative.
- If curl
F =
0 and the domain of
F is all of R
3 (or more generally: a simply-connected
region), then
F is certainly conservative.
- If curl
F =
0 and the domain of
F is not simply connected, then one cannot conclude:
F
could be conservative or not. One must work harder to answer the question.
