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Chapter 1
vector calculus
Vector calculus is the study of vector fields and related scalar functions. For the most part we
focus our attention on two or three dimensions in this study. However, certain theorems are easily
extended to R
n
. We explore these concepts in both Cartesian and the standard curvelinear coor-
diante systems. I also discuss the dictionary between the notations popular in math and physics
1
The importance of vector calculus is nicely exhibited by the concept of a force field in mechanics.
For example, the gravitational field is a field of vectors which fills space and somehow communi-
cates the force of gravity between massive bodies. Or, in electrostatics, the electric field fills space
and somehow communicates the influence between charges; like charges repel and unlike charges
attract all through the mechanism propagated by the electric field. In magnetostatics constant
magnetic fields fill space and communicate the magnetic force between various steady currents.
All the examples above are in an important sense static. The source charges
2 are fixed in space
and they cause a motion for some test particle immersed in the field. This is an idealization. In
truth, the influence of the test particle on the source particle cannot be neglected, but those sort
of interactions are far too complicated to discuss in elementary courses.
Often in applications the vector fields also have some time-dependence. The differential and inte-
gral calculus of time-dependent vector fields is not much different than that of static fields. The
main difference is that what was once a constant is now some function of time. A time-dependent
vector field is an assignment of a vectors at each point at each time. For example, the electric and
magnetic fields that together make light. Or, the velocity field of a moving liquid or gas. Many
other examples abound.
The calculus for vector fields involves new concepts of differentiation and new concepts of integra-
tion.
1 this version of my notes uses the inferior math conventions as to be consitent with earlier math courses etc...
2 the concept of a charge really allows for electric, magnetic or gravitational although isolated charges exist only
for two of the aformentioned
2 CHAPTER 1. VECTOR CALCULUS
For differentiation, we study gradients, curls and divergence. The gradient takes a scalar field and
generates a vector field (actually, this is not news for us). The curl takes a vector field and generates
a new vector field which says how the given vector field curls about a point. The divergence takes a
given vector field and creates a scalar function which quantifies how the given vector field diverges
from a point. Many novel product rules exist for these operations and the algebra which links these
operations is rich and interesting as we already saw at the beginning of this course as we studied
vectors at a point.
On the topic of integration there are two types of integration we naturally consider for a vector
field. We can integrate along an oriented curve, this type of integration is called a line integral
even when the curve is not a line. Also, for a given surface, we can calculate a surface integral
of a vector field. The line integral measures how the vector field lines up along the curve of inte-
gration, the measures something called the circulation. The surface integrals value depends on
how the vector field pokes through the surface of integration, it measures something called flux.
Critical to both linea and surface integrals are parametric equations for curves and surfaces. We’ll
see that we need parametrics to calculate anything yet the answers are completely indpendent of
the parameters utilized. In other words, the surface and line integrals are coordinate free objects.
This is in considerable contrast to the types of integrals we studied in the previous part of this
course.
Between the study of differentiation and integration of vector fields we find the unifying theorems of
Greene, Gauss and Stokes. We study Green’s theorem to begin since it is simple and merely a two-
dimensional result. However, it is evidence of something deeper. Both Gauss and Stokes reduce to
Greene’s in certain context. There is also a fundmental theorem of line integrals which helps validate
my claim that intuitively the ∇f is basically the derivative for a function of several variables. In
addition, we will learn about how these integral theorems relate to the path-independence of vector
fields. We saw earlier that certain vector fields could not be gradients because they violate Clairaut’s
Theorem on their components. On the other hand, just because a vector field has components
consistent with Clairaut’s theorem we also saw that it was not necessarily the case they were the
gradient of some scalar function on their whole domain. We’ll find sufficient conditions to make
the Clairaut test work. We will find how to say with certainty a given vector field is the gradient
of some scalar function. More than just that, we’ll find a method to calculate the scalar function.
4 CHAPTER 1. VECTOR CALCULUS
We need to integrate each component of the vector field to find this curve. Of course, given that
P, Q, R are typically functions of x, y, z the ”integration” requires thought. Even in differential
equations(334) the general problem of finding integral curves for vector fields is beyond our standard
techniques for all but a handful of well-behaved vector fields. That said, the streamlines for the
examples below are geometrically obvious so we can reasonably omit the integration.
Example 1.1.5. Suppose
F = ̂x then obviously x(t) = xo + t, y = yo, z = zo is the streamline
of x̂ through (x o
, y o
, z o
). Likewise, I think you can calculate the steamlines for ŷ and ̂z without
much trouble. In fact, any constant vector field
F = ~vo simply has streamlines which are lines with
direction vector ~v o
Example 1.1.6. The magnetic field around a long steady current in the postive z-direction is
conveniently written as
B(r, θ, z) =
μoI
2 πr
θ. The streamlines are circles which are centered on the
axis and point in the
θ direction.
Example 1.1.7. If a charge Q is distributed uniformly through a sphere of radius R then the
electric field can be show to me a function of the distance from the center of the sphere alone.
Placing that center at the origin gives
E(ρ) = ̂ρ
kρ
R
3
0 ≤ ρ ≤ R
k
ρ
2 ρ^ ≥^ R
The streamlines are simply lines which flow radially out from the origin in all directions.
Challenge: in electrostatics the density of streamlines (often called fieldlines in physics ) is used to
measure the magnitude of the electric field. Why is that reasonable?
Example 1.1.8. The other side of the thinking here is that given a differential equation we could
use the plot of the vector field to indicate the flow of solutions. We can solve numerically by playing
a game of directed connect the dots which is the multivariate analog of Euler’s method for solving
dy/dx = f (x, y).
dx
dt
= 2x(y
2
2 ),
dy
dt
= 2x
2 y,
dz
dt
= 2x
2 z
We’d look to match the curve up with the vector-field plot of
F = 〈 2 x(y
2
2 ), 2 x
2 y, 2 x
2 z〉. This
particular field is a gradient field with
F = ∇f for f (x, y, z) = x
2 (y
2
2 ). Solutions to the
differential equations describe paths which are orthogonal to the level surfaces of f since the paths
are parallel to ∇f.
Perhaps you can see how this way of thinking might be productive towards analyzing otherwise
intractable problems in differential equations. I merely illustrate here to give a bit more breadth to
the concept of a vector field. Of course Stewart
3 has pretty pictures with real world jutsu so you
should read that if this is not real to you without those comments.
3 we are covering chapter 17 from here on out
1.2. GRAD, CURL AND DIV 5
1.2 grad, curl and div
In this section we investigate a few natural derivatives we can construct with the operator ∇. Later
we will explain what these derivatives mean. First, the computation:
Definition 1.2.1.
Suppose f is a scalar function on R
3 then we defined the gradient vector field
grad(f ) = ∇f = 〈∂xf, ∂yf, ∂z f 〉
We studied this before, recall that we can compactly express this by
∇f =
3 ∑
i=
i
f )x̂ i
where ∂ i
= ∂/∂x i
and x 1
= x, x 2
= y and x 3
= z. Moreover, we have also shown previously in
notes or homework that the gradient has the following important properties:
∇(f + g) = ∇f + ∇g, & ∇(cf ) = c∇f, & ∇(f g) = (∇f )g + f (∇g)
Together these say that ∇ is a derivation of differentiable functions on R
n .
Definition 1.2.2.
Suppose
F = 〈F
1
, F
2
, F
3
〉 is a vector field. We define:
Div(
F ) = ∇ •^
F =
∂F
1
∂x
∂F
2
∂y
∂F
3
∂z
More compactly, we can express the divergence by
∇ •^
F =
3 ∑
i=
i
F
i
You can prove that the divergence satisfies the following important properties:
F +
G) = ∇
F + ∇
G & ∇
F ) = c∇
F , & ∇
G) = ∇f
G + f ∇
G
For example,
∇ •^ (
F + c
G) =
3 ∑
i=
i
(F
i
3 ∑
i=
i
F
i
3 ∑
i=
i
G
i
= ∇ •^
F + c∇ •^
G.
Linearity of the divergence follows naturally from linearity of the partial derivatives.
1.2. GRAD, CURL AND DIV 7
Proof: Consider (i.), let
F =
Fiei and
G =
Giei as usual,
F ×
G) =
k
[(
F ×
G)
k
]
k
[
ijk
F
i
G
j
]
ijk
[(∂
k
Fi)Gj + Fi(∂ k
Gj )]
ijk
i
F
j
)G
k
+ −F
j
ikj
i
G
k
G
k
(∇ ×
F ) − F
j
(∇ ×
G)
G · (∇ ×
F ) −
F · (∇ ×
G).
where the sums above are taken over the indices which are repeated in the given expressions. In
physics the
is often removed and the einstein index convention or implicit summation convention
is used to free the calculation of cumbersome summation symbols. The proof of the other parts of
this proposition can be handled similarly, although parts (viii) and (ix) require some thought so I
may let you do those for homework
5
.
Proposition 1.2.5.
If f is a differentiable R-valued function and
F is a differentiable vector field then
(i.) ∇ · (∇ ×
F ) = 0
(ii.) ∇ × ∇f = 0
(iii.) ∇ × (∇ ×
F ) = ∇(∇ ·
F ) − ∇
F
Before the proof, let me briefly indicate the importance of (iii.) to physics. We learn that in the
absence of charge and current the electric and magnetic fields are solutions of
∇ •^
E = 0, ∇ ×
E = −∂
t
B, ∇ •^
B = 0, ∇ ×
B = μ o
o
t
E
If we consider the curl of the curl equations we derive,
∇ × (∇ ×
E) = ∇ × (−∂
t
B) ⇒ ∇(∇ ·
E) − ∇
2 ~ E = −∂ t
(∇ ×
B) ⇒ ∇
2 ~ E = μ o
o
2
t
E.
∇ × (∇ ×
B) = ∇ × (μoo∂t
E) ⇒ ∇(∇ ·
B) − ∇
2 ~ B = μoo∂t(∇ ×
E) ⇒ ∇
2 ~ B = μoo∂
2
t
B.
These are wave equations. If you study the physics of waves you might recognize that the speed
of the waves above is v = 1/
μ o
o
. This is the speed of light. We have shown that the speed
of light apparently depends only on the basic properties of space itself. It is indpendent of the
x, y, z coordinates so far as we can see in the usual formalism of electromagnetism. This math was
only possible because Maxwell added a term called the displacement current in about 1860. Not
many years later radio and TV was invented and all because we knew to look for the possiblility
thanks to this mathematics. That said, the notation used above was not common in Maxwell’s
time. His original presentation of what we now call Maxwell’s Equations was given in terms of 20
scalar partial differential equations. Now we enjoy the clarity and precision of the vector formalism.
5 relax fall 2011 students this did not happen to you
8 CHAPTER 1. VECTOR CALCULUS
You might be interested to know that Maxwell (like many of the greatest 19-th century physicists)
was a Christian. Like Newton, they viewed their enterprise as revealing God’s general revelation.
Certainly their goal was not to remove God from the picture. They understood that the existence
of physical law does not relegate God to non-existence. Rather, we just get a clearer picture on
how He created the world in which we live. Just a thought. I have a friend who used to wear a
shirt with Maxwell’s equations and a taunt ”let there be light”, when he first wore it he thought
he was cleverly debunking God by showing these equations removed the need for God. Now, after
accepting Christ, he still wore the shirt but the equations don’t mean the same to him any longer.
The equations are evidence of God rather than his god.
Proof: I like to use parts (i.) and (ii.) for test questions at times. They’re pretty easy, I
leave them to the reader. The proof of (iii.) is a bit deeper. We need the well-known identity
∑ 3
j=
ikj lmj = δilδkm − δklδim
6
∇ × (∇ ×
F ) =
3 ∑
i,j,k=
ijk
i
(∇ ×
F )
j
x k
3 ∑
i,j,k=
ijk
i
3 ∑
l,m=
lmj
l
F
m
)̂ x k
3 ∑
i,j,k,l,m=
ijk
lmj
(∂i∂ l
Fm)x̂ k
3 ∑
i,j,k,l,m=
−ikj lmj (∂i∂lFm)̂xk
3 ∑
i,k,l,m=
(−δ il
δ km
δ im
i
l
F
m
)̂ x k
3 ∑
i,k,l,m=
(−δ il
δ km
i
l
F
m
) x̂ k
3 ∑
i,k,l,m=
(δ kl
δ im
i
l
F
m
) x̂ k
3 ∑
i,k=
i
i
(F
k
x k
3 ∑
i,k=
i
k
F
i
)̂x k
3 ∑
i=
i
i
3 ∑
k=
F
k
x k
3 ∑
k=
k
3 ∑
i=
i
F
i
)̂x k
2 ~ F + ∇(∇ •^
F ).
6 this is actually just the first in a whole sequence of such identities linking the antisymmetric symbol and the
kronecker deltas... ask me in advanced calculus, I’ll show you the secret formulas
10 CHAPTER 1. VECTOR CALCULUS
~γ(a + b − t). Clearly we have ~γreverse(a) = ~γ(b) = Q whereas ~γreverse(b) = ~γ(a) = Q. Perhaps it is
interesting to compare these paths at a common point,
γ(t) = ~γ reverse
(a + b − t)
The velocity vectors naturally point in opposite directions, (by the chain-rule)
d~γ
dt
(t) = −
d~γreverse
dt
(a + b − t).
Example 1.3.3. Suppose ~γ(t) = 〈cos(t), sin(t)〉 for π ≤ t ≤ 2 π covers the oriented curve C. If we
wish to parametrize −C by
β then we can use
β(t) = ~γ(3π − t) = 〈cos(3π − t), sin(3π − t)〉
Simplifying via trigonometry yields
β(t) = 〈− cos(t), − sin(t)〉 for π ≤ t ≤ 2 π. You can easily verify
that
β covers the lower half of the unit-circle in a CW-fashion, it goes from (1.0) to (− 1 , 0)
What I have just described is a general method to reverse a path whilst keeping the same domain
for the new path. Naturally, you might want to use a different domain after you change the
parametrization of a given curve. Let’s settle the general idea with a definition. This definition
describes what we allow as a reasonable reparametrization of a curve.
Definition 1.3.4.
Let ~γ 1 : [a 1 , b 1 ] → R
3 be a path. We say another path ~γ 2 : [a 2 , b 2 ] → R
3 is a
reparametrization of ~γ 1
if there exists a bijective (one-one and onto), continuous func-
tion u : [a 1 , b 1 ] → [a 2 , b 2 ] with continuous inverse u
− 1 : [a 2 , b 2 ] → [a 1 , b 1 ] such that
~γ 1
(t) = ~γ 2
(u(t)) for all t ∈ [a 1
, b 1
]. If the given curve is smooth or k-times differentiable then
we also insist that the transition function u and its inverse be likewise smooth or k-times
differentiable.
In short, we want the allowed reparametrizations to capture the same curve without adding any
artificial stops, starts or multiple coverings. If the original path wound around a circle 10 times
then we insist that the allowed reparametrizations also wind 10 times around the circle. Finally,
let’s compare the a path and its reparametrization’s velocity vectors, by the chain rule we find:
~γ 1
(t) = ~γ 2
(u(t)) ⇒
d~γ 1
dt
(t) =
du
dt
d~γ 2
dt
(u(t)).
This calculation is important in the section that follows. Observe that:
- if du/dt > 0 then the paths progress in the same direction and are consitently oriented
- if du/dt < 0 then the paths go in opposite directions and are oppositely oriented
Reparametrizations with du/dt > 0 are said to be orientation preserving.
1.3. LINE INTEGRALS 11
1.3.2 line-integral of scalar function
These are also commonly called the integral with respect to arclength. In lecture we framed
the need for this definition by posing the question of finding the area of a curved fence with height
f (x, y). It stood to reason that the infinitesimal area dA of the curved fence over the arclength ds
would simply be dA = f (x, y)ds. Then integration is used to sum all the little areas up. Moreover,
the natural calculation to accomplish this is clearly as given below:
Definition 1.3.5.
Let ~γ : [a, b] → C ⊂ R
n be a differentiable path and suppose that C ⊂ dom(f ) for a
continuous function f : dom(f ) → R then the scalar line integral of f along C is
C
f ds ≡
b
a
f (~γ(t)) ||~γ
′
(t)|| dt.
We should check to make sure there is no dependence on the choice of parametrization above. If
there was then this would not be a reasonable definition. Suppose ~γ 1
(t) = ~γ 2
(u(t)) for a 1
≤ t ≤ b 1
where u : [a 1 , b 1 ] → [a 2 , b 2 ] is differentiable and strictly monotonic. Note
b 1
a 1
f (~γ 1
(t))
d~γ 1
dt
dt =
b 1
a 1
f (~γ 2
(u(t)))
du
dt
d~γ 2
dt
(u(t))
dt
b 1
a 1
f (~γ 2 (u(t)))
d~γ 2
dt
(u(t))
du
dt
dt
If u is orientation preserving then du/dt > 0 hence u(a 1
) = a 2
and u(b 1
) = b 2
and thus
b 1
a 1
f (~γ 1
(t))
d~γ 1
dt
dt =
b 1
a 1
f (~γ 2
(u(t)))
d~γ 2
dt
(u(t))
du
dt
dt
b 2
a 2
f (~γ 2
(u)
d~γ 2
du
du.
On the other hand, if du/dt < 0 then |du/dt| = −du/dt and the bounds flip since u(a 1
) = b 2
and
u(b 1 ) = a 2
b 1
a 1
f (~γ 1
(t))
d~γ 1
dt
dt = −
b 1
a 1
f (~γ 2
(u(t)))
d~γ 2
dt
(u(t))
du
dt
dt
a 2
b 2
f (~γ 2
(u)
d~γ 2
du
du.
b 2
a 2
f (~γ 2
(u)
d~γ 2
du
du.
1.3. LINE INTEGRALS 13
C is given by
L =
C
ds =
4 π
0
x˙
2
2
2 dt
4 π
0
R
2
= 4π
R
2
Definition 1.3.7.
Let C be a curve with length L then the average of f over C is given by
f avg
L
C
f ds.
Example 1.3.8. The average mass per unit length of the helix with dm/dz = z as studied above
is given by
m avg
L
C
f ds =
4 π
R
2
8 π
2
R
2
Since z = t and 0 ≤ t ≤ 4 π over C this result is hardly surprising.
Another important application of the scalar line integral is to find the center of mass of a wire.
The idea here is nearly the same as we discussed for volumes, the difference is that the mass is
distributed over a one-dimensional space so the integration is one-dimensional as opposed to two-
dimensional to find the center of mass for a planar laminate or three-dimensional to find the center
of mass for a volume.
Definition 1.3.9.
Let C be a curve with length L and suppose dM/ds = δ is the mass-density of C. The total
mass of the curve found by M =
c
δds. The centroid or center of mass for C is found
at (¯x, y,¯ ¯z) where
x¯ =
M
C
xδ ds, ¯y =
M
C
yδ ds, z¯ =
M
C
zδ ds.
Often the centroid is found off the curve.
Example 1.3.10. Suppose x = R cos(t), y = R sin(t), z = h for 0 ≤ t ≤ π for a curve with δ = 1.
Clearly ds = Rdt and thus M =
C
δds =
π
0
Rdt = πR. Consider,
x ¯ =
πR
C
xds =
πR
π
0
R
2 cos(t)dt = 0
14 CHAPTER 1. VECTOR CALCULUS
whereas,
y ¯ =
πR
C
yds =
πR
π
0
R
2 sin(t)dt =
πR
(−R
2 cos(t)
π
0
2 R
π
The reader can easily verify that z¯ = h hence the centroid is at (0,
2 R
π
, h).
Of course there are many other applications, but I believe these should suffice for our current
purposes. We will eventually learn that
C
F •^
T ds and
C
F •^
N ds are also of interest, but we
should cover other topics before returning to these. Incidentally, it is pretty obvious that we have
the following properties for the scalar-line integral:
C
(f + cg)ds =
C
f ds + c
C
gds &
C∪
¯ C
f ds =
C
f ds +
¯ C
f ds
in addition if f ≤ g on C then
C
f ds ≤
C
gds. I leave the proof to the reader.
1.3.3 line-integral of vector field
For those of you who know a little physics, the motivation to define this integral follows from our
desire to calculate the work done by a variable force
F on some particle as it traverses C. In
particular, we expect the little bit of work dW done by
F as the particle goes from ~r to ~r + d~r is
given by
F •^ d~r. Then, to find total work, we integrate:
Definition 1.3.11.
Let ~γ : [a, b] → C ⊂ R
3 be a differentiable path which covers the oriented curve C and
suppose that C ⊂ dom(
F ) for a continuous vector field
F on R
3 then the vector line
integral of
F along C is denoted and defined as follows:
C
F •^ d~r =
b
a
F (~γ(t)) •
d~γ
dt
dt.
This integral measures the work done by
F over C. Alternatively, this is also called the circulation
of
F along C, however that usage tends to appear in the case that C is a loop. A closed curve is
defined to be a curve which has the same starting and ending points. We can indicate the line-
integral is taken over a loop by the notation
C
F •^ d~r. As with the case of the scalar line integral
we ought to examine the dependence of the definition on the choice of parametrization for C. If
we were to find a dependence then we would have to modify the definition to make it reasonable.
Once more consider the reparametrization γ~ 2 of γ~ 1 by a strictly monotonic differentiable function
16 CHAPTER 1. VECTOR CALCULUS
gives x
2
2 = 4 and the CCW direction. To find z we use the plane equation,
z = 1 − x − y = 1 − 2 cos(t) − 2 sin(t)
Therefore,
~r(t) = 〈2 cos(t), 2 sin(t), 1 − 2 cos(t) − 2 sin(t)〉
thus
d~r
dt
− 2 sin(t), 2 cos(t), 2[sin(t) − cos(t)]
Evaluate
F (x, y, z) = 〈y, z − 1 + x, 2 − x〉 at x = 2 cos(t), y = 2 sin(t) and z = 1 − 2 cos(t) − 2 sin(t)
to find
F (~r(t)) =
2 sin(t), −2 sin(t), 2[1 − cos(t)]
Now put it together,
C
F •^ d~r =
2 π
0
2 sin(t), −2 sin(t)), 2[1 − cos(t)]
− 2 sin(t), 2 cos(t), 2[sin(t) − cos(t)]
dt
2 π
0
[
−4 sin
2 (t) − 4 sin(t) cos(t) + 4[1 − cos(t)][sin(t) − cos(t)]
]
dt
2 π
0
[
−4 sin
2 (t) − 4 sin(t) cos(t) + 4 sin(t) − 4 cos(t) − 4 cos(t) sin(t) + 4 cos
2 (t)
]
dt
2 π
0
[
cos
2
(t) − sin
2
(t)
]
dt
2 π
0
[
(1 + cos(2t) −
(1 − cos(2t))
]
dt
2 π
0
[
(1 + cos(2t) −
(1 − cos(2t))
]
dt
2 π
0
[
cos(2t)
]
dt
The example above indicates how we apply the definition of the line-integral directly. Sometimes
it is convenient to use differential notation. If C is parametrized by ~r = 〈x, y, z〉 for a ≤ t ≤ b we
define the integrals of the differential forms P dx, Qdy and Rdz in the following way:
Definition 1.3.13.
Let ~r : [a, b] → C ⊂ R
3 be a differentiable path which covers the oriented curve C and
suppose that C ⊂ dom(〈P, Q, R〉) for a continuous vector field 〈P, Q, R〉 on R
3 then we
define
C
P dx =
b
a
P (~r(t))
dx
dt
dt,
C
Qdx =
b
a
Q(~r(t))
dy
dt
dt,
C
Rdx =
b
a
R(~r(t))
dz
dt
dt,
1.4. CONSERVATIVE VECTOR FIELDS 17
These are not basic calculations and in and of themselves they are not terribly interesting. I suppose
that the
C
P dx measures the work done by the x-vector-component of
F = 〈P, Q, R〉 whereas the
C
Qdyl and the
C
Rdz measure the work done by the y and z vector components of
F = 〈P, Q, R〉.
Primarily, these are interesting since when we add them we obtain the full line-integral:
C
〈P, Q, R〉 •^ d~r =
C
P dx + Qdy + Rdz
I invite the reader to verify the formula above. I will illustrate its use in many examples to follow.
It should be emphasized that these are just notation to organize the line integral.
Example 1.3.14. Calculate
C
F •^ d~r for
F (x, y, z) = 〈y, z−1+x, 2 −x〉 given that C is parametrized
by x = cos(t), y = sin(t), z = 1 for 0 ≤ t ≤ 2 π. Note that
dx = − sin(t)dt, dy = cos(t)dt, dz = 0
Thus, for P = y = sin(t) , Q = z − 1 + x = cos(t) and R = 2 − x = 2 − cos(t) we find
C
F •^ d~r =
C
P dx + Qdy + Rdz
2 π
0
− sin
2 (t)dt + cos(t) cos(t)dt =
2 π
0
cos(2t)dt = 0.
You might wonder if the integral around a closed curve is always zero.
Example 1.3.15. Let
F = 〈y, −x〉 and suppose x = R cos(t), y = R sin(t) parametrizes C for
0 ≤ t ≤ 2 π. Calculate,
P dx + Qdy = −yR sin(t)dt − xR cos(t)dt = −R
2
sin
2
(t)dt − R
2
cos
2
(t)dt = −R
2
dt
Thus,
C
F •^ d~r = −
2 π
0
R
2 dt = − 2 πR
2 .
Apparently just because we integrate around a loop it does not mean the answer is zero. I suspect
that there are loops for which
F (x, y, z) = 〈y, z − 1 + x, 2 − x〉. We will return to that example
once more in the next section after we learn a test to determine if the
C
F •^ d~r = 0 without direct
calculation. In conclusion, I should mention that the properties below are easily proved by direct
calculation on the defintion,
C
F + c
G) •^ d~r =
C
F •^ d~r + c
C
G •^ d~r &
C∪
¯ C
F •^ d~r =
C
F •^ d~r +
¯ C
F •^ d~r.
1.4 conservative vector fields
In this section we discuss how to identify a conservative vector field and how to use it. There are
about 5 equivalent ideas and our job in this section is to explore how these concepts are connected.
We also make a few connections with physics and it should be noted that part of the terminology
is certainly borrowed from classical mechanics. Let us begin with the fundamental theorem for line
integrals.
1.4. CONSERVATIVE VECTOR FIELDS 19
If
F is the net-force on a mass m then Newton’s Second Law states
F = m~a therefore, if C is a
curve from ~r 1
to ~r 2
C
F •^ d~r =
t 2
t 1
F •
d~r
dt
dt =
t 2
t 1
m~a •^ ~v
dt =
t 2
t 1
d
dt
[
mv
2
]
dt = K(t 2
) − K(t 1
where K =
1
2
mv
2 is the kinetic energy. This result is known as the work-energy theorem. It does
not require that
F be conservative. If
F is conservative then it is traditional to choose a potential
energy function U such that
F = −∇U. In this case we can use the FTC for line-integrals to once
more calculate the work done by the net-force,
C
F •^ d~r = −
C
∇U •^ d~r = −U (~r 2
) + U (~r 1
It follows that we have, for a conservative force, K 2 − K 1 = −U 2 + U 1 hence K 1 + U 1 = K 2 + U 2.
The quantity E = U + K is the total mechanical energy and it is a constant of the motion when
only conservative forces comprise the net-force. This is the reason I call a vector field which is a
gradient field of some pontential a conservative vector field. When viewed as a net-force it provides
the conservation of energy
9
. It turns out that usually we can find portions of the domain of an
arbitrary vector field on which the vector field is conservative. The obstructions to the existence of
a global potential are the interesting part.
Definition 1.4.4. path-independence
Suppose U ⊆ R
n then we say
F is path-independent on U iff
C 1
F •^ d~r =
C 2
F •^ d~r for
each pair of curves C 1
, C
2
⊂ U beginning at P and terminating at Q.
9 it is worth noticing that while physically this is most interesting to three dimensions, the math allows for more
20 CHAPTER 1. VECTOR CALCULUS
Proposition 1.4.5.
Suppose U is an open connected subset of R
n then the following are equivalent
F is conservative;
F = ∇f on all of U
F is path-independent on U
C
F •^ d~r = 0 for all closed curves C in U
- (add precondition n = 3 and U be simply connected) ∇ ×
F = 0 on U.
Proof: We postpone the proof of (4.) ⇒ (1.). However, we can show that (1.) ⇒ (4.). Suppose
F = ∇f. Note that ∇ ×
F = ∇ × ∇f = 0. I included this here since we can quickly test to see if
Curl(
F ) 6 = 0. When the curl is nontrivial then we can be certain the given vector field is not conser-
vative. On the other hand, vanishing curl is only useful if it occurs over a simply connected domain
10
(1.) ⇒ (2.). Assume
F = ∇f. Suppose C 1 , C 2 are two curves which both start at P and end at Q
in the set U. Apply the FTC for line-integrals in what follows:
C 1
F •^ d~r =
C 1
∇f •^ d~r = f (Q) − f (P ).
Likewise,
C 2
F •^ d~r =
C 2
∇f •^ d~r = f (Q) − f (P ). Therefore (2.) holds true.
(2.) ⇒ (1.). Assume
F is path-independent. Pick some point A ∈ U and let C be any curve in U
from A to B = (x, y, z). We define f (x, y, z) =
C
F
- d~r. This is single-valued since we assume
F
is path-independent. We need to show that ∇f =
F. Denote
F = 〈P, Q, R〉. We begin by isolating
the x-component. We need to show
∂
∂x
C
F •^ d~r = P (x, y, z). We can write C as curve C x
from
A to B x
= (x o
, y, z) with x o
< x pasted togther with the line-segment L x
from B x
to B. Observe
that the curve Cx has no dependence on x (of the B point)
∂x
C
F
∂x
[∫
Cx
F
Lx
F
]
∂x
[∫
Lx
F
]
The line segment L x
has parametrization ~r(t) = 〈t, y, z〉 for x o
≤ t ≤ x. We calculate that
Lx
F
x
xo
F (t, y, z)
x
xo
P (t, y, z)dt
Therefore,
∂x
C
F •^ d~r =
∂x
x
xo
P (t, y, z)dt = P (x, y, z).
10 a simply connected domain is a set with no holes, any loop can be smoothly shrunk to a point, it has a boundary
which is a simple curve. A simple curve is a curve with no self-intersections but perhaps one in the case it is closed.
A circle is simple a figure 8 is not.
