16.3 The Fundamental Theorem of Line Integrals
Recall the Fundamental Theorem of Calculus for a single-variable function f :
Z
b
a
f
0
(x) dx = f (b) − f (a)
It says that we may evaluate the integral of a derivative simply by knowing the values of the function
at the endpoints of the interval of integration [a, b].
The Fundamental Theorem of Line Integrals is a precise analogue of this for multi-variable functions.
The primary change is that gradient ∇ f takes the place of the derivative f
0
in the original theorem.
Theorem (Fundamental Theorem of Line Integrals). Suppose that C is a smooth curve from points A to
B parameterized by r(t) for a ≤ t ≤ b. Let f be a differentiable function whose domain includes C and whose
gradient vector ∇ f is continuous on C. Then
Z
C
∇ f · dr = f (r(b)) − f (r(a)) = f (B) − f (A)
The long caveats about differentiability and continuity are merely so that the original Fundamen-
tal Theorem of Calculus can be invoked in the proof.
Proof. (n = 2 or 3 for the purposes of this course)
Z
C
∇ f · dr =
Z
C
f
x
1
.
.
.
f
x
n
·
dx
1
.
.
.
dx
n
=
Z
C
∂ f
∂x
1
dx
1
+ · · · +
∂ f
∂x
n
dx
n
=
Z
b
a
∂ f
∂x
1
dx
1
dt
+ · · · +
∂ f
∂x
n
dx
n
dt
dt
=
Z
b
a
d
dt
f (x
1
( t), . . . , x
n
( t)) dt =
Z
b
a
d
dt
f (r(t)) dt (chain rule)
= f (r(b)) − f (r(a))
where we applied FTC in the final step.
The Theorem can be alternatively stated: if F is a conservative vector field with potential function f
then
Z
C
F · dr = f (end of C) − f (start of C)
We say that a line integral in a conservative vector field is independent of path.
Examples
1. Let C be the curve parameterized by r(t) =
1+sin
2
t
t+sin t
for 0 ≤ t ≤ 2π. Then
Z
C
∇(x
2
y
3
) · dr = x
2
y
3
(1,2π)
(1,0)
= 8π
3
− 0 = 8π
3
1
2. Let C be the curve parameterized by r(t) =
t
3
−1
t−t
−1
for 2 ≤ t ≤ 3. Then
Z
C
sin y dx + x cos y dy =
Z
C
∇(x sin y) · dr = x sin y
(26,
8
3
)
(7,
3
2
)
= 26 sin
8
3
− 7 sin
3
2
3. Evaluate the line integrals
R
C
i
y dx + x dy where C
1
is the
straight line from (0,0) and (1,1), and C
2
is the parabola y = x
2
between the same points.
For the first curve we have r(t) =
(
t
t
)
, so
Z
C
1
y dx + x dy =
Z
1
0
2t dt = 1
For the second curve we have r(t) =
t
t
2
, so
0
1
y
0 1
x
C
1
C
2
Z
C
2
y dx + x dy =
Z
1
0
t
2
dt + 2t
2
dt = 1
We expected the two solutions to be the same since
(
y
x
)
= ∇(xy) is conservative. We could
simply have applied the Fundamental Theorem:
Z
C
i
y
x
· dr =
Z
C
i
∇(xy) · dr = xy
(1,1)
(0,0)
= 1 − 0 = 1
4. Evaluate
R
C
y
2
z dx + 2xyz dy + xy
2
dz along any curve joining (1, 0, 0) and (2, 1, −1).
The integral is
R
C
F · dr where F = ∇(xy
2
z), so the path is irrelevant and we obtain
Z
C
y
2
z dx + 2xyz dy + xy
2
dz = xy
2
z
(2,1,−1)
(1,0,0)
= −2
Conservation of Energy The terminology (conservative, potential, etc.) all comes from Physics.
There are two primary forms of energy: potential (stored) and kinetic (motion).
Suppose that a particle of mass m follows a curve C through a conservative force field F = −∇ f .
We parameterize the curve so that the particle is at position r(t) at time t. Its velocity vector is then
v(t) = r
0
( t)
The particle has kinetic energy K =
1
2
m
|
v
|
2
and is said to have potential energy f .
Now we evaluate the line integral
R
C
F · dr in two ways.
1. Newton’s second law (F = ma = mv
0
) says that
1
Z
C
F · dr = m
Z
t
1
t
0
v
0
( t) · v(t) dt = m
Z
t
1
t
0
d
dt
1
2
|
v
|
2
dt =
1
2
m
|
v
|
2
t
1
t
0
= 4K
is the change in kinetic energy over the path.
1
By the product rule,
d
dt
v · v = v
0
· v + v · v
0
= 2v
0
· v = 2
|
v
|
2
.
2
2. Alternatively we may use the Fundamental Theorem:
Z
C
F · dr = −
Z
C
∇ f · dr = − f (r(t))
t
1
t
0
= −4 f
is negative the change in potential energy of the particle over the path.
Therefore 4 f + 4K = 0, and so total energy is conserved. Since Physicists always want energy
to be conserved, they typically choose potential functions to have a negative sign: F = −∇ f . In math-
ematics, we omit the negative.
Path Independence
The Fundamental Theorem has the amazing interpretation that line integrals
in conservative vector fields depend only on a curve’s endpoints. We want to
turn this idea on its head. Is it the case that a line integral (or integrals?) being
independent of path forces a vector field to be conservative?
Definition. A line integral
R
C
F · dr is independent of path if
R
C
F · dr =
R
ˆ
C
F · dr
for any curve
ˆ
C with the same endpoints as C
C
ˆ
C
A
B
Before we can state the relevant theorems, we need to understand the meaning of several terms.
Definition. A region D is open if it contains no
boundary points.
Open
Not Open
For example, the inside of the unit disk D = {(x, y) : x
2
+ y
2
< 1} is open.
Definition. A region D is (path-)connected if every pair of points A, B in D can be joined by a curve lying
entirely in D.
D
A
B
C
D
A
B
C?
A connected region with curve C joining A, B
A disconnected region: cannot join A, B with
a curve lying in D.
3
The adjectives open and connected apply only to domains/regions in this course.
The final adjective applies only to curves.
Definition. A curve is closed if it starts and finishes at the same point.
C
A
Independence of Path and Closed Curves The following important Theorem relates being inde-
pendent of path to line integrals round closed curves.
Theorem. Let C be a curve in a connected region D. Then
R
C
F · dr is independent of path if and only if
R
S
F · dr = 0 for every closed path S in D.
Read the Theorem carefully: if even one curve C has
R
C
F · dr independent of path, then the line
integrals over all curves must be independent of path. This says that independence of path is really a
property of the vector field F rather than a specific curve.
Proof. Suppose first that
R
C
F · dr is independent of path and let
S be a closed curve in D. Since D is connected, we may join the
starting point A of C to some fixed point P on S by a curve J lying
in D. Writing −J for the same curve travelled in reverse, we see
that the composite curve
C
1
= J ∪ S ∪ (−J) ∪ C
has the same endpoints as C. By independence of path, it follows
that
D
A
B
C
P
J
S
Z
C
F · dr =
Z
C
1
F · dr =
Z
J
+
Z
S
+
Z
−J
+
Z
C
F · dr
However
R
−J
F · dr = −
R
J
F · dr, since travelling in reverse changes the sign of a work integral.
Cancelling these and the
R
C
F · dr terms from both sides forces us to conclude that
R
S
F · dr = 0.
Conversely, suppose that the line integral round any closed curve
S evaluates to zero. Let C and C
1
be two curves with the same
endpoints. Then C ∪ ( −C
1
) is a closed curve S, whence
Z
C
F · dr −
Z
C
1
F · dr = 0.
Thus
R
C
F · dr is independent of path, regardless of C.
D
A
B
C
C
1
4
Finally, we see that independence of path is equivalent to the vector field being conservative.
Theorem. Let F be a continuous vector field on, and C a curve in, an open, connected region. Then
R
C
F · dr
is independent of path if and only if F is conservative.
Proof of Theorem (for 2 dimensions). If F is conservative with potential function f then the Fundamen-
tal Theorem tells us that
R
C
F · dr depends only on the endpoints of C.
Conversely, suppose that
R
C
F · dr is independent of path. By the previous Theorem,
R
C
F · dr is in-
dependent of path for any curve C in D. Choose a point A and define f (x, y) =
R
C
F · dr where C is
any curve joining A to (x, y). The fucntion f is well-defined because
R
C
F · dr is independent of path.
We claim that f is a potential function for F.
Choose (x, y) ∈ D. Since D is open, there exists a point (x
1
, y) ∈ D
such that x
1
< x. Let C
1
be a path from A to (x
1
, y) and C
2
the line
segment thence to (x, y). Then
f (x, y) =
Z
C
1
F · dr +
Z
C
2
F · dr
Since x
1
is constant, the first integral is independent of x and so
∂ f
∂x
=
∂
∂x
Z
C
2
F · dr
D
A
(x, y)(x
1
, y)
C
1
C
2
Now let F =
P
Q
. Along the curve C
2
we have y constant, hence dy = 0. Therefore
∂ f
∂x
=
∂
∂x
Z
(x,y)
(x
1
,y)
P dx + Q dy =
∂
∂x
Z
(x,y)
(x
1
,y)
P dx =
d
dx
Z
x
x
1
P(t, y) dt = P(x, y)
by the Fundamental Theorem of Calculus.
A similar argument (choose (x, y
1
) ∈ D with y
1
< y) shows that Q(x, y) =
∂ f
∂y
. Putting this together
we see that ∇ f = F and so F is conservative.
The proof in three dimensions requires a similar third argument for
∂ f
∂z
.
Example Evaluate the line integrals
R
C
i
2y
x
· dr over the same
line and parabola as before.
For the first curve we have r(t) =
(
t
t
)
, whence
Z
C
1
2y dx + x dy =
Z
1
0
3t dt =
3
2
For the second curve we have r(t) =
t
t
2
, and so
Z
C
2
2y dx + x dy =
Z
1
0
2t
2
dt +
Z
1
0
2t
2
dt =
4
3
6=
3
2
0
1
y
0 1
x
C
1
C
2
Considering the strength of the arrows in the picture it should be clear why
R
C
2
<
R
C
1
. The fact that
these integrals give different values tells us that F =
2y
x
is not conservative.
5
When is F conservative? Serching for a potential function can involve a lot of work. It is useful to
know if a vector field is conservative before you start looking.
Theorem. Suppose that P, Q have continuous first derivatives on a region D, then
F =
P
Q
conservative =⇒
∂Q
∂x
=
∂P
∂y
throughout D.
Equivalently:
∂Q
∂x
6=
∂P
∂y
=⇒ F =
P
Q
not conservative.
Proof. Suppose F = ∇ f =
f
x
f
y
. Then f
xy
= f
yx
, hence result.
While the Theorem can easily tell us that a vector field is not conservative, it doesn’t quite work
in reverse. We first need some more topology.
Simply-connected regions
Definition. A connected region D is simply-connected if and only if every closed curve in D may be shrunk
to a point without leaving D.
D simply-connected
D connected but not simply-connected: can-
not shrink C to a point due to the ‘hole’ in D
Theorem (Proof requires Green’s Theorem). Suppose F =
P
Q
has continuous partial derivatives on a
simply-connected region D and suppose that P, Q have continuous partial derivatives. Then F is conservative
if and only if
∂Q
∂x
=
∂P
∂y
throughout D.
Examples
1. Find a potential function, if there is one, for the vector field on F =
2xy
x
2
on D = R
2
.
You can dive straight in to computing, but it is helpful to use the Theorem first so that you don’t
waste time.
∂Q
∂x
= 2x =
∂P
∂y
=⇒ F is conservative.
6
C
D
(click)
C
D
(click)
Now that we know a potential function f exists, we can solve for it:
f
x
= P = 2xy =⇒ f (x, y) = x
2
y + g(y)
f
y
= Q = x
2
=⇒ f (x, y) = x
2
y + h(x)
for unknown functions g, h. Choosing g(y) = h(x) = c (constant) yields all possible potential
functions f (x, y) = x
2
y + c.
2. F(x, y) =
y sin x
x sin x
has domain D = R
2
. We quickly see that
∂Q
∂x
= sin x + x cos x and
∂P
∂y
= sin x
∂Q
∂x
6=
∂P
∂y
=⇒ F is non-conservative.
3. Let F(x, y) =
(2x+x
2
y)e
xy
x
3
e
xy
. Prove that
R
C
F · dr = 0 where C is any closed curve in the plane
We calculate the mixed partial derivatives:
∂
∂x
x
3
e
xy
= 3x
2
e
xy
+ x
3
ye
xy
= x
2
e
xy
(3 + xy)
∂
∂y
(2x + x
2
y)e
xy
= x
2
e
xy
+ (2x + x
2
y)xe
xy
= x
2
e
xy
(3 + xy)
These are equal, hence F is conservative and consequently all line integrals over closed paths C
evaluate to zero.
What changes in three dimensions? Essentially nothing!
A connected volume E is still simply-connected if every
curve can be shrunk to a point without leaving E.
The interior of a solid torus E is non-simply-connected since
a curve can be drawn inside it which cannot be shrunk to a
point
The primary difference is that the corresponding Theorem is not quite as easy to use as the 2D ver-
sion. . .
Theorem. Suppose F = Pi + Qj + Rk is a vector field on a simply-connected region E ⊆ R
3
and suppose
that P, Q, R have continuous partial derivatives. Then F is conservative iff
∂Q
∂x
=
∂P
∂y
,
∂R
∂y
=
∂Q
∂z
, and
∂P
∂z
=
∂R
∂x
throughout E.
The proof requires Stokes’ Theorem and will be much easier to remember once we’ve introduced
Curl. At the present it is not worth remembering.
7
Example Let F(x, y) =
2x+2y+6xyz+z
2x+3x
2
z
3x
2
y+x
. Writing F = Pi + Qj + Rk we can laboriously check that
P
y
= Q
x
, etc., to see that F is indeed conservative.
Writng F = ∇ f we can partially integrate:
f
x
= P = 2x + 2y + 6xyz + z
=⇒ f (x, y, z) = x
2
+ 2xy + 3x
2
yz + xz + g(y, z)
f
y
= Q = 2x + 3x
2
z
=⇒ f (x, y, z) = 2xy + 3x
2
yz + h(x, z)
f
z
= R = 3x
2
y + x
=⇒ f (x, y, z) = 3x
2
yz + xz + j(x, y)
for unknown functions g, h, j. Suitable choices yield
f (x, y, z) = x
2
+ 2xy + 3x
2
yz + xz
Does Simple-Connectedness Really Matter? Let us analyze the vector field F =
1
x
2
+ y
2
−y
x
and try to decide if it is conservative.
1. First compute the partial derivatives:
∂Q
∂x
=
∂
∂x
x
x
2
+ y
2
=
1
x
2
+ y
2
−
2x
2
(x
2
+ y
2
)
2
=
y
2
− x
2
(x
2
+ y
2
)
2
∂P
∂y
=
∂
∂y
−y
x
2
+ y
2
=
−1
x
2
+ y
2
+
2y
2
(x
2
+ y
2
)
2
=
y
2
− x
2
(x
2
+ y
2
)
2
Since
∂Q
∂x
=
∂P
∂y
we want to say that F is conservative.
2. Now calculate the line integral of F around the unit circle:
r(t) =
(
cos t
sin t
)
, 0 ≤ t ≤ 2π
Z
C
F · dr =
Z
2π
0
− sin t
cos t
·
− sin t
cos t
dt = 2π
Since
R
C
F · dr 6= 0 we conclude that F is non-conservative.
−1
1
y
−1 1
x
C
At least one of these arguments has to be false, but which? The answer, strangely, depends on
your choice of domain.
• With no additional information, you should assume that the domain of the vector field is the
largest set possible, in this case the punctured plane D = R
2
\ {(0, 0)}, the plane with the
8
origin removed. This is non-simply-connected; indeed the unit circle cannot be shrunk to a
point without some part of it passing through the origin! The easy Q
x
= P
y
theorem does not
apply and argument 1. is false. The vector field is conservative.
• Suppose that the domain was restricted so that it was simply
connected. For instance, we could exclude the positive x-
axis and choose the domain
D = R
2
\ {(x, 0) : x ≥ 0}
Now D is simply-connected and F is conservative. Moreover
R
C
F · dr = 0 for every closed curve in D. Note that the orig-
inal unit circle is no longer a curve in D, whence argument
2. is now false.
y
x
D
In the picture, the line integrals round the solid curves are zero, while around the blue curve
(which is not in D) the line integral evaluates to 2π. You can easily check that a suitable poten-
tial function for F on D is f = θ in polar-coordinates; otherwise said,
f (x, y) =
tan
−1
y
x
y > 0
π y = 0
π + tan
−1
y
x
y < 0
The problem in extending this potential function to the entire punctured plane is that θ is not
continuous everywhere; typically it is discontinuous on the positive x-axis.
Summary If F(x, y) = Pi + Qj is defined on a connected region D ⊆ R
2
, where P, Q have continu-
ous first derivatives then the following are equivalent:
1. F is conservative
2. F is a gradient field (= ∇ f for some potential function f )
3.
Z
C
F · dr is independent of path for any curve C in D
4.
Z
C
F · dr = 0 for every piecewise smooth closed curve C in D
In addition, if D is simply-connected we also have
5.
∂Q
∂x
=
∂P
∂y
If F = Pi + Qj + Rk on a volume E ⊆ R
3
then the fifth condition becomes
∂Q
∂x
=
∂P
∂y
,
∂R
∂y
=
∂Q
∂z
,
∂P
∂z
=
∂R
∂x
9
