LINE INTEGRALS
LINE INTEGRALS
10.1 Introduction
first for real-valued functions defined and bounded on finite intervals, and then for unbounded functions and infinite intervals. The concept was later extended to vector-valued functions and, in Chapter 7 of Volume II, to matrix functions.
In Volume I we discussed the integral
f(x)
This chapter extends the notion of integral in another direction. The interval [a, is replaced by a curve in n-space described by a vector-valued function a, and the integrand is a vector fieldfdefined and bounded on this curve. The resulting integral is called a line
integral, a curvilinear integral, or a contour integral, and is denoted by f da or by some similar symbol. The dot is used purposely to suggest an inner product of two vectors. The curve is called a path of integration.
Line integrals are of fundamental importance in both pure and applied mathematics. They occur in connection with work, potential energy, heat flow, change in entropy, circulation of a fluid, and other physical situations in which the behavior of a vector or
scalar field is studied along a curve.
10.2 Paths and line integrals
Before defining line integrals we recall the definition of curve given in Volume 1. Let a
be a vector-valued function defined on a finite closed interval [a, As runs through the function values a(t) trace out a set of points in n-space called
of the function. If a is continuous on J the graph is called a curve; more specifically, the curve described
by a.
In our study of curves in Volume I we found that different functions can trace out the same curve in different ways, for example, in different directions or with different velocities. In the study of line integrals we are concerned not only with the set of points on a curve but
with the actual manner in which the curve is traced out, that is, with the function a itself. Such a function will be called a continuous path.
which is continuous on J is called a continuous path in n-space. The path is called smooth if the derivative a’ exists and is continuous in the open interval (a,
DEFINITION.
Let J = [a, b] be
closed interval in
A function a:
The path is
Line integrals
smooth if the interval [a, b] can be partitioned into number of subintervals in each of which the path is smooth.
Figure 10.1 shows the graph of a piecewise smooth path. In this example the curve has a tangent line at all but a finite number of its points. These exceptional points subdivide the curve into arcs, along each of which the tangent line turns continuously.
F IGURE 10.1 The graph of a piecewise smooth path in the plane.
Let a be a piecewise smooth path in n-space on an interval [a, b], and let be a vector
DEFINITION OF LINE INTEGRAL.
defined and bounded on the graph of a. The line integral
f along a is denoted by the symbol
da and is defined by the equation
da =
. a’(t)
whenever the integral on the right exists, either as a proper or improper integral. Note: In most examples that occur in practice the dot
bounded on [a,
and continuous except possibly at a finite number of points, in which case the integral exists as a proper integral.
10.3 Other notations for line integrals
If C denotes the graph of a, the line integral
da is also written as
da and is
called the off along C.
If a = a(a) and = a(b) denote the end points of C, the line integral is sometimes written as
da and is called the line integral off from to along a. When the notation
for as
is used it should be kept in mind that the integral depends not only on the end points a and but also on the path a joining them.
When = the path is said to be closed. The symbol is often used to indicate integra- tion along a closed path. When
and a are expressed in terms of their components, say =
the integral on the right of (10.1) becomes a sum of integrals,
In this case the line integral is also written as
da, + +
da,.
325 In the two-dimensional case the path is usually
Other notations for line integrals
by a pair of parametric equations,
y=
and the line integral
dx In the three-dimensional case we use three parametric equations,
da is written as
and write the line integral
f da as
be a two-dimensional vector field given by
y) = Jy +
(0, 0) to (1, 1) along each of the following paths : (a) the line with parametric equations = , = , 0
for all (x, y) withy
0. Calculate the line integral
(b) the path with parametric equations =
0 1. Solution. For the path in part (a) we take a(t) = ti +
Then a’(t) = i + and +
and a’(t) is equal to + t and we find
+ t)j. Therefore the dot product of
da =
(Ji + + t) dt =
For the path in part (b) we take a(t) =
Then a’(t) = 2ti + and +
Therefore
a’(t) =
This example shows that the integral from one point to another may depend on the path joining the two points.
Now let us carry out the calculation for part (b) once more, using the same curve but with a different parametric representation. The same curve can be described by the function
where 0
Line integrals
This leads to the relation .
+ the integral of which from 0 to 1 is
(i
as before. This calculation illustrates that the value of the integral is independent of the parametric representation used to describe the curve. This is a general property of line integrals which is proved in the next section.
10.4 Basic properties of line integrals
Since line integrals are defined in terms of ordinary integrals, it is not surprising to find that they share many of the properties of ordinary integrals. For example, they have a linearity property with respect to the integrand,
and an additiveproperty with respect to the path of integration:
where the two curves
and
make up the curve C. That is, C is described by a function
are those traced out by a(t) as varies over subintervals [a, c] and. [c, b], respectively, for some c satisfying a < c < b. The proofs of these properties follow immediately from the definition of the line integral; they are left as exercises for the reader.
a defined on an interval [a, b], and the curves
and
Next we examine the behavior of line integrals under a change of parameter. Let a be a continuous path defined on an interval [a, b], let be a real-valued function that is differen- tiable, with
never zero on an interval [c, d], and such that the range of is [a, b]. Then the function defined on [c, d] by the equation
is a continuous path having the same graph as a. Two paths a and so related are called equivalent. They are said to provide different parametric representations of the same curve. The function is said to define a change of parameter.
the derivative of is always positive on [c, d] the function is increasing and we say that the two paths a and trace out C in the same direction. If the derivative of is always negative we say that a and trace out C in opposite directions. In the first case the function is said to be orientation-preserving;
Let C denote the common graph of two equivalent paths a and
in the second case is said to be orientation-reversing. An example is shown in Figure 10.2.
The next theorem shows that a line integral remains unchanged under a change of param- eter that preserves orientation; it reverses its sign if the change of parameter reverses orientation. We assume both intergals
da and
exist.
Basic properties of fine integrals
F IGURE 10.2 A change of parameter defined by = In (a), the function h preserves orientation. In (b), the function h reverses the orientation.
T H EO REM BEH A V IO R O F A LIN E IN T EG RA L U N D ER A C H A N G E O F PA RA M ET ER.
Let
a and be equivalent piecewise smooth paths. Then
if a and trace out C in the same direction; and
a and trace out C in opposite directions. P r o o f , It suffices to prove the theorem for smooth paths; then we invoke
the additive property with respect to the path of integration to deduce the result for wise smooth paths.
The proof is a simple application of the chain rule. The paths a and are related by an equation of the form
where is defined on an interval [c, d] and a is defined on an interval [a, b]. From the chain rule we have
Therefore we find
In the last integral we introduce the substitution = u(t), dv = u’(t) dt to obtain
. a’(v)
. a’(v)
Line integrals
where the + sign is used if a =
and b = u(d), and the
sign is used if a = u(d) and
b = u(c). The first case occurs if a and trace out C in the same direction, the second if they trace out C in opposite directions.
10.5 Exercises
In each of Exercises 1 through 8 calculate the line integral of the vector the path described.
1. = 2xy)i +
from
1) to (1, 1) along the parabola y =
xj, along the path described by
along the path described by a(t) = ti + +
4. y) = +
y )j, from to (2,0) along the curve y = 1 .
= in a counterclock- wise direction.
5. = (x + +
once around the ellipse
6. z) = 2xyi +
+ z)j + yk, from (1,
to
1) along a line segment.
along a line segment.
8. z) = xi +
along the path described by a(t) =
In each of Exercises 9 through 12, compute the value of the given line integral.
9. dx + (y ”
dy , where C is a path from
to (1, 1) along the parabola
y (x
(x
traversed once in a +
10. , where C is the circle +
clockwise direction. dx + dy where C is the square with vertices (1, 0), (0,
1 , 0), and (0, traversed
once in a counterclockwise direction.
12. + zdy + xdz, where (a) C is the curve of intersection of the two surfaces x + y = 2 and +
+ = 2(x + y) . The curve is to be traversed once in a direction that appears clockwise when viewed from the origin.
= traversed once in a direction that appears counterclockwise when viewed from high above the xy-plane.
(b) C is the intersection of the two surfaces z = xy and
10.6 The concept of work as a line intregal
Consider a particle which moves along a curve under the action of a force If the
curve is the graph of a piecewise smooth path a, the w ork done by f is defined to be the line
integral
f da. The following examples illustrate some fundamental properties of work. EXAMPLE 1. W ork done by a constantforce. If f is a constant force, say f= it can be
shown that the work done by f in moving a particle from a point to a point b along any
piecewise smooth path joining a and b is
(b
a), the dot product of the force and the
displacement vector b a. We shall prove this in a special case.
We let a = ...,
be a path joining a and
say a(a) = a and a(b) = b, and we
Line integrals with respect to arc length
write = (c,, . . . , c,). Assume a’ is continuous on [a, b]. Then the work done by f is
equal to
da =
dt =
a,(a)] = c . [a(b) a(a)] c .
For this force field the work depends only on the end points a and b and not on the curve joining them. Not all force fields have this property. Those which do are called conservative. The example on p. 325 is a nonconservative force field. In a later section we shall determine all conservative force fields.
EXAMPLE 2. The principle of work and energy. A particle of mass moves along a curve
under the action of a force field f. If the speed of the particle at time is v(t), its kinetic
energy is defined to be Prove that the change in kinetic energy in any time interval
is equal to the work done by f during this time interval.
Solution. Let r(t) denote the position of the particle at time t. The work done by f
during a time interval [a, b] is
f dr . We wish to prove that
dr =
From Newton’s second law of motion we have
= mu’(t),
where denotes the velocity vector at time The speed is the length of the velocity vector, v(t) =
. Therefore
(u(t) u(t)) = Integrating from a to b we obtain
f [r(t)] r’(t) = f [r(t)] . u(t) = mu’(t) u(t) =
f dr =
f [r(t)] r’(t) dt =
as required.
10.7 Line integrals with respect to arc length
Let a be a path with a’ continuous on an interval [a, b]. The graph of a is a rectifiable curve. In Volume I we proved that the corresponding arc-length function is given by the integral
The derivative of arc length is given by
Line integrals
Let be a scalar field defined and bounded on C, the graph of a. The line integral of with respect to arc length along C is denoted by the symbol
and is defined by the equation
ds =
whenever the integral on the right exists. Now consider a scalar field given by
the dot product of a vector fieldfdefined on C and the unit tangent vector T(t) =
. In this case the line integral
is the same as the line integral
da because
. a’(t) =
When f denotes a velocity field, the dot product T is the tangential component of
integral off along C. When C is a closed curve the flow integral is called the circulation off along C. These terms are commonly used in the theory of fluid flow.
velocity, and the line integral
f T ds is called
10.8 Further applications of line integrals
Line integrals with respect to arc length also occur in problems with mass distribution along a curve. For example, think of a curve C in 3-space as a wire made of a thin material of varying density. Assume the density is described by a scalar field
where is the
z) of C. Then the total mass of the wire is defined to be the line integral of with respect to arc length:
unit length at the point (x,
The center of mass of the wire is defined to be the point whose coordinates are determined by the equations
A wire of constant density is called uniform. In this case the center of mass is also called the centroid.
EXAMPLE 1. Compute the mass of one coil of a spring having the shape of the helix whose vector equation is
if the density at (x, z) is
Solution. The integral for A4 is
ds =
(a”
t+
dt .
331 Since s’(t) =
Exercises
we have s’(t) and hence
and a’(f) = -a sin
i + a cos tj
(a’ +
In this example the z-coordinate of the center of mass is given by
+ z”)
The coordinates and are requested in Exercise 15 of Section 10.9.
Line integrals can be used to define the moment of inertia of a wire with respect to an axis. If
z) represents the perpendicular distance from a point (x, z) of C to an axis L, the moment of inertia
is defined to be the line integral
where z) is the density at (x, z). The moments of inertia about the coordinate
axes are denoted by I,, I,, and I,.
EXAMPLE 2. Compute the moment of inertia of the spring coil in Example 1. Solution. Here
+ + so we have
ds =
+ z”) ds =
where is the mass, as computed in Example 1.
10.9 Exercises
. Compute the work done by this force in moving a particle from (0,
1. A force field f in
is given by
z) = xi + + (xz
along the line segment joining these two points.
to (1,
in moving a particle (in a counterclockwise direction) once around the square bounded by the coordinate axes and thelinesx =aandy
2. Find the amount of work done by the
>O.
3. A two-dimensional force fieldfis given by the y) = cxyi + where c is a positive constant. This force acts on a particle which must move from (0, 0) to the line x = 1 along a curve of the form
where a> 0
and
b >O .
Find a value of a (in terms of c) such that the work done by this force is independent of b.
4. A force
is given by the
z) = yzi +
+ + Calculate
the work done by fin moving a particle once around the triangle with vertices (0, 0, 0), (1, 1,
(-1, 1, -1) in that order.
Line integrals
+ (x along the curve of intersection of the sphere
5. Calculate the work done by the force
+ (z
+ z = 4 and the plane z = tan where 0< <
The path is transversed in a direction that appears counterclockwise when viewed from high above the xy-plane. 6. Calculate the work done by the force field
+ along the curve of intersection of the sphere
z) =
= ax, where z 0 and a > 0. The path is traversed in a direction that appears clockwise when viewed from high above the xy-plane.
and the cylinder
Calculate the line integral with respect to arc length in each of Exercises 7 through IO. 7. (x +
traversed in a counterclockwise direction. 8. where C has the vector equation
where C is the triangle with vertices (0, 0), (1, 0), and (0,
a(t) =
sin
cos
9. + where C has the vector equation a(t) =
t + t sin
t cos
10. z ds, where C has the vector equation
a(t) = tcosti + tsintj + tk,
11. Consider a uniform semicircular wire of radius (a) Show that the centroid lies on the axis of symmetry at a distance
from the center. (b) Show that the moment of inertia about the diameter through the end points of the wire
is where is the mass of the wire. 12. A wire has the shape of the circle
Determine its mass and moment of inertia about a diameter if the density at (x,
is
13. Find the mass of a wire whose shape is that of the curve of intersection of the sphere + + = 1 and the plane x + + z = 0 if the density of the wire at (x,
is 14. A uniform wire has the shape of that portion of the curve of intersection of the two surfaces
Find the z-coordinate of its centroid. 15. Determine the coordinates
= x connecting the points (0,
and (1, 1,
of the center of mass of the spring coil described in Example 1 of Section 10.8. 16. For the spring coil described in Example 1 of Section 10.8, compute the moments of inertia and .
10.10 Open connected sets. Independence of the path
Let S be an open set in R”. The set S is called connected if every pair of points in can be joined by a piecewise smooth path whose graph lies in S. That is, for every pair of points a and in S there is a piecewise smooth path a defined on an interval [a,
such that a(t) S for each in [a, b], with a(a)
a and a(b) =
Three examples of open connected sets in the plane are shown in Figure 10.3. Examples in 3-space analogous to these would be (a) a solid ellipsoid, (b) a solid polyhedron, and (c)
a solid torus; in each case only the interior points are considered.
333 An open set is said to be disconnected if is the union of two or more disjoint
The second fundamental theorem of calculus for line integrals
empty open sets. An example is shown in Figure 10.4. It can be shown that the class of open connected sets is identical with the class of open sets that are not disconnected.?
be a vector field that is continuous on an open connected set S. Choose two points and in S and consider the line integral off from to along some piecewise smooth path in S. The value of the integral depends, in general, on the path joining a to
Now let
For some vector fields, the integral depends only on the end points a and and not on the path which joins them. In this case we say the integral is independent of
from to We say the line integral off is independent of
in if it is independent of the path from
a to b for every pair of points a and b in S.
SS
F IGURE 10.3 Examples of open connected sets. F IGURE 10.4 A disconnected set S, the union of two disjoint circular disks.
Which have line integrals independent of the path? To answer this question we extend the first and second fundamental theorems of calculus to line integrals.
10.11 The second fundamental theorem of calculus for line integrals
The second fundamental theorem for real functions, as proved in Volume I (Theorem states that
provided that is continuous on some open interval containing both a and b. To extend this result to line integrals we need a slightly stronger version of the theorem in which continuity of
is assumed only in the open interval (a, b).
THEOREM 10.2. Let be a realfunction that is continuous on a closed interval [a, b] and
assume that the integral y’(t) dt exists. If is continuous on the open interval (a, b), we
have
Proof. For each in [a, b] define f(x) = dt . We wish to prove that (10.2)
For a further discussion of connectedness, see Chapter 8 of the author’s Analysis,
Line integrals
By Theorem 3.4 of Volume continuous on the closed interval [a, b]. By Theorem 5.1 of Volume
differentiable on the open interval (a, b), = for each x in (a, b). Therefore, by the zero-derivative theorem (Theorem 5.2 of Volume I), the difference
is constant on the open interval (a, b). By continuity, f is also
constant on the closed interval [a, b]. In particular, f(b)
But = 0, this proves (10.2).
= f(a)
THEOREM
10.3. SECOND FUNDAMENTAL THEOREM OF CALCULUS FOR LINE INTEGRALS.
Let be a
with a continuous gradient
on an open connected set
S in R". Then for any two points a and b joined by a piecewise smooth path a in S we have
da =
Choose any two points a and b in S and join them by a piecewise smooth path a in S defined on an interval [a, b]. Assume first that a is smooth on [a, b]. Then the line integral of
Proof.
from a to b along a is given by
da =
a’(t) dt
By the chain rule we have
where g is the composite function defined on [a, b] by the formula
The derivative is continuous on the open interval (a, b) because is continuous on S and a is smooth. Therefore we can apply Theorem 10.2 to to obtain
This proves the theorem if a is smooth.
When a is piecewise smooth we partition the interval [a, b] into a finite number (say r)
of subintervals , in each of which a is smooth, and we apply the result just proved to each subinterval. This gives us
as required.
As a consequence of Theorem 10.3 we see that the line integral of a gradient is inde- pendent of the path in any open connected set S in which the gradient is continuous. For a closed path we have b = a, so
= 0. In other words, the line integral of a
335 continuous gradient is zero around every piecewise smooth closed path in S. In Section 10.14
Applications to mechanics
we shall prove (in Theorem 10.4) that gradients are the continuous vector fields with this property.
10.12 Applications to mechanics
If a vector field is the gradient of a scalar field then is called a potential function In 3-space, the level sets of are called equipotential surfaces; in 2-space they are called equipotential lines. (If
denotes temperature, the word “equipotential” is replaced by “isothermal”; if denotes pressure the word “isobaric” is used.)
+ For every integer we have
EXAMPLE 1. In
where r = xi + + (See Exercise of Section 8.14.) Therefore is a potential of the vector field
z) =
The equipotential surfaces of are concentric spheres centered at the origin. EXAMPLE 2. The Newtonian potential. Newton’s gravitation law states that the force
which a particle of mass exerts on another particle of mass m is a vector of length where G is a constant and is the distance between the two particles. Place the origin at the particle of mass M, and let r = xi +
be the position vector from the origin to the particle of mass m. Then =
and --Y/r is a unit vector with the same direction as
f, so Newton’s law becomes
Taking =
1 in Example 1 we see that the gravitational force is the gradient of the scalar field given by
z) =
This is called the Newtonian potential.
The work done by the gravitational force in moving the particle of mass m from
where = + +
If the two points lie on the same equipotential surface then
and =
and no work is done.
EXAMPLE 3. Theprinciple of conservation of mechanical energy. Let f be a continuous force
field having a potential in an open connected set S. Theorem 10.3 tells us that the work done by fin moving a particle from to along any piecewise smooth path in S is
Line integrals
the change in the potential function. In Example 2 of Section 10.6 we proved that this work is also equal to the change in kinetic energy of the particle, k(x)
k(a) where k(x) denotes the kinetic energy of the particle when it is located at x. Thus, we have
or (10.3)
The scalar is called the potential
of the particle.
If a is kept fixed and is allowed to vary over the set Equation (10.3) tells us that the sum of k(x) and
is a gradient, the sum of the kinetic andpotential energies of aparticle moving in
is constant. In other words, if a
is constant. In mechanics this is called the principle of conservation of (mechanical) energy. A force field with a potential function is said to be conservative because the total energy, kinetic plus potential, is conserved. In a conservative field, no net work is done in moving a particle around a closed curve back to its starting point. A force field will not be conservative if friction or viscosity exists in the system, since these tend to convert mechanical energy into heat energy.
10.13 Exercises
Determine which of the following open sets
in
are connected. For each connected set,
choose two arbitrary distinct points in and explain
curve in S connecting the two points.
2. Given a two-dimensional vector field
where the partial derivatives
are continuous on an open set S. If is the gradient of some potential
and
prove that
at each point of S.
a gradient. Then find a closed path C such that =
3. For each of the following vector fields, use the result of Exercise 2 to prove that is
xi. =
4. Given a three-dimensional vector field
Some authors refer to
as the potential function
that the potential energy at x will be equal to the that the potential energy at x will be equal to the
line integrals
where the partial derivatives
are continuous on an open set S. Iffis the gradient of some potential function prove that
at each point of
5. For each of the following vector fields, use the result of Exercise 4 to prove that is not a gradient. Then find a closed path C such that
+xj +xk.
(b) z) = +
6. A force field is defined in
by the equation
(a) Determine whether or not fis conservative. (b) Calculate the work done in moving a particle along the curve described by
a(t) = cos ti + sin
as t runs from 0 to
7. A two-dimensional force field F is described by the equation
(a) Show that the work done by this force in moving a particle along a curve
(b) Find the amount of work done = 1, f(b) = 2, g(a) = 3, g(b) = 4.
8. A force field is given in polar coordinates by the equation
= -4 sin
4 sin
Compute the work done in moving a particle from the point to the origin along the spiral whose polar equation is r =
9. A radial or “central” force field F in the plane can be written in the form
y)
where r = xi + and r = . Show that such a force field is conservative.
+ 2)i + 16xj. in moving a particle from 1, to (1, 0) along the upper half of the ellipse
10. Find the work done by force
Which ellipse (that is, which value of b) makes the work a minimum?
10.14 The first fundamental theorem of calculus for line integrals
Section 10.11 extended the second fundamental theorem of calculus to line integrals. This section extends the first fundamental theorem. We recall that the first fundamental theorem states that every indefinite integral of a continuous function
has a derivative
Line integrals
equal to f. That is, if
then at the points of continuity off we have
=f
To extend this theorem to line integrals we begin with a vector continuous on an open connected set S, and integrate it along a piecewise smooth curve C from a fixed point a in S to an arbitrary point x. Then we let denote the scalar field defined by the line integral
where a describes C. Since S is connected, each point in S can be reached by such a curve. For this definition of
to be unambiguous, we need to know that the integral depends only on and not on the particular path used to join a to x. Therefore, it is natural
to require the line integral off to be independent of the path in S. Under these conditions, the extension of the first fundamental theorem takes the following form:
THEOREM
10.4. FIRST FUNDAMENTAL THEOREM FOR LINE INTEGRALS. Let f be a vector that is continuous on an open connected set in
and assume that the line integral off is independent of the path in S. Let a be
a on S by the equation
of S and
where a is any piecewise smooth path in S joining a to x. Then the gradient of exists and is equal to f; that is,
= f(x)
for every in
Proof, We shall prove that the partial derivative exists and is equal to the kth component of f(x), for each k = 2, . . . , and each in S. Let
r) be an n-ball with center at and radius r lying in S. If is a unit vector, the point
also lies in S for every real h satisfying 0 < < r, and we can form the difference quotient
Because of the additive property of line integrals, the numerator of this quotient can be written as
da,
and the path joining to + can be any piecewise smooth path lying in S. In particular,
Necessary and
conditions for a
to be a gradient
we can use the line segment described by
a(t) = + thy,
where 0
Since a’(t) = , the difference quotient becomes
Now we take y = the kth unit coordinate vector, and note that the integrand becomes + thy) . y =
+ the,). Then we make the change variable = ht , = h dt , and we write (10.4) in the form
du
where g is the function defined on the open interval (-r, r) by the equation
the first fundamental theorem for ordinary integrals tells us that g’(t) exists for each in (-r, r) and that
Since each component
is continuous on
In particular, g’(0)
Therefore, if we let h
0 in (10.5) we find that
g’(0) =
This proves that the partial derivative
exists and
as asserted.
10.15 Necessary and sufficient conditions for a vector field to be a gradient
The first and second fundamental theorems for line integrals together tell us that a necessary and sufficient condition for a continuous vector field to be a gradient on an open connected set is for its line integral between any two points to be independent of the path.
We shall prove now that this condition is equivalent to the that the line integral is zero around every piecewise smooth closed path. All these conditions are summarized in the following theorem.
T H EO REM N EC ES S A RY A N D S U FFIC IEN T C O N D IT IO N S FO R A V EC T O R FIELD T O BE A GRAD I ENT .
Let f be a continuous on an open connected set S in Then the three statements are equivalent. (a) is the gradient of somepotentialfinction in S. (b) The line integral off is independent of the path in S. (c) The line integral off is zero around every piecewise smooth closedpath in S.
Line integrals
We shall prove that (b) implies (a), (a) implies (c), and (c) implies (b). State- ment (b) implies (a) because of the first fundamental theorem. The second fundamental theorem shows that (a) implies (c).
To complete the proof we show that (c) implies (b). Assume (c) holds and let and
be any two piecewise smooth curves in with the same end points. Let
be the graph of a function a defined on an interval [a,
and let
be the graph of a function defined on
Define a new function y as follows:
if =
if
Then y describes a closed curve C such that
Since
dy = because of (c), we have
f da =
f so the integral off is
independent of the path. This proves (b). Thus, (a), (b), and (c) are equivalent.
closed curve C, then f is not a gradient. However, if a line integral
Note: If f 0 for a
f is zero for a particular closed curve C or even for infinitely many closed curves, it does not necessarily follow that f is a gradient. For example, the reader
can easily verify that the line integral of the vector field f(x,
= xi +
is zero for every
circle C with center at the origin. Nevertheless, this particular vector field is not a gradient.
10.16 Necessary conditions for a vector field to be a gradient
The first fundamental theorem can be used to determine whether or not a given vector field is a gradient on an open connected set S. If the line integral off is independent of the path in
we simply define a scalar field by integrating f from some fixed point to an arbitrary point in along a convenient path in S. Then we compute the partial derivatives of and compare
= for every in and every k, then f is a gradient on and is a potential. If
the kth component off. If
for some k
and some x, then f is not a gradient on S. The next theorem gives another test for determining when a vector field f is not a gradient.
This test is especially useful in practice because it does not require any integration.
THEOREM 10.6. NECESSARY CONDITIONS FOR A VECTOR FIELD TO BE A GRADIENT. Let .,
vector field on an open set in R”. is a gradient on then the partial derivatives of the components off are related by the equations
be a continuously
for j = . . . , n and every in
Necessary conditions for a vector field to be a gradient
Proof.
a gradient, then for some potential function This means that
foreachj= Differentiating both members of this equation with respect to we find
Similarly, we have
Since the partial derivatives
are continuous on S, the two mixed partial derivatives
and
and must be equal on S. This proves (10.6).
EXAMPLE
Determine whether or not the vector field
is a gradient on any open subset of Solution. Here we have
The partial derivatives
and
are given by
y) =
this vector field is not a gradient on any open subset of
except when x = 0 or y =
The next example shows that the conditions of Theorem 10.6 are not always sufficient for a vector field to be a gradient.
be the vector field defined on S by the equation
EXAMPLE 2. Let S be the set of all (x, y)
in
and let
Show that
everywhere on S but that, nevertheless, f is not a gradient on S.
y) for all y) in S. (See Exercise 17 in Section 10.18.)
Solution.
The reader can easily verify that
y) =
To prove thatfis not a gradient on S we compute the line integral offaround the unit circle given by
Line integrals
We obtain
da =
a'(t)
Since the line integral around this closed path is not zero, f is not a gradient on S. Further
properties of this vector field are discussed in Exercise 18 of Section 10.18.
At the end of this chapter we shall prove that the necessary conditions of Theorem 10.6 are also sufficient if they are satisfied on an open
set. (See Theorem 10.9.)
10.17 Special methods for constructing potential functions
The first fundamental theorem for line integrals also gives us a method for constructing
potential functions. If f is a continuous gradient on an open connected set S, the line
integral off is independent of the path in S. Therefore we can find a potential simply by
using any piecewise smooth path lying in S. The scalar field so obtained depends on the choice of the initial point a. If we start from another initial point, say
integrating f from some fixed point a to an arbitrary point
in
we obtain a new potential But, because of the additive property of line integrals,
can differ only by a constant, this constant being the integral off from a to
and
The following examples illustrate the use of different choices of the path of integration.
F IGURE 10.5 Two polygonal paths from (a, to (x,
EXAMPLE 1. Construction of a potential on an open rectangle. If f is a continuous gradient
can be constructed by integrating from a fixed point to an arbitrary point along a set of line segments parallel to the coordinate axes.
on an open rectangle in
a potential
A two-dimensional example is shown in Figure 10.5. We can integrate first from (a, b) to (x,
along a vertical segment. Along the horizontal segment we use the parametric representation
along a horizontal segment, then from (x,
to (x,
343 and along the vertical segment we use the representation
Special methods for constructing potential functions
a(t) = xi + tj,
y) + the resulting formula for a potential y) is
We could also integrate first from (a, b) to (a, y) along a vertical segment and then from (a, y) to (x, y) along a horizontal segment as indicated by the dotted lines in Figure 10.5. This gives us another formula for
y),
Both formulas (10.7) and (10.8) give the same value y) because the line integral of a gradient is independent of the path.
EXAMPLE 2. Construction of a potentialfunction by the use of integrals. The use of indefinite integrals often helps to simplify the calculation of potential functions. For example, suppose a three-dimensional vector field
is the gradient of a potential function on an open set in
Then we
everywhere on S. Using indefinite integrals and integrating the first of these equations with respect to x (holding y and z constant) we find
where z) is a “constant of integration” to be determined. Similarly, if we integrate the equation
with respect to y and the equation
with respect to z we
obtain the further relations
and
=j
where and y) are functions to be determined. Finding means finding three functions
z), B(x, z), and y) such that all three equations for y, z) agree in their right-hand members. In many cases this can be done by inspection, as illustrated by
Line integrals
EXAMPLE 3. Find a potential function for the vector field defined on by the equation =
4xy)j +
+ 2)k.
Solution. Without knowing in advance whether or not has a potential function we try to construct a potential as outlined in Example 2, assuming that a potential
exists. Integrating the
with respect to x we find
Integrating with respect to
and
with respect to
we find
By inspection we see that the choices
z) = +x and y)
z) =
x will make all three equations agree; hence the function given by the equation
is a potential for f on
EXAMPLE 4. Construction of a potential on a convex set.
A set S in
is called convex
if every pair of points in S can be joined by a line segment, all of whose points lie in S. An example is shown in Figure 10.6. Every open convex set is connected.
Iff is a continuous gradient on an open convex set, then a potential can be constructed by integratingf from a fixed point in
along the line segment joining
to an arbitrary point
to The line segment can be parametrized by the function
1. This gives us a’(t) =
a(t)
where 0
a, so the corresponding potential is given by the integral
F IGURE 10.6 In a convex set S, the segment joining a and is in for all points a and
345 If S contains the origin we can take a = 0 and write (10.9) more simply as
Exercises
10.18 Exercises In each of Exercises 1 through 12, a vector fieldfis defined by the formulas given. In each case
determine whether or not is the gradient of a scalar field. When is a gradient, find a correspond- ing potential function
1. = xi
2. y) = +
3. f(x, y) =
5. f(x, y) = [sin (xy) + xy cos
13. A fluid flows in the xy-plane, each particle moving directly away from the origin. If a particle is at a distance r from the origin its speed is
where a and are constants. (a) Determine those values of a and for which the velocity vector field is the gradient of some scalar field.
(b) Find a potential function of the velocity whenever the velocity is a gradient. The case = -1 should be treated separately.
14. If both and are potential functions for a continuous vector field f on an open connected set S in
prove that is constant on S. 15. Let S be the set of all x 0 in
Let r =
, and let f be the vector field defined on S by
the equation
f(x) =
where
is a real constant. Find a potential function for f on S. The case = -2 should be treated separately. 16. Solve Exercise 15 for the vector field defined by the equation
where is a real function with a continuous derivative everywhere on
The following exercises deal with the vector field f defined on the set S of all points (x, y)
in by the equation
Line integrals
In Example 2 of Section 10.16 we showed that f is not a gradient on S, even though
everywhere on S.
17. Verify that for every point (x,
in S we have
18. This exercise shows that f a gradient on the set
consisting of all points-in the xy-plane except those on the nonpositive x-axis. (a) If (x, y)
T, express and in polar coordinates,
sin
wherer > Oand < Prove that is given by the formulas
Y if > 0, if
if < 0.
t is the unique real number which satisfies the two conditions tan = t and
[Recall the definition of the arc tangent function: For any real t,
< < (b) Deduce that
ae Y
for all (x, y) in T. This proves that is a potential function for f on the set T.
This exercise illustrates that the existence of a potential function depends not only on the vector field f but also on the set in which the vector field is defined.
10.19 Applications to exact differential equations of first order
Some differential equations of the first order can be solved with the aid of potential functions. Suppose we have a first-order differential equation of the form
If we multiply both sides by a nonvanishing factor y) we transform this equation to the form
y) for y) and use the Leibniz notation for derivatives, writing
y) = 0. If we then write
for y’, the differential equation takes the form
Applications to exact
equations
order 347
We assume that P and Q are continuous on some open connected set S in the plane. With each such differential equation we can associate a vector field V, where
The components and Q are the coefficients of dx and dy in Equation (10.11). The differ- ential equation in (10.11) is said to be exact in S if the vector field V is the gradient of a potential; that is, if
where is some scalar field. When such a exists we have
y) =
y) for each point (x, y) in
= Q, and the differential equation in (10.11) becomes
= P and
We shall prove now that each solution of this equation satisfies the relation y)
C, where C is a constant. More precisely, assume there is a solution Y of the
differential equation (10.11) defined on an open interval (a,
such that the point (x, Y(x)) is in for each x in (a, b). We shall prove that
for some constant C. For this purpose we introduce the composite function defined on
(a,
by the equation
By the chain rule, the derivative of g is given by
y)+ = 0, so g’(x) = 0 for each x in (a,
where y = Y(x) and y’ = Y’(x). If y satisfies
then
and, therefore, is constant on (a, b). This proves that every solution y satisfies the equation
y) = C.
Now we may turn this argument around to find a solution of the differential equation. Suppose the equation
defines y as a differentiable function of x, say y Y(x) for x in an interval (a, b), and let g(x) =
Y(x)]. Equation (10.13) implies that is constant on (a, b). Hence, by y) +
= 0, so y is a solution. Therefore, we have proved the following theorem :
THEOREM 10.7. Assume that the
equation
Line integrals
is exact in an open connected set and let be a satisfying
and
-Q
everywhere in S. Then every solution y = Y(x) of (10.14) whose graph lies in
the equation
Y(x)] = C for some constant C. Conversely, the equation
y implicitly as a differentiable function of x, then this function is a solution of the
equation (10.14). The foregoing theorem provides a straightforward method for solving exact differential
equations of the first order. We simply construct a potential function and then write the equation.
y) = C, where C is a constant. Whenever this equation defines y implicitly as a function of the corresponding y satisfies (10.14). Therefore we can use Equation (10.13) as a representation of a one-parameter family of integral curves. Of course, the only admissible values of C are those for which
= C for some , in S.
EXAMPLE 1. Consider the differential equation
Clearing the fractions we may write the equation as
dx +
This is now a special case of (10.14) with
and y) = + Integrating P with respect to and Q with respect toy and comparing results, we find that
y) = +
a potential function is given by the formula
y)=
By Theorem 10.7, each solution of the differential equation satisfies
for some C. This provides an implicit representation of a family of integral curves. In this particular case the equation is quadratic in and can be solved to give an explicit formula for y in terms of and C.
EXAMPLE 2. Consider the first-order differential equation (10.15)
Exercises
Here y) = y and
= 2, this differential equation is not exact. However, if we multiply both sides by
y) = 2x. Since
= a.nd
we obtain an equation that is exact:
+ 2xy dy = 0.
and every solution of (10.16) satisfies the relation
A potential of the vector field
is
y) =
= C. This relation also represents a family of integral curves for Equation (10.15).
The multiplier y which converted (10.15) into an exact equation is called an integrating
fact o r. In general, if multiplication of a first-order equation by a factor y) results in an exact equation, the multiplier
y) is called an integrating factor of the original equation. A differential equation may have more than one integrating factor. For example,
y) = is another integrating factor of (10.15). Some special differ- ential equations for which integrating factors can easily be found are discussed in the following set of exercises.
10.20 Exercises
Show that the differential equations in Exercises through are exact, and in each case find a one-parameter family of integral curves.
6. Show that a linear first-order equation, y’ +
has the integrating factor = Use this to solve the equation.
Let y) be an integrating factor of the differential equation
y) dx +
y ) dy = 0.
Show that
Use this equation to deduce the following rules for finding integrating factors: (a) If
is an integrating factor. (b) If
is a function of x alone, say j”(x), then
is a function ofy alone,
then
is an integrating factor.
Use Exercise 7 to find integrating factors for the following equations, and determine a parameter family of integral curves.
dx + dy } is an integrating factor of the differential equation
If show
exp
y) dy = 0. Find such an integrating factor for each of the following equations and obtain a one-parameter family of integral curves. (a)
tany)dx + dy = 0.
Line integrals
10. The following differential equations have an integrating factor in common. Find such an integrating factor and obtain a one-parameter family of integral curves for each equation.
10.21 Potential functions on convex sets In Theorem 10.6 we proved that the conditions
are necessary for a continuously differentiable vector
,... to be a gradient on an open set S in R”. We then showed, by an example, that these conditions are not always sufficient. In this section we prove that the conditions are sufficient whenever the set S is convex. The proof will make use of the following theorem concerning differentiation under the integral sign.
THEOREM 10.8. Let S be a closed interval in
with
interior and let J = [a, b]
be a closed interval in
Write each point in as (x, t), where S and t J,
Let
be the closed interval S x J in
Assume that is a scalarjeld
on
such that the partial derivative is continuous
wherekisoneof
n.
a scalarjeld on S by the equation
Then the partial derivative exists at each interior point of S and is given by the formula
In other words, we have
Note: This theorem is usually described by saying that we can differentiate under the integral sign.
Proof. Choose any in the interior of S. Since int S is open, there is an r > 0 such that + he, int S for all h satisfying 0 <
is the kth unit coordinate vector in R”. For such h we have
< r. Here
+ he,, t)
t)} dt .
351 Applying the mean value theorem to the integrand we have
Potential functions on convex sets
where z lies on the line segment joining and + he,. Hence (10.17) becomes
The last integral has absolute value not exceeding
where the maximum is taken for all z on the segment joining to + he,, and all in [a, b]. Now we invoke the uniform continuity of
on S (Theorem 9.10) to conclude that for every > 0 there is a > 0 such that this maximum is
a), whenever 0<
Therefore
wlhenever 0 < < 6.
h -a This proves that
t) dt, as required. Now we shall use this theorem to give the following necessary and sufficient condition for
exists and equals
a vector field to be a gradient on a convex set.
vector on an open convex set S in
be a continuously
Then is a gradient on S and only we have (10.18)
n.
Proof. We know, from Theorem 10.6, that the conditions are necessary. To prove sufficiency, we assume (10.18) and construct a potential on S. For simplicity, we assume that S contains the origin. Let
be the integral the line segment from 0 to an arbitrary point in S. As shown earlier in Equation (10.10) we have
dt =
t)
Line integrals
where
t) = f(tx)
. There is a closed n-dimensional subinterval T of S with
empty interior such that satisfies the hypotheses of Theorem 10.8 in T x
where = [0,
Therefore the partial derivative exists for each = 2, . . . , n and can be computed by differentiating under the integral sign,
To compute
we differentiate the dot
. and obtain
where in the last step we used the relation (10.18). Therefore we have
Now let =
. By the chain rule we have
so the last formula for
t) becomes
Integrating this from 0 to 1 we find
dt . We integrate by parts in the last integral to obtain
g(t) dt. Therefore (10.19) becomes
This shows that
= f on S, which completes the proof.