The origin of the notion of line integral (really a path integral) comes from the ...
definition of what a path and line integrals are and see some examples of how to
...
Line Integrals and Vector Fields The origin of the notion of line integral (really a path integral) comes from the physical notion of work. We will see that particular application presently. Our first task is to give a definition of what a path and line integrals are and see some examples of how to compute them.
1
Path Integrals
We start with a definition; later on we will try to justify the definition in terms of appropriate Riemann sums. The problem that we want to treat is that of integrating a real-valued function of two or three variables along a curve in two- or three-dimensional space. We will give our definitions in terms of R3 . We start with a given a function f defined on a domain D ⊂ R3 so that f : D −→ R, and a curve c with parameter domain an interval of the real line, c : [a, b] −→ D, c(t) = (x(t), y(t), z(t))> , a ≤ t ≤ b. Since for all t ∈ [a, b], c(t) ∈ D, we can form the composition f ◦ c so that (f ◦ c)(t) = f (x(t), y(t), z(t)), a ≤ t ≤ b. The path integral of the function f along the curve c is then defined as the integral of the scalar-valued function
Rb
f (x(t), y(t), z(t)) kc0 (t)k dt.
(1)
a
Notice that the factor kc0 (t)k is the speed of traversal of the curve as t runs between endpoints of the interval a and b. To give a simple interpretation in a special case, recall that, if f ≡ 1, then the integral of a function f over a domain D ⊂ R2 , ZZ
ZZ f (x, y) dx dy =
D
1 dx dy , D
1
gives just the area of the domain D. In the present case of the path integral, if f ≡ 1, the path integral is simply Zb
kc0 k dt.
a
which is just the length of the curve! Let us begin by giving a couple of examples. √ Example 1.1 We start with an example in R2 . Let f (x, y) = x y along the segment of the parabola y = x2 lying between the points (0, 0) and (1, 1). Given the curve, we can parameterize it in any way that is convenient provided that the direction along the curve remains the same. This statement requires some proof and we will remark on that later. In the case of the parabola, a particularly easy parameterization is (x, y) = (t, t2 ), 0 ≤ t ≤ 1. √ In this case c(t) = (t, t2 ) and kc0 (t)k = 1 + 4t2 . We can form and compute this path integral as indicated, using the substitution u = 1 + 4t2 :
Z1
Z f (x, y)ds
=
c
√
Z
2t 1 +
4 t2 dt
= 2 1
0
=
5
1 1 u 2 du 8
5 1 2 3 1 √ u 2 = (5 5 − 1). 4 3 6 1
Example 1.2 We look at the integral of the function f (x, y, z) = x2 + y 2 + z 2 along the helix c(t) = (cos (t), sin (t), t)> , 0 ≤ t ≤ 2π. Here, c0 = (− sin (t), cos (t), 1)> . The speed is then given by 0
kc (t)k =
q
(− sin2 (t)) + (cos2 (t) + 1 =
√
2,
while, f (x(t), y(t), z(t)) = cos2 (t) + sin2 (t) + t2 = 1 + t2 , so that
Z2π
Z f (x, y, z) ds c
=
2
1+t 0
√
2 dt =
√
2π t3 2 t+ 3 0
√ Exercise 1.3 Let f (x, y) = 8 y + 2z and c(t) := (1, t2 , e2t )> . Compute the path integral of f along the curve. 2
In our discussion of curves in the previous chapter, we saw that the natural order of numbers on the real line induces a direction of motion along the curve, what we call an orientation of the curve. For example, if we parameterize the line between the point P and Q in R3 by (1 − t)P + tQ, 0 ≤ t ≤ 1 then a point moving along the line, starts at the point P when t = 0, and ends at the point Q when t = 1. On the other hand, if we move from t = 1 to t = 0, then the point moves from Q to P . What happens to the path integral in such a case? We consider the following example: Example 1.4 Let f (x, y, z) = x + y + z, and let c be the straight line segment joining the points P = (0, 0, 0) and Q = (1, 1, 1). Then if c(t) = (1 − t)(0, 0, 0)> + t(1, 1, 1)> , 0 ≤ 1, we can compute Z1 0
1 3 3 3 2 3 t dt = t = (12 − 02 ) = , 2 0 2 2
while if we set τ = −t and parameterize the straight line by (1 − τ )(0, 0, 0)> + τ (1, 1, 1)> then when t = 0, τ = 0, and when t = 1, τ = −1, Z−1 0
−1 3 2 3 3 τ dτ = τ = (02 − (−1)2 ) = − . 2 0 2 2
So in this example, moving in the opposite direction along the curve, that is, reversing the orientation, changes the sign of the integral. As another example, look at the Example (1.1), but now starting at (1, 1) to (0, 0). Example 1.5 In Example (1.1), we claimed that we could use any convenient parameterization of the curve provided that the orientation is not changed. Let us look at three cases: (a) Consider the curve parameterized by (cos (2t), sin (2t))> . For 0 ≤ t ≤ π2 , the vector (cos (2t), sin (2t))> traces out the portion of the unit circle in the first quadrant. As a simple example, we integrate the function f (x, y) = 2xy along this arc. Hence c(t) = (cos (2t), sin (2t))> and so
kc0 (t)k
= k(−2, sin (2t), 2 cos (2t))> k q = 4 sin2 (2t) + 4 cos2 (2t) = 2.
The path integral is then 3
π
Z
Z4
π 4
(2 sin (t) cos (t)) 2 dt =
2
sin (2t)dt
0
0
=
π h i 4 π π 1 2 − cos (2t) = − cos ( ) − 1 2 2 2 0
=
1 r
π describes the same path, 2 that is the first quarter of the unit circle. In this case, the relevant quantities are
(b) The curve given by c(τ ) = (cos (τ 2 ), sin (τ 2 ))> , 0 ≤ τ ≤
kc0 (τ )k = k(−2τ, sin (τ 2 ), 2τ cos (τ 2 ))> k q = 4τ 2 sin2 (τ 2 ) + 4τ 2 cos2 (τ 2 ) = 2τ. Then the path integral is
Z √π 2
0
(2 cos (τ 2 ) sin (τ 2 )) 2τ dτ
√π Z 2 =
2
sin (2τ 2 ) τ dτ
0
√ π 2 π 1 1 1 2 cos (2 ) − = − cos (2 τ ) = − = 1 2 2 2 2 0 Exercise 1.6 In the example immediately above, start with the integral in part (a) and 2 make the change of variable t = τ2 . Carry out the change of variable including the limits of integration and show that you get the same integral as in part(b) above.
(c) Looking at our present example, suppose that we parameterize the quarter circle by π c(t) = (sin (t), cos (t))> . Note that in this case, as t runs from 0 to , the point tracing 2 out the quarter circle runs between the points (0, 1) and (1, 0), that is, in the opposite direction from the direction of traversal in the preceeding two parameterizations. If c is given by this parameterization, then note that c0 (t) = (cos (t), − sin (t))> and so
4
Z0
0
kc (t)k
=
(2 sin (t) cos (t)) 1 dt
1, while π 2
1 π = − cos (0) − cos ( ) = −1 2 2 We should now give a definition of the path integral by introducing the appropriate Riemann sums. All of the properties of the path integral follow easily from the definition in terms of these sums in a way completly analogous to the properties of the usual Riemann integral. To motivate the definition of the path integral we consider sums SN in the same way that is done in defining arc length. For simplicity, let c(t) be a continuously differentiable curve on an interval [a, b]. This interval may be subdivided by introducing a partition a = to < t1 < t2 < . . . < tN = b. This leads to a decomposition of the curve into subcurves, which we denote by ci , each defined on the appropriate subinterval [ti , ti+1 ] for i = 1, 2, . . . , N − 1. Denote the arc length of the subcurve ci by ∆ si . Then Z
ti+1
∆ si =
kc0 (t)kdt.
ti
When N is large, the arc length ∆ si is small and f (x, y, z) is approximately constant for points on ci . We consider the sums
SN =
N −1 X
f (x(t∗i ), y(t∗i ), z(t∗i )) ∆ si ,
i=0
t∗i
where ∈ [ti , ti+1 ]. These sums are basically Riemann sums. We then make the following definition: Definition 1.7 Suppose that a function f : D → R3 and a differentiable curve c with parameter domain [a, b] whose graph lies in D are given and that f ◦ c is continuous on [a, b]. If, as N → ∞ and max(ti+1 , ti ) → 0, limN →∞ SN exists, then this limit is called the path integral and is written Zb lim SN =
0
Z
f (x(t), y(t), z(t))kc (t)k dt =
N →∞
a
f (x, y, z) ds. c
It is often useful to introduce the notion of the average value of a function defined over a curve. If we look at the definition of the path integral in terms of Riemann sums, we know that the partial sum has the form 5
SN :=
N X
f (x(t∗i ), y(t∗i ))(si − si−1 ),
i=1
where we have (si − si−1 ) =
tR i+1
kc0 (t))k dt. Then
ti N X
(si − si−1 ) = L(c)
i=1
and so the quotient N P
N P
f (x(t∗i ), y(t∗i ))(si − si−1 )
i=1 N P
=
f (x(t∗i ), y(t∗i ))(si − si−1 )
i=1
(si − si−1 )
L(c)
,
i=1
is the approximate average obtained by considering f (x(t∗i ), y(t∗i )) to be constant along each of the arcs ∆si . The limit as N → ∞ of the sequence {SN }∞ N =1 is the average value of f along c. We then can state
Average Value of a Function on a Curve Rb
f (x(t), y(t), z(t)) kc0 (t)k dt
a
Rb
kc0 (t)k dt
a
Example 1.8 If we consider the example (1.2), then we might think of the helix as a wire which is heated and interpret the function f (x, y, z) = x2 + y 2 + z 2 as giving the temperature at each point of the helical wire. Then the average temperature on the wire is given by
6
R2π 2 (x + y 2 + z 2 ) kc0 (t)k dt 0
R2π
Rb =
√ (cos2 (t) + sin2 (t) + t2 ) 2 dt
a
R2π √
kc0 (t)k dt
0
2 dt
0
√ R2π (1 + t2 ) 2 dt =
0
R2π √
2 dt
i 2π √ h t3 2 t+ 3 0 √ = 2(2π)
0
=
√ 2 2π 3 + 4π 2 1 3√ = (3 + 4π 2 ). 3 2(2π)
π Exercise 1.9 Define the path c in the xy-plane by c(t) = (30 cos3 (t), 30 sin3 (t))> , − ≤ 2 y π t ≤ . Consider the function f (x, y) = 1 + . Find the average value of the function over 2 3 the given curve.
2
Vector Fields
The next integral we study is the line integral which is an integral which involves vector fields. Before we consider the integrals, we look at the idea of vector fields which we contrast with scalar fields. An example of the former is a region D ⊂ R2 at each point of which the acceleration of a particle moving in D is given. In the latter case, a scalar field is illustrated by giving the temperature (a scalar!) at each point of D. Vector and scalar fields are of great importance in applications in thermodynamics, fluid dynamics, acoustics, and electromagnetics among other applied areas. For us, vector fields are not completely new. We start with a definition, and then turn to some familiar examples. Definition 2.1 Let D ⊂ Rn . A vector field is a mapping F : D −→ Rn . In this definition, the domain of the vector field is considered a subset of Rn while the range of F is considered as a subset of an n-dimensional vector space which we take to be Rn . Usually, this distinction is skipped over. In contrast to a vector field, a scalar field is a function f : D → R. These definitions immediately give us an entire class of vector fields, the so-called gradient fields. 7
Example 2.2 Suppose that we are given a scalar field f : R2 → R defined by f (x, y) = x exy + 3y 2 , Then the gradient of f defines a vector field on all of R2 . Specifically, the vector field is given by exy + x y exy ∇f =
!
x y exy + 6 y
As another simple example consider Example 2.3 Let D be the unit disk in R2 and F : D − {o} → R2 be given by y
F(x, y) = p
x2
+
y2
ˆi − p
x x2
+
y2
ˆj =
y p
x2 + y 2
!
x
, −p
x2 + y 2
.
Remark: Note that the norm of each element in the range of F, is v !2 u u y t p + kF(x, y)k = x2 + y 2
−p
x
!2
x2 + y 2
We can compute the corresponding vectors at each point in D, so, for example, along the y-axis, the vector field is a vector of increasing magnitude as we approach the origin. For the y positive y-axis, 0 ≤ y ≤ 1, F(0, y) = ( p , 0)> = (1, 0)> , while along the negative y-axis, y2 p y y F(0, y) = ( p , 0)> = (−1, 0)> (remember that y 2 = |y| so that = sign(y)!). |y| y2 x Now looking along the x-axis, for x > 0, F(x, 0) = (0, − √ )> = (0, −1)> while for x < x2 > 0, F(x, 0) = (0, 1) . If we look at the line x = y, then y y sign(y) p √ p √ 2 y2 2 2 y2 F(x, y) = = x x = sign(x) −√ √ −√ − √ x2 2 2 x2 2
Exercise 2.4 Sketch this vector field on the unit disk.
8
Example 2.5 A more familiar example is the vector field defined by Newton’s Law of Gravitation. Recall that the magnitude of the gravitational force between two masses, taking the center of mass of the larger body as the origin of coordinates, is given by kF(x, y, z)k = mM G . The direction of the force is in the direction from the point (x, y, z) toward the r2 r or, writing r = krk = center of the larger body of mass M i.e. in the direction − krk p r r x2 + y 2 + z 2 , − = − . Hence, the gravitation vector field is given by krk r F(x, y, z) = −
mM G r. r3
One important thing that should be pointed out, if we define a scalar field by V (x, y, z) = mM G 1 then we can look at the gradient of V . If we differentiate p with respect r x2 + y 2 + z 2 to the variable x, we obtain, ∂ ∂x
!
1
= −
p
x2 + y 2 + z 2
Considering the symmetry of the function
∇V
=
(m M G)
=
(m M G)
∂V ∂x ∂V ∂y ∂V ∂z x r3 y − 3 r z − 3 r −
2x x 1 3 = − 3. 2 (x2 + y 2 + z 2 ) 2 r
1 , we have r
= −(m M G) (x, y, z) = − (m M G) r. r3 r3
If we look at this result, we see that the vector field of the gravitational force is the gradient of a vector field. The scalar field V is called the potential of the field F. Whenever a vector field can be expressed as the gradient of a scalar field, we call that vector field conservative. This use of terminology will be justified in the next example. From this example, we see that it is important to understand just when a vector field is conservative and moreover how to find the scalar potential field.
9
Example 2.6 The total energy E of a particle of mass m, moving in a conservative force field F is defined to be the total of the kinetic and potential energies of the particle. Specifically, if F(r(t)) = −∇ V (r(t)), we have 1 m kr0 (t)k2 + V (r(t)). 2 We have called such a force field conservative. From this expression for the energy, it follows that E(t) =
d E(t) dt
=
1 d d 1 m [r00(t) · r0 (t)] + V (r(t)) = m 2 [r00 (t) · r0 (t)] + ∇ V (r(t)) · r0 (t) 2 dt dt 2
=
(m r00 (t) · r0 (t)) − F(r(t)) · r0 (t) = F(r(t)) · r0 (t) − F(r(t)) · r0 (t) = 0.
dE Since the derivative vanishes, the energy is constant. This is exactly what is meant by dt conservation of energy.
Now line integrals are integrals of the restriction of a vector field, f , to an oriented path. To define them, we go back to the notion of Riemann sum. Begin by dividing the curve c into pieces. Each point of the subdivision is represented by a vector ri , i = 0, 1, . . . , n, and we write ∆ri = ri+1 − ri . Now we form the dot product f (ri ) · ∆ri which compares ∆ri with the value of the vector field at ri . Then we sum over all i to get a Riemann sum n X
f (ri ) · ∆ri .
i=1
Taking limits as n → ∞ and k∆ri k → 0 we have, provided the limit exists, n X
Z f · dr :=
lim
n→∞
f (ri ) · ∆ri .
k∆ri k→0 i=1
c
Example 2.7 Let f = (1, 2)> and let c be the path consisting of the line segment `1 joining (1, 1) to (10, 1) and `2 that joining (10, 1) to (10, 10). To compute the integral, look at the line segments separately. On `1 , ∆ri = (∆x, 0)> and f ·∆ri = (1, 2)> ·(∆x, 0)> = ∆x. Hence Z10
Z f · dr =
1
`1 >
dx = 9 .
>
On `2 , ∆ri = (0, ∆y) and f · ∆ri = (1, 2) · (0, ∆y)> = 2∆y. Hence 10
Z10
Z f · dr =
1
`2
Hence
R
2 dy = 18 .
f · dr = 9 + 18 = 27.
c
A more complicated example is the following. Example 2.8 Let f be as in the previous example, but take for c the parabolic path y = (1/9) (x2 − 2x + 10). We think of the curve as parameterized by x so that c : (t, t2 /9 − 2t/9 + 10/9)> 1 ≤ t ≤ 10. This is a curve with the same endpoints at the curve in the preceeding example. Now, r˙ (t) = (1, (2t − 2)/9)> or, in differential form, dr = (1, (2t − 2)/9)> dt. So Z
Z f ·r =
c(1, 2)> · (1, (2t − 2)/9)> dt
c
Z10 =
4t 5 + 9 9
dt =
200 50 + 9 9
−
243 7 = . 9 9
1
This example shows that the value of the integral may well depend on the path. The line integral of a vector field along a path has a physical interpretation in the case that f is a field due to a force and the curve is the path of a particle. Then the integral has the interpretation of work done by the force along the path c. Example 2.9 Consider a mass on a flat table attatched to a spring with spring constant k and whose other end is attatched to a wall. We consider the frictionless case. Suppose that the spring is stretched 20 cm beyond the equilibrium position. We want to compute the work done by the spring. The force exerted by the spring is f = −k(x, 0)> where x is the displacement from equilibrium. In this experiment, the path is just the line from x = 20 to x = 0 and ∆x is negative. Hence f · ∆r = −kx∆x so that Z0
Z f · dr = c
20
0 −kx2 k(20)2 (−k x) dx = = = 200 k . 2 20 2
Another physical example is given by
11
Example 2.10 Compute the work done by gravity as a mass moves 8000 km along a circular arc at a height 10,000 km above the surface of the earth. In this case the arc of the circle is everywhere perpendicular to the radius vector. Since the gravitational force acts along the radius vector from the center of the earth, this vector field is always perpendicular to the the arc. Hence f (r) · ∆r = 0 along the path. Hence the work done is zero. This is why satteliets can remain in orbit without expending fuel once they are in orbit. In the Example (2.8), we somewhat cavalerly used the parameterization of a curve to compute the line integral. How can we justify the computation. Suppose that we have a curve parameterized by r(t) = (x(t), y(t), z(t))> , a ≤ t ≤ b. We start with the velocity vector r˙ (t) and partition the parameter interval by points {ti }ni=0 . Then, on the interval [ti , ti+1 ] , ∆ri = (∆xi , ∆yi , ∆zi )> ,where, for example ∆yi := y(ti+1 ) − y(ti ). Assuming the continuity of the vector function t 7→ r˙ (t), we can use the Mean Value Theorem to write that ∆ri = r(ti+1 ) − r(ti ) = r˙ (ξi )∆ti where ti < ξ < ti+1 . We form the Riemann sum n X
f (r(ξi )) · ∆ri =
i=1
n X
f (r(ξi )) · r˙ (ξi ) ∆ti ,
i=1
which, in the limit as k∆ti k → 0, yields the integral
Rb a
f (r(t)) · r˙ (t) dt.
Example 2.11 Let f (r) = (x+y, y)> and take c to be the quarter circle in the first quadrant with positive orientation. Use the parameterization r(t) = (cos (t), sin (t))> , 0 ≤ t ≤ π/2. Then
Z f · dr =
Zπ/2 (cos (t) + sin (t), sin (t))> · (− sin (t), cos (t))> dt 0
c
=
Zπ/2 − cos (t) sin (t) − sin2 (t) + sin (t) cos (t)) dt 0
Zπ/2 π = − sin2 (t) dt = − . 4 0
Example 2.12 A particle is traveling along a circular helix biven in parametric form by r(t) = (cos (t), sin (t), 2t)> , 0 ≤ t ≤ 3π, subject to the force field f = (x, y, −xy)> . What is the work done by the force on the particle in moving it from the point P : (1, 0, 0) to the point Q : (1, 0, 6π)?
12
In this case Z
Z3π f · dr =
c
f (r(t)) · r˙ (t) dt 0
Z3π = (cos (t), 2t, − cos (t) sin (t))> · (− sin (t), cos (t), 2)> dt 0
Z3π = (− cos (t) sin (t) + 2 t cos (t) − 2 cos (t) sin (t)) dt 0
Z3π (−3 cos (t) sin (t) + 2 t cos (t)) dt = −4 . = 0
REMARK: It is very common to see a different notation for a line integral, particularly in the plane R2 . Give functions P (x, y) and Q(x, y), defined in some region D ⊂ R2 , and an oriented curve c, defined in the same region, we write Z
Z f · dr =
(P (x, y)dx + Q(x, y) dy , c
c >
where we have taken f = (P, Q) and dr = (dx, dy)> . As an example, we have R Example 2.13 Evaluate the line integral c = xydx−y 2 dy where c is the line segment from (0, 0) to (2, 6). To do this, parameterize the curve as (1 − t)(0, 0)> + t(2, 6)> = (2t, 6t)> , 0 ≤ t ≤ 1. Then Z
Z
2
x y dx − y dy
1
(x y, −y ) · dr =
=
c
Z
2 >
(12 t2 , −36 t2 )> · (2, 6)> dt
0
c
Z =
1
(24 t2 − 216 t2 ) dt =
0
Z1
1 −24 t2 dt = −64 t3 = −64 . 0
0
Note that we could, as well, parameterized the line segment by r(t) = (t, 3t) , 0 ≤ t ≤ 2. Then r˙ (t) = (1, 3)> and Z
2
x y dx − y dy
Z
Z
2
(3 t2 , −9 t2 )> · (1, 3)> dt
(x y, −y ) · dr =
=
c
2 >
0
c
Z = 0
2
2 (−24 t ) dt = −8 t = −64 . 2
2
0
13
This last computation agrees with the fact that the result is independent of the parameterization of the curve. We now ask what happens when the vector field f has a potential. If we want to compute the line integral of f · dr along the curve c and if we assume that we have a parameterization (for example, we could parameterize the curve with respect to arc length), then the potential function g, evaluated along the curve gives rise to a scalar function G(t) := g(x(t), y(t), z(t)) , a ≤ t ≤ b. First note that G(a) = g(P ) and G(b) = g(Q) as a result of the parameterization of the curve. It is easy to compute the derivative of the function g by using the chain rule. Indeed, we have, writing as usual, r(t) = (x(t), y(t), z(t)), ∂g dx ∂g dy ∂g dz dG = (r(t)) + (r(t)) + (r(t)) = grad (g)(r(t)) · r˙ (t) = f (r(t)) · r˙ (t) . dt ∂x dt ∂y dt ∂z dt With this result we have Zb
Z f · dr = c
Zb f (r(t)) · r˙ (t) dt =
a
dG dt dt
a
b = G(t) = G(b) − G(a) = g(Q) − g(P ) . a
or, in summary, we have an analog in this case of the Fundamental Theorem of Calculus
Rb
grad (g(r(t)) r˙ (t) dt = g(Q) − g(P ) ,
(2)
a
It is important to understand that this last equation tells us that R if f has a potential, or what is the same thing, if f is conservative, then the line integral c f · dr is independent of the path of integration. Example 2.14 Let P (x, y) = 2x + 3y and Q(x, y) = 3x − 2y , and note that if g(x, y) = x2 + 3xy − y 2 then gx (x, y) = P (x, y) and gy = Q(x, y). So if we want to compute the line integral over a curve c joining the points (1, 3) and (−2, 5) we do not have to produce a parameterization of the path. Indeed Z P dx + Q dy g(−2, 5) − g(1, 3) = −51 − 1 = −52 . c
14
Example 2.15 A particle moves along the curve y = x2 from the point (1, 1) to the point (3, 4). If the motion is caused by the force f = (x2 + 2 y, 2 x + 2 y)> applied to the particle, what is the work done by f on the particle? The work is given by the line integral Z W =
[(x2 + 2 y) dx + 2 x + 2 y dy] .
c 3
2
Since f = grad (x /3 + 2xy + y ) we have W =
(3,4) 10 137 x3 2 + 2xy + y = 49 − = . 3 3 3 (1,1)
REMARK: If we are given a closed curve with positive orientation, we can choose any two distinct point P and Q on the curve. Then the two portions of the close curve represent two paths, both joining P with Q and both having positive orientation. If the line integral of f is independent of the path, then the values of these two line integrals along these two paths are exactly the same. Let us call this common value V . On the other hand, if we denote these two curves by c1 and c2 we traverse c by first traversing c1 with positive orientation and then c2 in the NEGATIVE orientation. Hence Z
I f ·r = c
Z f ·r+
c1
Note that we have used the notation a close curve.
f · r = V − V = 0. c2
H
to emphasize that the integral is being taken around
As an example, consider Example 2.16 Let f = (2x+4y, 4(x−y))> and let c be the closed curve consisting of the arc of the parabola y = x2 joining (0, 0) with (1, 1), which we call c1 , and the line segment from (1, 1) to (0, 0). We introduce the parameterization r1 (t) for c1 given by r1 (t) = (t, t2 )> , 0 ≤ t ≤ 1, so that r˙ 1 (t) = (1, 2t))> . For c2 we take, as usual, r2 (t) = (1 − t, 1 − t)> , 0 ≤ t ≤ 1. Then r˙ 2 (t) = (−1, −1)> . Note that the curve c = c1 + c2 . Then, writing I
Z f · dr =
c
Z P (x, y) dx + Q(x, y) dy +
c1
P (x, y) dx + Q(x, y) dy , c2
we have
15
Z1
Z P (x, y) dx + Q(x, y) dy
=
[(2t + 4t2 ) 1 + (4t − 4t2 ) 2t] dt
0
c1
Z1 =
1 [2t + 12t − 8t ] dt = [t + 4t − 2t = 3 , 2
3
2
3
4
0
0
and
Z0
Z P (x, y) dx + Q(x, y) dy
[2(1 − t) + 4(1 − t)] 1 + [4t − 4t] 1 dt
=
c2
1
Z
0
Z 6[(1 − t) dt = −
= 1
0
1
1 6(1 − t) dt = (6t − 3t2 = −3 . 0
H
Hence c f · dr = 3 − 3 = 0. We could anticipate this result by noting that grad(x2 + 4xy − 2y 2 ) = (2x + 4y, 4x − 4y)> so that the vector field is a gradient field. Remark: Note that we could parameterize the straight line segment in the opposite direction by (t, t)> , 0 ≤ t ≤ 1. In this case Z1
Z P (x, y) dx + Q(x, y) dy = −c1
0
1 6t dt = 3t = 3 . 2
0
But since the orientation is opposite to the actual orientation of the closed path, this latter value should be subtracted from the first result.
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