Chapter 14
Line Integral and Curl
Last chapter introduced the concept of flux and the surface integral associated
with it. Flux uses the directional property of a vector field to have it pierce an
element of area. The directional property can also naturally assign a varying
direction along a line. One can consider how a vector field changes direction
as it moves along a curve in space. This change can also lead to a new kind of
integration and differentiation of vector fields. The integration leads to the notion of a line integral and the associated differentiation to the concept of curl.
14.1
The Line Integral
The prime example of a line integral is the work done by a force. Consider
the force field F(r) acting on an object and imagine the object being moved
by a small displacement ∆r. Then the work done by the force in effecting this
displacement is defined as
∆W = F(r) · ∆r,
where it is assumed that F(r) is (approximately) constant during the displacement.
To calculate the work for a finite displacement, such as the one shown
in Figure 14.1, we break up the displacement into N small segments, calculate
!N the work for each segment, and add all contributions to obtain W ≈
i=1 F(ri ) · ∆ri . The approximation sign can be removed by taking ∆ri as
small as possible and N as large as possible. Then we have
" P2
"
W =
F(r) · dr ≡
F · dr,
(14.1)
P1
C
where C stands for the particular curve on which the force is displaced. This
equation is, by definition, the line integral of the force field F. In this particular case it is the work done by F in moving from P1 to P2 . Of course, we can
apply the line integral to any vector field, not just force. In electromagnetic
line integral
defined
388
Line Integral and Curl
P2
∆ rN
F(xi, yi, zi)
∆ ri
P1
∆ r3
∆ r2
∆ r1
Figure 14.1: The line integral of a vector field F from P1 to P2 .
theory, for example, the line integrals of the electric and magnetic fields play
a central role.
The most general way to calculate a line integral is through parametric
equation of the curve. Thus, if the Cartesian set of parametric equations of
the curve is
x = f (t),
y = g(t),
z = h(t),
then the components of the vector field A will be functions of a single variable
t obtained by substitution:
!
"
Ax (x, y, z) = Ax f (t), g(t), h(t) ≡ F(t),
!
"
Ay (x, y, z) = Ay f (t), g(t), h(t) ≡ G(t),
!
"
Az (x, y, z) = Az f (t), g(t), h(t) ≡ H(t),
and the components of dr are
dx = f ′ (t) dt,
line integral in
terms of the
parametric
equations of the
curve
dy = g ′ (t) dt,
dz = h′ (t) dt.
Then the line integral of A can be written as
#
#
A · dr =
(Ax dx + Ay dy + Az dz)
C
=
C
b
#
a
$
%
F(t)f ′ (t) + G(t)g ′ (t) + H(t)h′ (t) dt,
(14.2)
where t = a and t = b designate the initial and final points of the curve,
respectively. Other coordinate systems can be handled similarly. Instead of
giving a general formula for these coordinate systems, we present an example
using cylindrical coordinates.
Example 14.1.1. Consider the vector field given by
A = c1 zϕêρ + c2 ρzêϕ + c3 ρϕêz ,
where c1 , c2 , and c3 are constants. We want to calculate the line integral of this
field, starting at z = 0, along one turn of a uniformly wound helix of radius a whose
14.1 The Line Integral
389
z
b
y
x
Figure 14.2: The helical path for calculating the line integral.
windings are separated by a constant value b (see Figure 14.2 ). The parametric
equation of this helix in cylindrical coordinates is
ρ ≡ f (t) = a, ϕ ≡ g(t) = t, z ≡ h(t) =
b
t.
2π
Notice that as ϕ = t changes by 2π, the height (i.e., z) changes by b as required.
Substituting for the three coordinates in terms of t in the expression for A, we obtain
"
!
b
b
A ≡ ⟨F(t), G(t), H(t)⟩ = c1 t2 , c2 a t, c3 at .
2π
2π
Similarly,
dr = ⟨dρ, ρ dϕ, dz⟩ = ⟨f ′ (t), f (t)g ′ (t), h′ (t)⟩dt =
so that
#
C
#
b
!
0, a,
b
2π
"
dt,
%
F(t)f ′ (t) + G(t)g ′ (t) + H(t)h′ (t) dt
a
'
# 2π &
b
b
0 + c2 a2 t + c3 at dt = πab(c2 a + c3 ).
=
2π
2π
0
A · dr =
$
!
Example 14.1.2. Consider the vector field A = K(xy 2 êx + x2 yêy ). We want
to evaluate the line integral of this field from the origin to the point (a, a) in the
xy-plane along three different paths (i), (ii), and (iii), as shown in Figure 14.3. Since
the vector field is independent of z and the paths are all in the xy-plane, we ignore
z completely.
The first path is the straight line y = x. A convenient parameterization is x = at,
y = at with 0 ≤ t ≤ 1. Along this path the components of A become
Ax = Kxy 2 = K(at)(at)2 = Ka3 t3 ,
Ay = Kx2 y = K(at)2 (at) = Ka3 t3 .
Furthermore, taking the differentials of x and y, we obtain dx = a dt and dy = a dt.
Thus,
#
# (a,a)
# 1
() 3 3 *
)
*
+
A · dr =
(Ax dx + Ay dy ) = K
a t a dt + a3 t3 a dt
C
(0,0)
= 2Ka4
#
0
1
0
t3 dt =
4
Ka
.
2
390
Line Integral and Curl
y
(iv)
(i)
(a, a)
(ii)
(iii)
x
Figure 14.3: The three paths joining the origin to the point (a, a). Path (iv) is to
illustrate the importance of parameterization.
Although parameterization is very useful, systematic, and highly recommended,
it is not always necessary. We calculate the line integral along path (ii)—given by
y = x2 /a—without using parameterization. All we have to notice is that all the y’s
are to be replaced by x2 /a [and therefore, dy by (2x/a) dx]. Thus,
! 2 "2
! 2"
x
x5
x4
x
Ax = Kxy 2 = Kx
=K 2,
Ay = Kx2 y = Kx2
=K .
a
a
a
a
The line integral can now be evaluated easily:
"%
! 4"!
# (a,a)
# a $! 5 "
2x
x
x
dx
(Ax dx + Ay dy ) = K
dx
+
a2
a
a
(0,0)
0
# a 5
4
x
Ka
= 3K
dx =
.
2
2
0 a
Finally, we calculate the line integral along the quarter of a circle. For this calculation, we return to the parameterization technique, because it eases the integration.
A simple parameterization is
π
x = a − a cos t, y = a sin t,
0≤t≤ ,
2
with dx = a sin t dt and dy = a cos t dt. This yields
Ax dx + Ay dy = K[(a − a cos t)a2 sin2 t]a sin t dt + K[(a − a cos t)2 a sin t]a cos t dt
= Ka4 [(1 − cos t)(1 − cos2 t) + (1 − cos t)2 cos t] sin t dt
= Ka4 (1 − 3 cos2 t + 2 cos3 t) sin t dt.
This is now integrated to give the line integral:
# π/2
# (a,a)
(Ax dx + Ay dy ) = Ka4
(1 − 3 cos2 t + 2 cos3 t) sin t dt
(0,0)
= Ka
4
$
0
&π/2
&π/2
&
&
+ cos3 t&
−
− cos t&
0
0
1
2
&π/2 %
Ka4
&
cos4 t&
.
=
2
0
The fact that the three line integrals yield the same result may seem surprising.
However, as we shall see shortly, it is a property shared by a special group of vector
fields of which A is a member.
!
14.2 Curl of a Vector Field and Stokes’ Theorem
Many a time parameterization makes life a lot easier! Suppose we want
to calculate the line integral of a vector field along path (iv) of Figure 14.3.
First let us attempt to calculate the line integral using the coordinates. Along
path (iv) dr = −êx dx; so A · dr = −Ax dx. Then
!
(0,a)
(a,a)
A · dr = −
!
0
Ax dx =
a
!
a
Ax dx.
0
Thus, if Ax > 0 (try Ax = x2 ), the integral will be positive. But this is wrong:
A positive Ax should yield a negative A · dr because the two vectors are in
opposite directions!
With parameterization, this problem is alleviated. A parameterization
that represents path (iv) is
x = a(1 − t),
y = a,
0 ≤ t ≤ 1.
Clearly, t = 0 corresponds to the beginning of path (iv) and t = 1 to its
endpoint. The parameterization automatically gives dx = −a dt and dy = 0.
For instance, the vector field of Example 14.1.2 yields
!
(0,a)
(a,a)
A · dr =
!
0
1
a(1 − t)a2 (−a dt) = −a4
!
0
1
(1 − t) dt = − 21 a4 .
This has the correct sign because Ax is positive and the direction of integration
negative. The other method would have given a positive result!
14.2
Curl of a Vector Field and Stokes’
Theorem
Line integrals around a closed path are of special interest. For example, if
the velocity vector of a fluid has a nonzero integral around a closed path, the
fluid must be turning around that path and a whirlpool must reside inside
the closed path. It is remarkable that such a mundanely concrete idea can be
applied verbatim to much more abstract and sophisticated concepts such as
electromagnetic fields with proven success and relevance. Thus, for a vector
field, A, and a closed path, C, we denote the line integral as
"
A · dr
C
where the circle on the integral sign indicates that the path is closed and C
denotes the particular path taken.
In our discussion of divergence and flux, we encountered Equation (13.11)
where an integral (over volume V ) was related to an integral over its boundary
(surface S). This remarkable property has an analog in one lower dimension:
Any closed curve bounds a surface inside it. Is it possible to connect the
391
parameterization
is essential for
obtaining the
correct sign for
some line
integrals!
392
Line Integral and Curl
S2
S1
C
Figure 14.4: There is no “the” surface having C as its boundary. Both S1 and S2 —as
well as a multitude of others—are such surfaces.
Right-hand rule
(RHR) rules here!
line integral over the closed curve to a surface integral over the surface? The
answer is yes, but we have to be careful here. What do we mean by “the” surface? A given closed curve may bound many different surfaces, as Figure 14.4
shows. It turns out that this freedom, which was absent in the divergence
case,1 is irrelevant and the relation holds for any surface whose boundary is
the given curve.
Let us now develop the analog of the divergence theorem for closed line
integrals. To begin, we consider a small closed rectangular path with a unit
normal ên , which is related to the direction of traversing the path by the
right-hand rule (RHR):
Box 14.2.1. (The Right-Hand Rule). Curl the fingers of your right
hand in the direction of integration along the curve, your thumb should
then point in the direction of ên .
Without loss of generality we assume that the rectangle is parallel to the xyplane with sides parallel to the x-axis and the y-axis and that ên is parallel
to the z-axis (see Figure 14.5). The line integral can be written as
!
C
A · dr =
"
a
b
A · dr +
"
c
b
A · dr +
"
c
d
A · dr +
"
d
a
A · dr.
We do the first integral in detail; the rest are similar. Along ab the element
of displacement dr is always in the positive x-direction and has magnitude dx,
1 It should be clear that we cannot change the shape of the volume enclosed in S without
changing S itself. This rigidity is due to the maximality of the dimension of the enclosed
region: A volume is a three-dimensional object, and three is the maximum dimension we
have. Theories with higher dimension than three will allow a deformability similar to the
one discussed above.
14.2 Curl of a Vector Field and Stokes’ Theorem
z
ên
A
a
b
393
c
∆y
d
∆x
z
x
x
y
O
y
Figure 14.5: A closed rectangular path parallel to the xy-plane with center at (x, y, z).
so it can be written as dr = êx dx. Thus, the first integral on the RHS above
becomes
! b
! b
! b
! b
A · dr ≡
A1 · dr1 =
A1 · (êx dx) =
A1x dx,
a
a
a
a
where, as before, the subscript 1 indicates that we have to evaluate A at the
midpoint of ab and the subscript x denotes the x-component. Now, since ab
is small and the angle between A and dr does not change appreciably on ab,2
we can approximate the integral with A1x ab and write
#
"
! b
∆y
, z ∆x
A · dr ≈ A1x ab = A1x ∆x = Ax x, y −
2
a
$
%&
'
coordinates of
midpoint of ab
≈
(
)
∆y ∂Ax
Ax (x, y, z) −
∆x,
2 ∂y
where in the last line we used the Taylor expansion of Ax . Similarly, we can
write
! d
! d
! d
! d
A · dr =
A2 · dr2 =
A2 · (−êx dx) = −
A2x dx
c
c
c
c
#
"
∆y
, z ∆x
≈ −A2x cd = −A2x ∆x = −Ax x, y +
2
$
%&
'
coordinates of
midpoint of cd
(
)
∆y ∂Ax
≈ − Ax (x, y, z) +
∆x.
2 ∂y
Adding the contributions from sides ab and cd yields
! b
! d
∂Ax
∆x ∆y.
A·d r+
A · dr ≈ −
∂y
a
c
2 This condition is essential, because a rapidly changing angle implies a rapidly changing
component A1x which is not suitable for the approximation to follow.
394
Line Integral and Curl
The contributions from the other two sides of the rectangle can also be
calculated:
! a
! c
A · dr +
A · dr ≈ A3y ∆y − A4y ∆y
b
d
#
#
"
"
∆x
∆x
, y, z ∆y − Ay x −
, y, z ∆y
= Ay x +
2
2
$
%
$
%
∆x ∂Ay
∆x ∂Ay
≈ Ay (x, y, z) +
∆y − Ay (x, y, z) −
∆y
2 ∂x
2 ∂x
∂Ay
∆x ∆y.
=
∂x
The sum of these two equations gives the total contribution:
"
#
&
∂Ax
∂Ay
−
A·d r ≈
∆x ∆y.
∂x
∂y
C
(14.3)
Let us look at Equation (14.3) more closely. The expression in parentheses
can be interpreted as the z-component of the cross product of the gradient
operator ∇ with A. In fact, using the mnemonic determinant form of the
vector product, we can write
⎞
⎛
êx êy êz
⎟
⎜
⎜ ∂
∂
∂ ⎟
⎟
⎜
∇ × A = det ⎜
⎟
⎜ ∂x ∂y ∂z ⎟
⎠
⎝
=
curl of a vector
field defined
"
Ax
Ay
∂Ay
∂Az
−
∂y
∂z
#
Az
êx +
"
∂Ax
∂Az
−
∂z
∂x
#
êy +
"
∂Ay
∂Ax
−
∂x
∂y
#
êz .
This cross product is called the curl of A and is an important quantity in
vector analysis. We will look more closely at it later. At this point, however,
we are interested only in its definition as applied in Equation (14.3). The
RHS of that equation can be written as
#
"
∂Ax
∂Ay
−
∆x ∆y = (∇ × A)z ∆x ∆y = (∇ × A) · êz ∆a,
∂x
∂y
where ∆a = ∆x ∆y is the area of the rectangle. Noting that êz is in the
direction normal to the area, we can replace it with ên . Therefore, we can
write Equation (14.3) as
&
A · dr ≈ (∇ × A) · ên ∆a = (∇ × A) · ∆a.
(14.4)
C
Equation (14.4) states that for a small rectangular path C the closed line
integral is equal to the normal component of the curl of A evaluated at the
center of the rectangle times the area of the rectangle. This statement does
14.2 Curl of a Vector Field and Stokes’ Theorem
not depend on the choice of coordinate system. In fact, any rectangle (or any
closed planar loop) defines a plane and we are at liberty to designate that
plane the xy-plane. Thus, we can define the curl of a vector field this way:
Definition 14.2.1. Given a small closed curve C, calculate the line integral
of A around it and divide the result by the area enclosed by C. The component
of the curl of A along the unit normal to the area is given by
!
A · dr
.
(14.5)
Curl A · ên ≡ ∇ × A · ên = lim C
∆a→0
∆a
395
coordinate
independent
definition of curl
The direction of ên is related to the sense of integration via the right-hand
rule.
In Equation (14.5) we are assuming that the area is flat. This is always
possible by taking the curve small enough. Definition 14.2.1 is completely
independent of the coordinate system and we shall use it to derive expressions
for the curl of vector fields in spherical and cylindrical coordinates as well.
The reader should be aware that the notation ∇ × A is just that, a notation,
and—except in Cartesian coordinates—should not be considered as a cross
product.
What happens with a large closed path? Figure 14.6 shows a closed path C
with an arbitrary surface S, whose boundary is the given curve. We divide S
into small rectangular areas and assign a direction to their contours dictated
by the direction of integration around C.3 If we sum all the contributions
from the small rectangular paths, we will be left with the integration around
C because the contributions from the common sides of adjacent rectangles
cancel.4 This is because the sense of integration along their common side is
C
ê n ∆a
S
Figure 14.6: An arbitrary surface with the curve C as its boundary. The sum of the
line integrals around the rectangular paths shown is equal to the line integral around C.
3 The direction of the contour with one side on the curve C is determined by the direction
of the integration of C. The direction of a distant contour is determined by working one’s
way to it one (small) rectangle at a time.
4 This situation is completely analogous to the calculation of the total flux in the derivation of the divergence theorem.
from small
rectangles to large
loops
396
Line Integral and Curl
opposite for two adjacent rectangles (see Figure 14.6). Thus, the macroscopic
version of Equation (14.4) is
!
C
the most
important Stokes’
theorem
A · dr ≈
N
"
i=1
(∇ × A)i · êni ∆ai =
N
"
i=1
(∇ × A)i · ∆ai ,
where (∇ × A)i is the curl of A evaluated at the center of the ith rectangle,
which has area ∆ai and normal êni , and N is the number of rectangles on
the surface S. If the areas become smaller and smaller as N gets larger and
larger, we can replace the summation by an integral and obtain
Theorem 14.2.1. (Stokes’ Theorem). The line integral of a vector field
A around a closed path C is equal to the surface integral of the curl of A on
any surface whose only edge is C. In mathematical symbols, we have
##
!
A · dr =
∇ × A · da.
(14.6)
C
S
The direction of the normal to the infinitesimal area da of the surface S is
related to the direction of integration around C by the right-hand rule.
Example 14.2.2. In this example we apply the concepts of closed line integral
and the Stokes’ theorem to a concrete vector field. Consider the vector field
A = K(x2 yêx + xy 2 êy )
obtained from the vector field of Example 14.1.2 by switching the x- and y-components.
We want to calculate the line integral around the two closed loops (the circle and
the rectangle) of Figure 14.7 and verify the Stokes’ theorem.
A convenient parameterization for the circle is
x = a cos t,
y = a sin t,
0 ≤ t ≤ 2π,
with dx = −a sin t dt and dy = a cos t dt. Thus,
A · dr = K(a cos t)2 (a sin t)(−a sin t dt) + K(a cos t)(a sin t)2 (a cos t dt) = 0,
y
2b
a
b
x
Figure 14.7: Two loops around which the vector field of Example 14.2.2 is calculated.
14.2 Curl of a Vector Field and Stokes’ Theorem
and the LHS of the Stokes’ theorem
vector.
!
! êx
!
!
!
! ∂
∇ × A = K ! ∂x
!
!
! 2
!x y
397
is zero. For the RHS, we need the curl of the
!
êy
êz !!
!
!
∂
∂ ! = K(y 2 − x2 )ê .
z
∂y
∂z !!
!
!
xy 2
0 !
It is convenient to use cylindrical coordinates for integration over the area of the
circle. Moreover, the right-hand rule determines the unit normal to the area of the
circle to be êz . Thus,
""
∇ × A · da = K
S
"
a
0
"
2π
0
(ρ2 sin2 ϕ − ρ2 cos2 ϕ)ρ dρ dϕ = 0
by the ϕ integration. Thus the two sides of the Stokes’ theorem agree.
The two sides of the rectangular loop sitting on the axes will give zero because
A = 0 there. The contribution of the side parallel to the y-axis can be obtained by
noting that x = 2b and dx = 0, so that
A · dr = Ax dx + Ay dy = 0 + 2bKy 2 dy
and
"
(2b,b)
(2b,0)
A · dr = 2bK
"
b
0
y 2 dy = 23 Kb4 .
To avoid ambiguity,5 we employ parameterization for the last line integral. A convenient parametric equation would be
x = 2b(1 − t),
y = b,
0 ≤ t ≤ 1,
which gives dx = −2b dt, dy = 0, and for which the line integral yields
" 1
" 1
" (2b,0)
A · dr = K
[2b(1 − t)]2 (b)(−2b dt) = −8b4 K
(1 − t)2 dt = − 83 Kb4 .
(2b,b)
0
0
So, the line integral for the entire loop (the LHS of the Stokes’ theorem) is
#
A · dr = 23 Kb4 − 83 Kb4 = −2Kb4 .
C
We have already calculated the curl of A. Thus, the RHS of the Stokes’ theorem
becomes
""
""
∇ × A · da = K
(y 2 − x2 ) dx dy
S
S
=K
and the two sides agree.
5 See
"
$0
2b
" b
" 2b
" b
dx
y 2 dy −K
x2 dx
dy = −2Kb4
0
0
0
%&
'
%&
'
$
=2b(b3 /3)
the discussion following Example 14.1.2.
(8b3 /3)b
!
398
Line Integral and Curl
Historical Notes
George Gabriel
Stokes 1819–1903
George Gabriel Stokes published papers on the motion of incompressible fluids in
1842–43 and on the friction of fluids in motion, and on the equilibrium and motion
of elastic solids in 1845.
In 1849 Stokes was appointed Lucasian Professor of Mathematics at Cambridge,
and in 1851 he was elected to the Royal Society and was secretary of the society
from 1854 to 1884 when he was elected president.
He investigated the wave theory of light, named and explained the phenomenon
of fluorescence in 1852, and in 1854 theorized an explanation of the Fraunhofer lines
in the solar spectrum. He suggested these were caused by atoms in the outer layers
of the Sun absorbing certain wavelengths. However, when Kirchhoff later published
this explanation, Stokes disclaimed any prior discovery.
Stokes developed mathematical techniques for application to physical problems
including the most important theorem which bears his name. He founded the science
of geodesy, and greatly advanced the study of mathematical physics in England. His
mathematical and physical papers were published in five volumes, the first three of
which Stokes edited himself in 1880, 1883, and 1891. The last two were edited by
Sir Joseph Larmor in 1887 and 1891.
14.3
conservative
vector fields
defined
Conservative Vector Fields
Of great importance are conservative vector fields, which are those vector fields that have vanishing line integrals around every closed path. An
immediate result of this property is that
Box 14.3.1. The line integral of a conservative vector field between two
arbitrary points in space is independent of the path taken.
To see this, take any two points P1 and P2 connected by two different directed
paths C1 and C2 as shown in Figure 14.8(a). The combination of C1 and the
negative of C2 forms a closed loop [Figure 14.8(b)] for which we can write
!
!
A · dr +
A · dr = 0
−C2
C1
because A is conservative by assumption. The second integral is the negative
of the integral along C2 . Thus, the above equation is equivalent to
!
!
!
!
A · dr −
A · dr = 0 ⇒
A · dr =
A · dr
C1
C2
C1
C2
which proves the above claim.
Now take an arbitrary reference point P0 and connect it via arbitrary paths
to all points in space. At each point P with Cartesian coordinates (x, y, z),
define the function Φ(x, y, z) by
! P
!
Φ(x, y, z) = −
A · dr ≡ −
A · dr,
(14.7)
P0
C
14.3 Conservative Vector Fields
399
C2
− C2
P2
P1
P2
P1
C1
C1
(a)
(b)
Figure 14.8: (a) Two paths from P1 to P2 , and (b) the loop formed by them.
where C is any path from P0 to P and the minus sign is introduced for
historical reasons only. Φ is a well-defined function because its value does not
depend on C and is called the potential associated with the vector field A.
We note that the potential at P0 is zero. That is why P0 is called the potential
reference point.
Now consider two arbitrary points P1 and P2 , with Cartesian coordinates
(x1 , y1 , z1 ) and (x2 , y2 , z2 ), connected by some path C. We can also connect
these two points by a path that goes from P1 to P0 and then to P2 (see
Figure 14.9). Since A is conservative, we have
!
P2
P1
A · dr =
!
P0
P1
A · dr +
!
P2
P0
A · dr = Φ(x1 , y1 , z1 ) − Φ(x2 , y2 , z2 )
or
Φ(x2 , y2 , z2 ) − Φ(x1 , y1 , z1 ) = −
!
P2
P1
(14.8)
A · dr,
which expresses the potential difference between the two points.
If P1 and P2 are displaced infinitesimally by dr, then their infinitesimal
potential difference will be
dΦ = −A · dr.
On the other hand, Φ, being a scalar differentiable function of x, y, and z, has
infinitesimal increment
dΦ =
∂Φ
∂Φ
∂Φ
dx +
dy +
dz = (∇Φ) · dr,
∂x
∂y
∂z
C
the function Φ, so
defined, has the
mathematical
property expected
of a function,
namely, that for
every point P , the
function has only
one value that we
may denote as
Φ(P ).
P2
P1
P0
Figure 14.9: Any path C from P1 to P2 is equivalent to the path P1 → P0 → P2 .
potential of a
conservative
vector field
400
Line Integral and Curl
so we have
−A · dr = (∇Φ) · dr.
But this is true for an arbitrary dr. Taking dr to be êx dx, êy dy, and êz dz
in turn, we obtain the equality of the three components of ∇Φ and −A.
Therefore, we have
A = −∇Φ,
(14.9)
which states that
Theorem 14.3.1. A conservative vector field can be written as the negative
gradient of a potential function defined as
Φ(x, y, z) = −
!
P
P0
A · dr,
where (x, y, z) are the coordinates of P , and the integral is taken along any
path connecting P0 and P .
Another property of a conservative vector field can be obtained by rewriting Equation (14.4), which is true for an arbitrary infinitesimal closed path:
"
A · dr ≈ (∇ × A) · ên ∆a.
C
the curl of a
conservative
vector field is zero.
∇ × A = 0 does
not necessarily
imply that A is
conservative!
However, the LHS is zero because A is conservative. Thus we have
(∇ × A) · ên ∆a = 0.
This is true for arbitrary ∆a and ên . Therefore, we have the important
conclusion that ∇ ×#A = 0 for a conservative vector field. It is important to
note that although C A · dr is zero and C is small, we cannot deduce that
A · dr = 0 and, therefore, A = 0. (Why?)
A conservative vector field demands the vanishing of the curl. But is
∇ × A = 0 sufficient for A to be conservative? The answer, in general, is
no! (See Example 14.3.3 below.) If the vector field is well defined and well
behaved (smoothly varying, #differentiable, etc.) in a region of space U , then
∇× A = 0 in U implies that C A·dr = 0 for all closed curves C lying entirely
in U . In modern mathematical jargon such a region is said to be contractible
to zero, which means that any closed curve in U can be contracted to a point
(or “zero” closed curve) without encountering any singular point of the vector
field (where it is not defined or well behaved). We state this result as follows:
Box 14.3.2. Let the region U in space be contractible to zero for the vector
field #A. Then for any closed curve C in U , the two relations ∇ × A = 0
and C A · dr = 0 are equivalent.
14.3 Conservative Vector Fields
401
Example 14.3.2. The line integral of the vector field of Example 14.1.2 was
independent of the three paths examined there. Could it be that the vector field is
conservative? The vector field is clearly well behaved everywhere. Therefore, the
vanishing of its curl proves that it is conservative. But
!
!
! êx
êy
êz !!
!
!
!
!
!
!
! ∂
∂
∂
∇ × A = K ! ∂x
∂y
∂z !! = (0)êx + (0)êy + (2xy − 2xy)êz = 0.
!
!
!
!
! 2
!xy
x2 y
0 !
So, A is indeed conservative.
Next we find the potential of A at a point (x0 , y0 ) in the xy-plane.6 Let the
reference point be the origin. Since it does not matter what path we take, we choose
a straight line joining the origin and (x0 , y0 ). A convenient parametric equation is
x = x0 t,
y = y0 t,
0 ≤ t ≤ 1,
which gives dx = x0 dt and dy = y0 dt. We now have
" (x0 ,y0 )
A · dr
Φ(x0 , y0 ) = −
(0,0)
= −K
"
1
[(x0 t)(y0 t)2 (x0 dt) + (x0 t)2 (y0 t)(y0 dt)]
0
= −2Kx20 y02
"
1
0
t3 dt = − 12 Kx20 y02 .
We can now substitute (x, y) for (x0 , y0 ) to obtain
Φ(x, y) = − 12 Kx2 y 2 .
The reader may verify that A = −∇Φ.
!
It should be clear that ∇×A ̸= 0 always implies that A is not conservative.
However, ∇ × A = 0 implies that A is conservative only if the region in
question is contractible to zero.
Example 14.3.3. Consider the vector field
A=
kx
ky
êx − 2
êy ,
x2 + y 2
x + y2
where k is a constant. Since the components of this vector are independent of z, the
curl of the vector can have only a z-component:
!
!
! êx
êy
êz !!
!
! #
!
$
!
!
∂Ay
∂Ax
!
! ∂
∂
∂
−
êz .
∇ × A = ! ∂x ∂y ∂z ! =
!
!
∂x
∂y
!
!
!
!
! Ax Ay
0 !
6 We
completely ignore the z-coordinate because A has no component in that direction.
402
Line Integral and Curl
The reader may easily verify that
2x2
∂Ay
k
+k 2
,
=− 2
2
∂x
x +y
(x + y 2 )2
so that
∂Ax
2y 2
k
+k 2
,
= 2
2
∂y
x +y
(x + y 2 )2
∂Ax
2k
2(x2 + y 2 )
∂Ay
−
=− 2
+k 2
=0
2
∂x
∂y
x +y
(x + y 2 )2
and ∇ × A = 0.
Now take a circle of radius a about the origin and calculate the line integral of
A on this circle. For integration, use the parameterization
x = a cos t,
y = a sin t,
0 ≤ t ≤ 2π,
with dx = −a sin t dt and dy = a cos t dt. Then
A · dr = Ax dx + Ay dy =
and, therefore
!
k(a sin t)(−a sin t dt)
k(a cos t)(a cos t dt)
−
= −k dt
(a cos t)2 + (a sin t)2
(a cos t)2 + (a sin t)2
circ
A · dr = −k
"
2π
0
dt = −2πk.
This is an example of a vector field whose curl vanishes but yields a nonzero
result for a closed line integral. The reason is, of course, that the region inside the
circle is not contractable to zero: At the origin the vector is infinite.
!
If the vector field is conservative, in principle we can determine its potential
either by direct antidifferentiation or by integration. The following example
illustrates the former procedure.
Example 14.3.4. Consider the vector field
A = (2xy + 3z 2 )êx + (x2 + 4yz)êy + (2y 2 + 6xz)êz .
The reader may check that ∇ × A = 0. Thus, since A is well defined everywhere,
it is conservative. To find its potential Φ, we note that
∂Φ
= −Ax = −2xy − 3z 2 ⇒ Φ = −x2 y − 3z 2 x + g(y, z),
∂x
where we have simply antidifferentiated Ax with respect to x—assuming that y
and z are merely constants—and added a “constant” of integration: As far as x
differentiation is concerned, any function of y and z is a constant. Now differentiate
Φ obtained this way with respect to y and set it equal to −Ay :
−Ay = −(x2 + 4zy) =
This gives
$
∂ # 2
∂Φ
∂g
=
−x y − 3z 2 x + g(y, z) = −x2 +
.
∂y
∂y
∂y
∂g
= −4yz ⇒ g(y, z) = −2y 2 z + h(z)
∂y
Note that our second “constant” of integration has no x-dependence because g(y, z)
does not depend on x. Substituting this back in the expression for Φ, we obtain
Φ = −x2 y − 3z 2 x + g(y, z) = −x2 y − 3z 2 x − 2y 2 z + h(z).
14.3 Conservative Vector Fields
403
Finally, differentiating this with respect to z and setting it equal to −Az , we obtain
−Az = −(2y 2 + 6xz) =
This gives
"
∂ ! 2
∂Φ
dh
=
−x y − 3z 2 x − 2y 2 z + h(z) = −6xz − 2y 2 +
.
∂z
∂z
dz
dh
= 0 ⇒ h(z) = const. ≡ C.
dz
The final answer is therefore
Φ(x, y, z) = −x2 y − 3z 2 x − 2y 2 z + C.
The arbitrary constant depends on the potential reference point, and is zero if we
choose the origin as that point. It is easy to verify that −∇Φ is indeed the vector
field we started with.
!
There are various vector identities which connect gradient, divergence,
and curl. Most of these identities can be obtained by direct substitution. For
example, by substituting the Cartesian components of A × B in the Cartesian
expression for divergence, one can show that
∇ · (A × B) = B · ∇ × A − A · ∇ × B.
(14.10)
Similarly, one can show that
∇ · (f A) = A · ∇f + f ∇ · A,
∇ × (f A) = f ∇ × A + (∇f ) × A
A × (∇ × A) =
2
1
2 ∇|A|
(14.11)
− (A · ∇)A
We can use Equation (14.10) to derive an important vector integral relation
akin to the divergence theorem. Let B be a constant vector. Then the second
term on the RHS vanishes. Now apply the divergence theorem to the vector
field A × B:
##
###
A × B · da =
∇ · (A × B) dV.
S
V
Using Equation (14.10), the RHS can be written as
RHS =
###
B · ∇ × A dV = B ·
V
###
∇ × A dV.
V
Moreover, the use of the cyclic property of the mixed triple product (see
Problem 1.15) will enable us to write the LHS as
LHS =
##
S
(da × A) · B =
##
S
B · (da × A) = B ·
##
S
da × A.
404
Line Integral and Curl
Equating the new versions of the two sides, we obtain
!!!
∇ × A dV = B ·
!!!
!!
B·
V
or
⎛
B·⎝
!!
da × A
S
∇ × A dV −
V
S
⎞
da × A⎠ = 0.
Since the last relation is true of arbitrary B, the vector inside the parentheses
must be zero. This gives the result we are after:
!!!
∇ × A dV =
V
14.4
!!
da × A.
S
Problems
14.1. Evaluate the line integral of
A(x, y, z) = x2 êx + y 2 êy − z 2 êz
along the path given parametrically by
x = at2 ,
y = bt,
z = c sin (πt/2)
from the origin to (a, b, c).
14.2. Evaluate the line integral of
A(x, y, z) = xêx +
y2
z2
êy − êz
b
c
along the path given parametrically by
x = a cos(πt/2),
y = b sin(πt/2),
from (a, 0, 0) to (0, b, c).
14.3. Evaluate the line integral of
A(x, y) = xêx +
y2
êy
b
along the closed ellipse given parametrically by
x = a cos t,
14.4. Show that ∇ × (A × r) = 2A.
y = b sin t.
z = ct
(14.12)
14.4 Problems
405
14.5. Let
A(x, y) = Ax (x, y)êx + Ay (x, y)êy
B(x, y) = Bx (x, y)êx + By (x, y)êy
be vectors in two-dimensions.
(a) Apply the divergence theorem to A using a volume V enclosed by a cylinder whose bottom base is an arbitrary closed curve C in the xy-plane and
whose top base is the same curve in a plane parallel to the xy-plane, and
whose lateral side is parallel to the z-axis. Now conclude that
$
"" #
!
∂Ax
∂Ay
+
(Ax dy − Ay dx) =
dx dy
∂x
∂y
C
R
where R is the region enclosed by C in the xy-plane. This is the divergence
theorem in two dimensions.
(b) Apply Stokes’ theorem to B with C as above and S the region R defined
above. Show that
$
"" #
!
∂By
∂Bx
−
(Bx dx + By dy) =
dx dy
∂x
∂y
C
R
This is the Stokes’ theorem in two dimensions.
(c) Show that in two dimensions the Stokes’ theorem and divergence theorem
are the same.
14.6. Evaluate the line integral of
&
%
&
%
A(x, y) = x2 + 3y êx + y 2 + 2x êy
from the origin to the point (1, 2):
(a) along the straight line joining the two points; and
(b) along the parabola passing through the two points as well as the point
(−1, 2).
(c) Is A conservative?
2
2
14.7. Is the vector field A(x, y) = xex cos y êx − 12 ex sin y êy conservative?
If so, find its potential.
14.8. A vector field is given by
Φ0 ' (
x)
xy * (x+z)/b
A= 2 y 1+
êx + xêy +
êz e
,
b
b
b
where Φ0 and b are constants.
(a) Determine whether or not A is conservative.
(b) Find the potential of A if it is conservative.
14.9. The components of a vector field are given by
Ax = V0 k 3 yzek
2
xy
,
Ay = V0 k 3 xzek
2
xy
+ V0 k sin ky,
(a) Determine whether A is conservative or not.
(b) If it is conservative, find its potential.
Az = V0 kek
2
xy
.
406
Line Integral and Curl
14.10. The Cartesian components of a vector are given by
Ax = 2axekz ,
Ay = 2ayekz ,
Az = ka(x2 + y 2 )ekz ,
where a and k are constants.
(a) Test whether A is conservative or not.
(b) If A is conservative, find its potential.
14.11. Prove Equations (14.10) and (14.11).
14.12. Show that
∇(A · B) = (B · ∇)A + (A · ∇)B + B × (∇ × A) + A × (∇ × B)
and that
A × (∇ × B) = ∇(A · B) − (A · ∇)B
14.13. Verify the vector identity
∇ × (A × B) = (B · ∇)A − (A · ∇)B − B(∇ · A) + A(∇ · B)
14.14. Verify that for constant A and B
∇[A · (B × r)] = A × B