September 8, 2011 19:28 c15 Sheet number 1 Page number 1084 cyan magenta yellow black 15 TOPICS IN VECTOR CALCULUS
NASA Goddard Space Flight Center (NASA/GSFC)
Results in this chapter provide tools The main theme of this chapter is the concept of a “flow.” The body of mathematics that we for analyzing and understanding the will study here is concerned with analyzing flows of various types—the flow of a fluid or the behavior of hurricanes and other flow of electricity, for example. Indeed, the early writings of Isaac Newton on calculus are fluid flows. replete with such nouns as “fluxion” and “fluent,” which are rooted in the Latin fluere (to flow). We will begin this chapter by introducing the concept of a vector field, which is the mathematical description of a flow. In subsequent sections, we will introduce two new kinds of integrals that are used in a variety of applications to analyze properties of vector fields and flows. Finally, we conclude with three major theorems, Green’s Theorem, the Divergence Theorem, and Stokes’ Theorem. These theorems provide a deep insight into the nature of flows and are the basis for many of the most important principles in physics and engineering.
15.1 VECTOR FIELDS
In this section we will consider functions that associate vectors with points in 2-space or 3-space. We will see that such functions play an important role in the study of fluid flow, gravitational force fields, electromagnetic force fields, and a wide range of other applied problems.
VECTOR FIELDS To motivate the mathematical ideas in this section, consider a unit point-mass located at any point in the Universe. According to Newton’s Law of Universal Gravitation, the Earth exerts an attractive force on the mass that is directed toward the center of the Earth and has a magnitude that is inversely proportional to the square of the distance from the mass to the Earth’s center (Figure 15.1.1). This association of force vectors with points in space is called the Earth’s gravitational field. A similar idea arises in fluid flow. Imagine a stream in which the water flows horizontally at every level, and consider the layer of water at a specific depth. At each point of the layer, the water has a certain velocity, which we can represent by a vector at that point (Figure 15.1.2). This association of velocity vectors with points in the two-dimensional layer is called the velocity field at that layer. These ideas are Figure 15.1.1 captured in the following definition.
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15.1.1 definition A vector field in a plane is a function that associates with each point P in the plane a unique vector F(P ) parallel to the plane. Similarly, a vector field in 3-space is a function that associates with each point P in 3-space a unique vector F(P ) in 3-space.
Observe that in this definition there is no reference to a coordinate system. However, for Figure 15.1.2 computational purposes it is usually desirable to introduce a coordinate system so that vec- tors can be assigned components. Specifically, if F(P ) is a vector field in an xy-coordinate system, then the point P will have some coordinates (x, y) and the associated vector will have components that are functions of x and y. Thus, the vector field F(P ) can be expressed as Notice that a vector field is really just a F(x, y) = f(x,y)i + g(x,y)j vector-valued function. The term “vec- tor field” is commonly used in physics Similarly, in 3-space with an xyz-coordinate system, a vector field F(P ) can be expressed as and engineering. F(x,y,z)= f(x,y,z)i + g(x,y,z)j + h(x,y,z)k
GRAPHICAL REPRESENTATIONS OF VECTOR FIELDS A vector field in 2-space can be pictured geometrically by drawing representative field vec- tors F(x, y) at some well-chosen points in the xy-plane. But, just as it is usually not possible to describe a plane curve completely by plotting finitely many points, so it is usually not pos- sible to describe a vector field completely by drawing finitely many vectors. Nevertheless, such graphical representations can provide useful information about the general behavior of the field if the vectors are chosen appropriately. However, graphical representations of vector fields require a substantial amount of computation, so they are usually created using computers. Figure 15.1.3 shows four computer-generated vector fields. The vector field in part (a) might describe the velocity of the current in a stream at various depths. At the bottom of the stream the velocity is zero, but the speed of the current increases as the depth decreases. Points at the same depth have the same speed. The vector field in part (b) might TECHNOLOGY MASTERY describe the velocity at points on a rotating wheel. At the center of the wheel the velocity is If you have a graphing utility that can zero, but the speed increases with the distance from the center. Points at the same distance generate vector fields, read the relevant from the center have the same speed. The vector field in part (c) might describe the repulsive documentation and try to make rea- force of an electrical charge—the closer to the charge, the greater the force of repulsion. sonable duplicates of parts (a) and (b) Part (d ) shows a vector field in 3-space. Such pictures tend to be cluttered and hence are of Figure 15.1.3. of lesser value than graphical representations of vector fields in 2-space. Note also that the
Vectors not to scale Vectors not to scale −5 1 4 3 0 5 2 5 3 1 2 0 0 0 −1 1 − −2 5 −5 − 0 3 −1 0 −2 −1 0 1 2 −3 −2 −1 0 1 2 3 −1 0 1 5
F(x, y) = 1 √y i F(x, y) = −yi + xj xi + yj xi + yj + zk 5 F(x, y) = F(x, y, z) = − 10(x2 + y2)3/2 (x2 + y2 + z2)3/2
(a)(b)(c)(d) Figure 15.1.3 September 8, 2011 19:28 c15 Sheet number 3 Page number 1086 cyan magenta yellow black
1086 Chapter 15 / Topics in Vector Calculus
vectors in parts (b) and (c) are not to scale—their lengths have been compressed for clarity. We will follow this procedure throughout this chapter.
A COMPACT NOTATION FOR VECTOR FIELDS Sometimes it is helpful to denote the vector fields F(x, y) and F(x,y,z)entirely in vector notation by identifying (x, y) with the radius vector r = xi + y j and (x,y,z) with the radius vector r = xi + y j + zk. With this notation a vector field in either 2-space or 3- space can be written as F(r). When no confusion is likely to arise, we will sometimes omit the r altogether and denote the vector field as F.
INVERSE-SQUARE FIELDS According to Newton’s Law of Universal Gravitation, particles with masses m and M attract each other with a force F of magnitude GmM F = (1) r2 where r is the distance between the particles and G is a constant. If we assume that the particle of mass M is located at the origin of an xyz-coordinate system and r is the radius vector to the particle of mass m, then r = r , and the force F(r) exerted by the particle of mass M on the particle of mass m is in the direction of the unit vector −r/ r . Thus, it follows from (1) that GmM r GmM F(r) =− =− r (2) r 2 r r 3 If m and M are constant, and we let c =−GmM, then this formula can be expressed as c F(r) = r r 3 Vector fields of this form arise in electromagnetic as well as gravitational problems. Such fields are so important that they have their own terminology.
15.1.2 definition If r is a radius vector in 2-space or 3-space, and if c is a constant, c then a vector field of the form F(r) = r (3) r 3 is called an inverse-square field.
Observe that if c>0 in (3), then F(r) has the same direction as r, so each vector in the field is directed away from the origin; and if c<0, then F(r) is oppositely directed to r,so each vector in the field is directed toward the origin. In either case the magnitude of F(r) is inversely proportional to the square of the distance from the terminal point of r to the origin, since |c| |c| F(r) = r = r 3 r 2 We leave it for you to verify that in 2-space Formula (3) can be written in component form as c F(x, y) = (xi + y j) (4) (x2 + y2)3/2 and in 3-space as c F(x,y,z)= (xi + y j + zk) (5) (x2 + y2 + z2)3/2 [see parts (c) and (d ) of Figure 15.1.3]. September 8, 2011 19:28 c15 Sheet number 4 Page number 1087 cyan magenta yellow black
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Example 1 Coulomb’s law states that the electrostatic force exerted by one charged particle on another is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This has the same form as Newton’s Law of Universal Gravitation, so the electrostatic force field exerted by a charged particle is an inverse-square field. Specifically, if a particle of charge Q is at the origin of a coordinate system, and if r is the radius vector to a particle of charge q, then the force F(r) that the particle of charge Q exerts on the particle of charge q is of the form qQ F(r) = r 3 4π0 r
where 0 is a positive constant (called the permittivity constant). This formula is of form (3) with c = qQ/4π0.
GRADIENT FIELDS An important class of vector fields arises from the process of finding gradients. Recall that if φ is a function of three variables, then the gradient of φ is defined as ∂φ ∂φ ∂φ ∇φ = i + j + k ∂x ∂y ∂z This formula defines a vector field in 3-space called the gradient field of φ. Similarly, the gradient of a function of two variables defines a gradient field in 2-space. At each point in a gradient field where the gradient is nonzero, the vector points in the direction in which the rate of increase of φ is maximum.
y
6 Example 2 Sketch the gradient field of φ(x,y) = x + y.
4 Solution. The gradient of φ is ∂φ ∂φ 2 ∇φ = i + j = i + j ∂x ∂y x which is constant [i.e., is the same vector at each point (x, y)]. A portion of the vector field 246 is sketched in Figure 15.1.4 together with some level curves of φ. Note that at each point, Figure 15.1.4 ∇φ is normal to the level curve of φ through the point (Theorem 13.6.6).
CONSERVATIVE FIELDS AND POTENTIAL FUNCTIONS If F(r) is an arbitrary vector field in 2-space or 3-space, we can ask whether it is the gradient field of some function φ, and if so, how we can find φ. This is an important problem in various applications, and we will study it in more detail later. However, there is some terminology for such fields that we will introduce now.
15.1.3 definition A vector field F in 2-space or 3-space is said to be conservative in a region if it is the gradient field for some function φ in that region, that is, if
F = ∇φ
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Example 3 Inverse-square fields are conservative in any region that does not contain the origin. For example, in the two-dimensional case the function c φ(x,y) =− (6) (x2 + y2)1/2 is a potential function for (4) in any region not containing the origin, since ∂φ ∂φ ∇φ(x,y) = i + j ∂x ∂y cx cy = i + j (x2 + y2)3/2 (x2 + y2)3/2 c = (xi + y j) (x2 + y2)3/2 = F(x, y) In a later section we will discuss methods for finding potential functions for conservative vector fields.
DIVERGENCE AND CURL We will now define two important operations on vector fields in 3-space—the divergence and the curl of the field. These names originate in the study of fluid flow, in which case the divergence relates to the way in which fluid flows toward or away from a point and the curl relates to the rotational properties of the fluid at a point. We will investigate the physical interpretations of these operations in more detail later, but for now we will focus only on their computation.
15.1.4 definition If F(x,y,z)= f(x,y,z)i + g(x,y,z)j + h(x,y,z)k, then we define the divergence of F, written div F, to be the function given by
∂f ∂g ∂h div F = + + (7) ∂x ∂y ∂z
15.1.5 definition If F(x,y,z)= f(x,y,z)i + g(x,y,z)j + h(x,y,z)k, then we define the curl of F, written curl F, to be the vector field given by ∂h ∂g ∂f ∂h ∂g ∂f curl F = − i + − j + − k (8) ∂y ∂z ∂z ∂x ∂x ∂y
REMARK Observe that div F and curl F depend on the point at which they are computed, and hence are more properly written as div F(x,y,z)and curl F(x,y,z). However, even though these functions are expressed in terms of x, y, and z, it can be proved that their values at a fixed point depend only on the point and not on the coordinate system selected. This is important in applications, since it allows physicists and engineers to compute the curl and divergence in any convenient coordinate system.
Before proceeding to some examples, we note that div F has scalar values, whereas curl F has vector values (i.e., curl F is itself a vector field). Moreover, for computational September 8, 2011 19:28 c15 Sheet number 6 Page number 1089 cyan magenta yellow black
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purposes it is useful to note that the formula for the curl can be expressed in the determinant form ijk ∂ ∂ ∂ curl F = (9) ∂x ∂y ∂z fgh
You should verify that Formula (8) results if the determinant is computed by interpreting a “product” such as (∂/∂x)(g) to mean ∂g/∂x. Keep in mind, however, that (9) is just a mnemonic device and not a true determinant, since the entries in a determinant must be numbers, not vectors and partial derivative symbols.
Example 4 Find the divergence and the curl of the vector field
F(x,y,z)= x2yi + 2y3zj + 3zk
Solution. From (7) ∂ ∂ ∂ div F = (x2y) + (2y3z) + (3z) ∂x ∂y ∂z = 2xy + 6y2z + 3 and from (9) ijk TECHNOLOGY MASTERY ∂ ∂ ∂ curl F = ∂x ∂y ∂z Most computer algebra systems can 2 3 compute gradient fields, divergence, x y 2y z 3z and curl. If you have a CAS with these ∂ ∂ ∂ ∂ ∂ ∂ capabilities, read the relevant docu- = (3z) − (2y3z) i + (x2y) − (3z) j + (2y3z) − (x2y) k mentation, and use your CAS to check ∂y ∂z ∂z ∂x ∂x ∂y the computations in Examples 2 and 4. =−2y3i − x2k
Example 5 Show that the divergence of the inverse-square field c F(x,y,z)= (xi + y j + zk) (x2 + y2 + z2)3/2 is zero.
Solution. The computations can be simplified by letting r = (x2 + y2 + z2)1/2, in which case F can be expressed as cxi + cy j + czk cx cy cz F(x,y,z)= = i + j + k r3 r3 r3 r3 We leave it for you to show that ∂r x ∂r y ∂r z = , = , = ∂x r ∂y r ∂z r September 8, 2011 19:28 c15 Sheet number 7 Page number 1090 cyan magenta yellow black
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Thus ∂ x ∂ y ∂ z div F = c + + (10) ∂x r3 ∂y r3 ∂z r3
But 3 − 2 / 2 ∂ x = r x(3r )(x r) = 1 − 3x ∂x r3 (r3)2 r3 r5 2 ∂ y = 1 − 3y ∂y r3 r3 r5 2 ∂ z = 1 − 3z ∂z r3 r3 r5 Substituting these expressions in (10) yields 3 3x2 + 3y2 + 3z2 3 3r2 div F = c − = c − = 0 r3 r5 r3 r5
THE ∇ OPERATOR Thus far, the symbol ∇ that appears in the gradient expression ∇φ has not been given a meaning of its own. However, it is often convenient to view ∇ as an operator
∂ ∂ ∂ ∇ = i + j + k (11) ∂x ∂y ∂z
which when applied to φ(x,y,z) produces the gradient ∂φ ∂φ ∂φ ∇φ = i + j + k ∂x ∂y ∂z We call (11) the del operator. This is analogous to the derivative operator d/dx, which when applied to f(x) produces the derivative f (x). The del operator allows us to express the divergence of a vector field F = f(x,y,z)i + g(x,y,z)j + h(x,y,z)k in dot product notation as
∂f ∂g ∂h (div F = ∇ ؒ F = + + (12 ∂x ∂y ∂z
and the curl of this field in cross-product notation as ijk ∂ ∂ ∂ curl F = ∇ × F = (13) ∂x ∂y ∂z fgh
THE LAPLACIAN ∇2 The operator that results by taking the dot product of the del operator with itself is denoted by ∇2 and is called the Laplacian operator. This operator has the form
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When applied to φ(x,y,z) the Laplacian operator produces the function ∂2φ ∂2φ ∂2φ ∇2φ = + + ∂x2 ∂y2 ∂z2 Note that ∇2φ can also be expressed as div (∇φ). The equation ∇2φ = 0 or, equivalently, ∂2φ ∂2φ ∂2φ + + = 0 ∂x2 ∂y2 ∂z2 is known as Laplace’s equation. This partial differential equation plays an important role in a wide variety of applications, resulting from the fact that it is satisfied by the potential function for the inverse-square field.
✔QUICK CHECK EXERCISES 15.1 (See page 1093 for answers.)
1. The function φ(x,y,z) = xy + yz + xz is a potential for 3. An inverse-square field is one that can be written in the form the vector field F = . F(r) = . 2. The vector field F(x,y,z)= , defined for 4. The vector field (x,y,z) = (0, 0, 0), is always directed toward the origin F(x,y,z)= yzi + xy2j + yz2k and is of length equal to the distance from (x,y,z) to the origin. has divergence and curl .
Pierre-Simon de Laplace (1749–1827) French mathe- papers by so young a person in such a short time. Laplace had little matician and physicist. Laplace is sometimes referred interest in pure mathematics—he regarded mathematics merely as to as the French Isaac Newton because of his work in a tool for solving applied problems. In his impatience with math- celestial mechanics. In a five-volume treatise entitled ematical detail, he frequently omitted complicated arguments with Traité de Mécanique Céleste, he solved extremely dif- the statement, “It is easy to show that....” He admitted, however, ficult problems involving gravitational interactions be- that as time passed he often had trouble reconstructing the omitted tween the planets. In particular, he was able to show that our solar details himself! system is stable and not prone to catastrophic collapse as a result of At the height of his fame, Laplace served on many government these interactions. This was an issue of major concern at the time committees and held the posts of Minister of the Interior and Chan- because Jupiter’s orbit appeared to be shrinking and Saturn’s ex- cellor of the Senate. He barely escaped imprisonment and execution panding; Laplace showed that these were expected periodic anoma- during the period of the Revolution, probably because he was able lies. In addition to his work in celestial mechanics, he founded to convince each opposing party that he sided with them. Napoleon modern probability theory, showed with Lavoisier that respiration described him as a great mathematician but a poor administrator who is a form of combustion, and developed methods that fostered many “sought subtleties everywhere, had only doubtful ideas, and ...car- new branches of pure mathematics. ried the spirit of the infinitely small into administration.” In spite of Laplace was born to moderately successful parents in Normandy, his genius, Laplace was both egotistic and insecure, attempting to his father being a farmer and cider merchant. He matriculated in ensure his place in history by conveniently failing to credit mathe- the theology program at the University of Caen at age 16 but left maticians whose work he used—an unnecessary pettiness since his for Paris at age 18 with a letter of introduction to the influential own work was so brilliant. However, on the positive side he was mathematician d’Alembert, who eventually helped him undertake a supportive of young mathematicians, often treating them as his own career in mathematics. Laplace was a prolific writer, and after his children. Laplace ranks as one of the most influential mathemati- election to the Academy of Sciences in 1773, the secretary wrote cians in history. that the Academy had never received so many important research [Image: http://commons.wikimedia.org/wiki/File:Pierre-Simon-Laplace_%281749-1827%29.jpg] September 8, 2011 19:28 c15 Sheet number 9 Page number 1092 cyan magenta yellow black
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EXERCISE SET 15.1 Graphing Utility C CAS
FOCUS ON CONCEPTS 7. F(x, y) = yi − x j.[Note: Each vector in the field is per- = + 1–2 Match the vector field F(x, y) with one of the plots, and pendicular to the position vector r xi y j.] ■ xi + y j explain your reasoning. 8. F(x, y) = .[Note: Each vector in the field is 1. (a) F(x, y) = xi (b) F(x, y) = sin xi + j x2 + y2 a unit vector in the same direction as the position vector 2. (a) F(x, y) = i + j x y r = xi + y j.] (b) F(x, y) = i + j x2 + y2 x2 + y2 9–10 Use a graphing utility to generate a plot of the vector field. y y ■ 9. F(x, y) = i + cos y j 10. F(x, y) = yi − x j
11–14 True–False Determine whether the statement is true or x x false. Explain your answer. ■ 11. The vector-valued function F(x, y) = y i + x2 j + xyk is an example of a vector field in the xy-plane. I II 12. If r is a radius vector in 3-space, then a vector field of the form 1 y y F(r) = r r 2 is an example of an inverse-square field. 13. If F is a vector field, then so is ∇ × F. x x 14. If F is a vector field and ∇ ؒ F = φ, then φ is a potential function for F.
15–16 Confirm that φ is a potential function for F(r) on some region, and state the region. ■ 15. (a) φ(x,y) = tan−1 xy III IV y x F(x, y) = i + j + 2 2 + 2 2 3–4 Determine whether the statement about the vector field 1 x y 1 x y = 2 − 2 + 2 F(x, y) is true or false. If false, explain why. ■ (b) φ(x,y,z) x 3y 4z F(x,y,z)= 2xi − 6y j + 8zk 3. F(x, y) = x2i − y j. 16. (a) φ(x,y) = 2y2 + 3x2y − xy3 (a) F(x, y) →0as(x, y)→(0, 0). F(x, y) = (6xy − y3)i + (4y + 3x2 − 3xy2)j (b) If (x, y) is on the positive y-axis, then the vector (b) φ(x,y,z) = x sin z + y sin x + z sin y points in the negative y-direction. F(x,y,z)= (sin z + y cos x)i + (sin x + z cos y)j (c) If (x, y) is in the first quadrant, then the vector points + (sin y + x cos z)k down and to the right. x y 4. F(x, y) = i − j. 17–22 Find div F and curl F. ■ x2 + y2 x2 + y2 (a) As (x, y) moves away from the origin, the lengths of 17. F(x,y,z)= x2i − 2j + yzk the vectors decrease. 18. F(x,y,z)= xz3i + 2y4x2 j + 5z2yk (b) If (x, y) is a point on the positive x-axis, then the 19. F(x,y,z)= 7y3z2i − 8x2z5 j − 3xy4k vector points up. = xy − + 2 (c) If (x, y) is a point on the positive y-axis, the vector 20. F(x,y,z) e i cos y j sin zk 1 points to the right. 21. F(x,y,z)= (xi + y j + zk) x2 + y2 + z2 − 5–8 Sketch the vector field by drawing some representative 22. F(x,y,z)= ln xi + exyzj + tan 1(z/x)k nonintersecting vectors. The vectors need not be drawn to scale, ■ .(but they should be in reasonably correct proportion relative to 23–24 Find ∇ ؒ (F × G ■ each other. 23. F(x,y,z)= 2xi + j + 4yk 5. F(x, y) = 2i − j6.F(x, y) = y j,y>0 G(x,y,z)= xi + y j − zk September 8, 2011 19:28 c15 Sheet number 10 Page number 1093 cyan magenta yellow black
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24. F(x,y,z)= yzi + xzj + xyk 44. (a) Use part (a) of Exercise 43, Exercise 36, and Exercise G(x,y,z)= xy j + xyzk 41(a) to show that curl F = 0. (b) Use the result in part (a) of Exercise 43 and Exercises Find ∇ ؒ (∇ × F). ■ 35 and 42(a) to show that 26–25 = + − + f (r) 25. F(x,y,z) sin xi cos(x y)j zk ∇2f(r) = 2 + f (r) 26. F(x,y,z)= exzi + 3xey j − eyzk r 45. Use the result in Exercise 43(b) to show that the divergence of the inverse-square field F = r/ r 3 is zero. 27–28 Find ∇ × (∇ × F). ■ 46. Use the result of Exercise 43(b) to show that if F is a vec- 27. F(x,y,z)= xy j + xyzk tor field of the form F = f( r )r and if div F = 0, then F = 2 − + 28. F(x,y,z) y xi 3yzj xyk is an inverse-square field. [Suggestion: Let r = r and C 29. Use a CAS to check the calculations in Exercises 23, 25, multiply 3f(r) + rf (r) = 0 through by r2. Then write the and 27. result as a derivative of a product.] C 30. Use a CAS to check the calculations in Exercises 24, 26, 47. A curve C is called a flow line of a vector field F if F is a and 28. tangent vector to C at each point along C (see the accom- panying figure). = = 31–38 Let k be a constant, F F(x,y,z), G G(x,y,z), and (a) Let C be a flow line for F(x, y) =−yi + x j, and let = φ φ(x,y,z). Prove the following identities, assuming that all (x, y) be a point on C for which y = 0. Show that the derivatives involved exist and are continuous. ■ flow lines satisfy the differential equation 31. div(kF) = k div F 32. curl(kF) = k curl F dy =−x 33. div(F + G) = div F + div G dx y 34. curl(F + G) = curl F + curl G (b) Solve the differential equation in part (a) by separation div(φF) = φ div F + ∇φ ؒ F of variables, and show that the flow lines are concentric .35 circles centered at the origin. 36. curl(φF) = φ curl F + ∇φ × F 37. div(curl F) = 0 38. curl(∇φ) = 0 C 39. Rewrite the identities in Exercises 31, 33, 35, and 37 in an Flow lines of ؒ ∇ equivalent form using the notation for divergence and a vector field ∇ × for curl. Figure Ex-47 40. Rewrite the identities in Exercises 32, 34, 36, and 38 in an equivalent form using the notation ∇ ؒ for divergence and 48–50 Find a differential equation satisfied by the flow lines of ∇ × for curl. F (see Exercise 47), and solve it to find equations for the flow lines of F. Sketch some typical flow lines and tangent vectors. 41–42 Verify that the radius vector r = xi + y j + zk has the ■ stated property. ■ r 48. F(x, y) = i + x j 49. F(x, y) = xi + j,x>0 41. (a) curl r = 0 (b) ∇ r = r 50. F(x, y) = xi − y j,x>0 and y>0 1 r 42. (a) div r = 3 (b) ∇ =− 51. Writing Discuss the similarities and differences between r r 3 the concepts “vector field” and “slope field.” 52. Writing In physical applications it is often necessary to Let r = xi + y j + zk, let r = r , let f be a differen- 43–44 deal with vector quantities that depend not only on position tiable function of one variable, and let F(r) = f(r)r. ■ in space but also on time. Give some examples and discuss 43. (a) Use the chain rule and Exercise 41(b) to show that how the concept of a vector field would need to be modified f (r) ∇f(r) = r to apply to such situations. r (b) Use the result in part (a) and Exercises 35 and 42(a) to show that div F = 3f(r) + rf (r).
✔QUICK CHECK ANSWERS 15.1 c 1. (y + z)i + (x + z)j + (x + y)k 2. −r =−xi − y j − zk 3. r 4. 2xy + 2yz; z2i + y j + (y2 − z)k r 3 September 8, 2011 19:28 c15 Sheet number 11 Page number 1094 cyan magenta yellow black
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15.2 LINE INTEGRALS
In earlier chapters we considered three kinds of integrals in rectangular coordinates: single integrals over intervals, double integrals over two-dimensional regions, and triple integrals over three-dimensional regions. In this section we will discuss integrals along curves in two- or three-dimensional space.
LINE INTEGRALS The first goal of this section is to define what it means to integrate a function along a curve. To motivate the definition we will consider the problem of finding the mass of a Q very thin wire whose linear density function (mass per unit length) is known. We assume that we can model the wire by a smooth curve C between two points P and Q in 3-space (Figure 15.2.1). Given any point (x,y,z)on C, we let f(x,y,z) denote the corresponding value of the density function. To compute the mass of the wire, we proceed as follows:
• Divide C into n very small sections using a succession of distinct partition points
P = P0,P1,P2,...,Pn−1,Pn = Q
as illustrated on the left side of Figure 15.2.2. Let Mk be the mass of the kth section, and let sk be the length of the arc between Pk−1 and Pk.
= P Q Pn Figure 15.2.1 A bent thin wire modeled by a smooth curve Pn−1 Δ sk Pk
Pk−1 Δ Mass Mk P3
P2 P Δ 1 sk Pk = P*(x*, y*, z*) P P0 Pk−1 k k k k Δ ≈ * * * Δ Mk f (xk , yk, zk ) sk Figure 15.2.2
∗ ∗ ∗ ∗ • Choose an arbitrary sampling point Pk (xk ,yk ,zk ) on the kth arc, as illustrated on the right side of Figure 15.2.2. If sk is very small, the value of f will not vary much along the kth section and we can approximate f along this section by the value ∗ ∗ ∗ f(xk ,yk ,zk ). It follows that the mass of the kth section can be approximated by ≈ ∗ ∗ ∗ Mk f(xk ,yk ,zk )sk • The mass M of the entire wire can then be approximated by n n = ≈ ∗ ∗ ∗ M Mk f(xk ,yk ,zk )sk (1) k=1 k=1
• We will use the expression max sk →0 to indicate the process of increasing n in such a way that the lengths of all the sections approach 0. It is plausible that the error in (1) will approach 0 as max sk →0 and the exact value of M will be given by n = ∗ ∗ ∗ M lim f(xk ,yk ,zk )sk (2) max sk →0 k=1 September 8, 2011 19:28 c15 Sheet number 12 Page number 1095 cyan magenta yellow black
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The limit in (2) is similar to the limit of Riemann sums used to define the definite integral of a function over an interval (Definition 5.5.1). With this similarity in mind, we make the following definition.
15.2.1 definition If C is a smooth curve in 2-space or 3-space, then the line inte- gral of f with respect to s along C is Although the term “curve integrals” is n more descriptive, the integrals in Defi- = ∗ ∗ nition 15.2.1 are called “line integrals” f(x,y)ds lim f(xk ,yk )sk 2-space (3) max sk →0 for historical reasons. C k=1 or n = ∗ ∗ ∗ f(x,y,z)ds lim f(xk ,yk ,zk )sk 3-space (4) max sk →0 C k=1 provided this limit exists and does not depend on the choice of partition or on the choice of sample points.
It is usually impractical to evaluate line integrals directly from Definition 15.2.1. How- ever, the definition is important in the application and interpretation of line integrals. For example:
• If C is a curve in 3-space that models a thin wire, and if f(x,y,z) is the linear density function of the wire, then it follows from (2) and Definition 15.2.1 that the mass M of the wire is given by M = f(x,y,z)ds (5) C That is, to obtain the mass of a thin wire, we integrate the linear density function over the smooth curve that models the wire. • If C is a smooth curve of arc length L, and f is identically 1, then it immediately follows from Definition 15.2.1 that n ds = lim sk = lim L = L (6) max sk →0 max sk →0 C k=1
z • If C is a curve in the xy-plane and f(x,y) is a nonnegative continuous function defined on C, then C f(x,y)ds can be interpreted as the area A of the “sheet” that is swept out by a vertical line segment that extends upward from the point (x, y) to a height of f(x,y) and moves along C from one endpoint to the other (Figure 15.2.3). ∗ ∗ To see why this is so, refer to Figure 15.2.4 in which f(xk ,yk ) is the value of f at an y ∗ arbitrary point Pk on the kth arc of the partition and note the approximation f(x, y) ∗ ∗ Ak ≈ f(x ,y )sk C k k (x, y) It follows that n n x = ≈ ∗ ∗ Figure 15.2.3 A Ak f(xk ,yk )sk k=1 k=1 It is then plausible that n = ∗ ∗ = A lim f(xk ,yk )sk f(x,y)ds (7) max sk →0 k=1 C September 8, 2011 19:28 c15 Sheet number 13 Page number 1096 cyan magenta yellow black
1096 Chapter 15 / Topics in Vector Calculus
z
ximation pro Ap * * f(xk , yk ) Δ An ... y Δ Ak ... ΔA ΔA 3 ΔA 2 Pn−1 Pk 1 P* x . . . k P P3 k−1 P P2 Δ 1 sk Figure 15.2.4
Since Definition 15.2.1 is closely modeled on Definition 5.5.1, it should come as no sur- prise that line integrals share many of the common properties of ordinary definite integrals. For example, we have [f(x,y) + g(x,y)] ds = f(x,y)ds + g(x,y)ds C C C provided both line integrals on the right-hand side of this equation exist. Similarly, it can be shown that if f is continuous on C, then the line integral of f with respect to s along C exists.
EVALUATING LINE INTEGRALS Except in simple cases, it will not be feasible to evaluate a line integral directly from (3) or (4). However, we will now show that it is possible to express a line integral as an ordi- nary definite integral, so that no special methods of evaluation are required. For example, suppose that C is a curve in the xy-plane that is smoothly parametrized by r(t) = x(t)i + y(t)j (a ≤ t ≤ b)
Moreover, suppose that each partition point Pk of C corresponds to a parameter value of tk in [a,b]. The arc length of C between points P − and P is then given by k 1 k tk sk = r (t) dt (8) tk−1
(Theorem 12.3.1). If we let tk = tk − tk−1, then it follows from (8) and the Mean-Value ∗ Theorem for Integrals (Theorem 5.6.2) that there exists a point t in [tk−1,tk] such that k tk = = ∗ sk r (t) dt r (tk ) tk tk−1 y ∗ ∗ ∗ = ∗ ∗ ∗ ∗ We let Pk (xk ,yk ) Pk (x(tk ), y(tk )) correspond to the parameter value tk (Figure 15.2.5). Since the parametrization of C is smooth, it can be shown that max sk →0 if and only → Pk−1 if max tk 0 (Exercise 57). Furthermore, the composition f(x(t), y(t)) is a real-valued Δ sk function defined on [a,b] and we have r(tk−1) * Pk n = ∗ ∗ f(x,y)ds lim f(xk ,yk )sk Definition 15.2.1 r(t*) P max sk →0 k k C k=1 n ∗ ∗ ∗ r(tk) = C lim f(x(tk ), y(tk )) r (tk ) tk Substitution max sk →0 k=1 x n = ∗ ∗ ∗ lim f(x(tk ), y(tk )) r (tk ) tk max tk →0 k=1 Figure 15.2.5 b = f(x(t), y(t)) r (t) dt Definition 5.5.1 a September 8, 2011 19:28 c15 Sheet number 14 Page number 1097 cyan magenta yellow black
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Therefore, if C is smoothly parametrized by r(t) = x(t)i + y(t)j (a ≤ t ≤ b) then b f(x,y)ds = f(x(t), y(t)) r (t) dt (9) C a
Similarly, if C is a curve in 3-space that is smoothly parametrized by r(t) = x(t)i + y(t)j + z(t)k (a ≤ t ≤ b) then b Explain how Formulas (9) and (10) f(x,y,z)ds = f(x(t), y(t), z(t)) r (t) dt (10) confirm Formula (6) for arc length. C a