Line integrals

11
Line, surface and volume integrals
In the previous chapter we encountered continuously varying scalar and vector
fields and discussed the action of various differential operators on them. In
addition to these differential operations, the need often arises to consider the
integration of field quantities along lines, over surfaces and throughout volumes.
In general the integrand may be scalar or vector in nature, but the evaluation
of such integrals involves their reduction to one or more scalar integrals, which
are then evaluated. In the case of surface and volume integrals this requires the
evaluation of double and triple integrals (see chapter 6).
11.1 Line integrals
In this section we discuss line or path integrals, in which some quantity related
to the field is integrated between two given points in space, A and B, along a
prescribed curve C that joins them. In general, we may encounter line integrals
of the forms
φ dr,
a · dr,
a × dr,
(11.1)
C
C
C
where φ is a scalar field and a is a vector field. The three integrals themselves are
respectively vector, scalar and vector in nature. As we will see below, in physical
applications line integrals of the second type are by far the most common.
The formal definition of a line integral closely follows that of ordinary integrals
and can be considered as the limit of a sum. We may divide the path C joining
the points A and B into N small line elements ∆rp , p = 1, . . . , N. If (xp , yp , zp ) is
any point on the line element ∆rp then the second type of line integral in (11.1),
for example, is defined as
N
a · dr = lim
a(xp , yp , zp ) · ∆rp ,
C
N→∞
p=1
where it is assumed that all |∆rp | → 0 as N → ∞.
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LINE, SURFACE AND VOLUME INTEGRALS
Each of the line integrals in (11.1) is evaluated over some curve C that may be
either open (A and B being distinct points) or closed (the curve C forms a loop,
so that A and
/ B are coincident). In the case where C is closed, the line integral
is written C to indicate this. The curve may be given either parametrically by
r(u) = x(u)i + y(u)j + z(u)k or by means of simultaneous equations relating x, y, z
for the given path (in Cartesian coordinates). A full discussion of the different
representations of space curves was given in section 10.3.
In general, the value of the line integral depends not only on the end-points
A and B but also on the path C joining them. For a closed curve we must also
specify the direction around the loop in which the integral is taken. It is usually
taken to be such that a person walking around the loop C in this direction
always has the region R on his/her left; this is equivalent to traversing C in the
anticlockwise direction (as viewed from above).
11.1.1 Evaluating line integrals
The method of evaluating a line integral is to reduce it to a set of scalar integrals.
It is usual to work in Cartesian coordinates, in which case dr = dx i + dy j + dz k.
The first type of line integral in (11.1) then becomes simply
φ dr = i φ(x, y, z) dx + j φ(x, y, z) dy + k φ(x, y, z) dz.
C
C
C
C
The three integrals on the RHS are ordinary scalar integrals that can be evaluated
in the usual way once the path of integration C has been specified. Note that in
the above we have used relations of the form
φ i dx = i φ dx,
which is allowable since the Cartesian unit vectors are of constant magnitude
and direction and hence may be taken out of the integral. If we had been using
a different coordinate system, such as spherical polars, then, as we saw in the
previous chapter, the unit basis vectors would not be constant. In that case the
basis vectors could not be factorised out of the integral.
The second and third line integrals in (11.1) can also be reduced to a set of
scalar integrals by writing the vector field a in terms of its Cartesian components
as a = ax i + ay j + az k, where ax , ay , az are each (in general) functions of x, y, z.
The second line integral in (11.1), for example, can then be written as
a · dr = (ax i + ay j + az k) · (dx i + dy j + dz k)
C
C
= (ax dx + ay dy + az dz)
C
=
ax dx +
ay dy +
az dz.
(11.2)
C
C
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C
11.1 LINE INTEGRALS
A similar procedure may be followed for the third type of line integral in (11.1),
which involves a cross product.
Line integrals have properties that are analogous to those of ordinary integrals.
In particular, the following are useful properties (which we illustrate using the
second form of line integral in (11.1) but which are valid for all three types).
(i) Reversing the path of integration changes the sign of the integral. If the
path C along which the line integrals are evaluated has A and B as its
end-points then
A
B
a · dr = −
a · dr.
A
B
This implies that if the path C is a loop then integrating around the loop
in the opposite direction changes the sign of the integral.
(ii) If the path of integration is subdivided into smaller segments then the sum
of the separate line integrals along each segment is equal to the line integral
along the whole path. So, if P is any point on the path of integration that
lies between the path’s end-points A and B then
P
B
B
a · dr =
a · dr +
a · dr.
A
A
P
Evaluate the line integral I = C a · dr, where a = (x + y)i + (y − x)j, along each of the
paths in the xy-plane shown in figure 11.1, namely
(i) the parabola y 2 = x from (1, 1) to (4, 2),
(ii) the curve x = 2u2 + u + 1, y = 1 + u2 from (1, 1) to (4, 2),
(iii) the line y = 1 from (1, 1) to (4, 1), followed by the line x = 4 from (4, 1)
to (4, 2).
Since each of the paths lies entirely in the xy-plane, we have dr = dx i + dy j. We can
therefore write the line integral as
I=
a · dr = [(x + y) dx + (y − x) dy].
(11.3)
C
C
We must now evaluate this line integral along each of the prescribed paths.
Case (i). Along the parabola y 2 = x we have 2y dy = dx. Substituting for x in (11.3)
and using just the limits on y, we obtain
2
(4,2)
[(x + y) dx + (y − x) dy] =
[(y 2 + y)2y + (y − y 2 )] dy = 11 31 .
I=
(1,1)
1
Note that we could just as easily have substituted for y and obtained an integral in x,
which would have given the same result.
Case (ii). The second path is given in terms of a parameter u. We could eliminate u
between the two equations to obtain a relationship between x and y directly and proceed
as above, but it is usually quicker to write the line integral in terms of the parameter u.
Along the curve x = 2u2 + u + 1, y = 1 + u2 we have dx = (4u + 1) du and dy = 2u du.
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LINE, SURFACE AND VOLUME INTEGRALS
y
(4, 2)
(i)
(ii)
(iii)
(1, 1)
x
Figure 11.1 Different possible paths between the points (1, 1) and (4, 2).
Substituting for x and y in (11.3) and writing the correct limits on u, we obtain
(4,2)
[(x + y) dx + (y − x) dy]
I=
(1,1)
1
=
0
[(3u2 + u + 2)(4u + 1) − (u2 + u)2u] du = 10 32 .
Case (iii). For the third path the line integral must be evaluated along the two line
segments separately and the results added together. First, along the line y = 1 we have
dy = 0. Substituting this into (11.3) and using just the limits on x for this segment, we
obtain
(4,1)
4
[(x + y) dx + (y − x) dy] =
(x + 1) dx = 10 21 .
(1,1)
1
Next, along the line x = 4 we have dx = 0. Substituting this into (11.3) and using just the
limits on y for this segment, we obtain
(4,2)
2
[(x + y) dx + (y − x) dy] =
(y − 4) dy = −2 21 .
(4,1)
1
The value of the line integral along the whole path is just the sum of the values of the line
integrals along each segment, and is given by I = 10 21 − 2 12 = 8. When calculating a line integral along some curve C, which is given in terms
of x, y and z, we are sometimes faced with the problem that the curve C is such
that x, y and z are not single-valued functions of one another over the entire
length of the curve. This is a particular problem for closed loops in the xy-plane
(and also for some open curves). In such cases the path may be subdivided into
shorter line segments along which one coordinate is a single-valued function of
the other two. The sum of the line integrals along these segments is then equal
to the line integral along the entire curve C. A better solution, however, is to
represent the curve in a parametric form r(u) that is valid for its entire length.
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11.1 LINE INTEGRALS
Evaluate the line integral I =
x2 + y 2 = a2 , z = 0.
/
C
x dy, where C is the circle in the xy-plane defined by
Adopting the usual convention mentioned above, the circle C is to be traversed in the
anticlockwise direction. Taking the circle as a whole means x is not a single-valued
function of y. We must therefore divide the path intotwo parts with x = + a2 − y 2 for
the semicircle lying to the right of x = 0, and x = − a2 − y 2 for the semicircle lying to
the left of x = 0. The required line integral is then the sum of the integrals along the two
semicircles. Substituting for x, it is given by
0
a
−a I=
− a2 − y 2 dy
x dy =
a2 − y 2 dy +
C
−a
a
a
=4
a2 − y 2 dy = πa2 .
0
Alternatively, we can represent the entire circle parametrically, in terms of the azimuthal
angle φ, so that x = a cos φ and y = a sin φ with φ running from 0 to 2π. The integral can
therefore be evaluated over the whole circle at once. Noting that dy = a cos φ dφ, we can
rewrite the line integral completely in terms of the parameter φ and obtain
0
2π
I=
x dy = a2
cos2 φ dφ = πa2 . C
0
11.1.2 Physical examples of line integrals
There are many physical examples of line integrals, but perhaps the most common
is the expression for the total work done by a force F when it moves its point
of application from a point A to a point B along a given curve C. We allow the
magnitude and direction of F to vary along the curve. Let the force act at a point
r and consider a small displacement dr along the curve; then the small amount
of work done is dW = F · dr, as discussed in subsection 7.6.1 (note that dW can
be either positive or negative). Therefore, the total work done in traversing the
path C is
F · dr.
WC =
C
Naturally, other physical quantities can be expressed in such a way. For example,
the electrostatic potential
energy gained by moving a charge q along a path C in
an electric field E is −q C E · dr. We may also note that Ampère’s law concerning
the magnetic field B associated with a current-carrying wire can be written as
0
B · dr = µ0 I,
C
where I is the current enclosed by a closed path C traversed in a right-handed
sense with respect to the current direction.
Magnetostatics also provides a physical example of the third type of line
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LINE, SURFACE AND VOLUME INTEGRALS
integral in (11.1). If a loop of wire C carrying a current I is placed in a magnetic
field B then the force dF on a small length dr of the wire is given by dF = I dr×B,
and so the total (vector) force on the loop is
0
dr × B.
F=I
C
11.1.3 Line integrals with respect to a scalar
In addition to those listed in (11.1), we can form other types of line integral,
which depend on a particular curve C but for which we integrate with respect
to a scalar du, rather than the vector differential dr. This distinction is somewhat
arbitrary, however, since we can always rewrite line integrals containing the vector
differential dr as a line integral with respect to some scalar parameter. If the path
C along which the integral is taken is described parametrically by r(u) then
dr
du,
du
and the second type of line integral in (11.1), for example, can be written as
dr
du.
a · dr =
a·
du
C
C
dr =
A similar procedure can be followed for the other types of line integral in (11.1).
Commonly occurring special cases of line integrals with respect to a scalar are
φ ds,
a ds,
C
C
where s is the arc length along the curve C. We can always represent C parametrically by r(u), and from section 10.3 we have
ds =
dr dr
·
du.
du du
The line integrals can therefore be expressed entirely in terms of the parameter u
and thence evaluated.
Evaluate the line integral I = C (x − y)2 ds, where C is the semicircle of radius a running
from A = (a, 0) to B = (−a, 0) and for which y ≥ 0.
The semicircular path from A to B can be described in terms of the azimuthal angle φ
(measured from the x-axis) by
r(φ) = a cos φ i + a sin φ j,
where φ runs from 0 to π. Therefore the element of arc length is given, from section 10.3,
by
dr dr
·
dφ = a(cos2 φ + sin2 φ) dφ = a dφ.
ds =
dφ dφ
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