Chapter 16 Vector Calculus - University Of Cincinnati
Chapter 16 Vector Calculus Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16.2 Line Integrals Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals (1 of 20) In this section we define an integral that is similar to a single integral except that instead of integrating over an interval [a, b], we integrate over a curve C. Such integrals are called line integrals, although “curve integrals” would be better terminology. They were invented in the early 19th century to solve problems involving fluid flow, forces, electricity, and magnetism. Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals (2 of 20) We start with a plane curve C given by the parametric equations 1 x x (t ) y y (t ) a t b or, equivalently, by the vector equation r(t) x(t) i y(t) j, and we assume that C is a smooth curve. [This means that r is continuous and r (t ) 0.] Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals (3 of 20) If we divide the parameter interval [a, b] into n subintervals [ti 1, ti] of equal width and we let xi x(ti), and yi y(ti), then the corresponding points Pi(xi, yi) divide C into n subarcs with lengths Δs1, Δs2, . . . , Δsn. (See Figure 1.) Figure 1 Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals (4 of 20) We choose any point Pi ( xi , y i ) in the ith subarc. (This corresponds to a point t i in [ti 1, ti].) Now if f is any function of two variables whose domain includes the curve C, we evaluate f at the point ( xi , y i ), multiply by the length Δsi of the subarc, and form the sum n ( i 1 ) f xi , y i si which is similar to a Riemann sum. Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals (5 of 20) Then we take the limit of these sums and make the following definition by analogy with a single integral. 2 Definition If f is defined on a smooth curve C given by Equations 1, then the line integral of f along C is c f ( x, y ) ds lim n n ( ) xi , y i si i 1 if this limit exists. We have found that the length of C is L b a 2 2 dx dy dt dt dt Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals (6 of 20) A similar type of argument can be used to show that if f is a continuous function, then the limit in Definition 2 always exists and the following formula can be used to evaluate the line integral: 3 c f ( x, y ) ds a f ( x (t ) , y (t ) ) b 2 2 dx dy dt dt dt The value of the line integral does not depend on the parametrization of the curve, provided that the curve is traversed exactly once as t increases from a to b. Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals (7 of 20) If s(t) is the length of C between r(a) and r(t), then 2 ds dx dy dt dt dt 2 So the way to remember Formula 3 is to express everything in terms of the parameter t: Use the parametric equations to express x and y in terms of t and write ds as 2 2 dx dy ds dt dt dt Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals (8 of 20) In the special case where C is the line segment that joins (a, 0) to (b, 0), using x as the parameter, we can write the parametric equations of C as follows: x x, y 0, a x b. Formula 3 then becomes b C f ( x, y ) ds a f ( x, 0) dx and so the line integral reduces to an ordinary single integral in this case. Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals (9 of 20) Just as for an ordinary single integral, we can interpret the line integral of a positive function as an area. In fact, if f ( x, y ) 0, C f ( x, y ) ds represents the area of one side of the “fence” or “curtain” in Figure 2, whose base is C and whose height above the point (x, y) is f(x, y). Figure 2 Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Example 1 Evaluate C ( ) 2 x 2 y ds, where C is the upper half of the unit circle x 2 y 2 1. Solution: In order to use Formula 3, we first need parametric equations to represent C. Recall that the unit circle can be parametrized by means of the equations x cos t y sin t and the upper half of the circle is described by the parameter interval 0 t π. (See Figure 3.) Figure 3 Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Example 1 – Solution Therefore Formula 3 gives ( 2 x y ) ds 2 C ( 2 cos ( 2 cos ( 2 cos 2 t sin t 2 t sin t 0 0 0 2 ) 2 2 dx dy dt dt dt ) t sin t ) dt sin2 t cos2 t dt cos t 2t 3 0 2 2 3 3 Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals (10 of 20) Suppose now that C is a piecewise-smooth curve; that is, C is a union of a finite number of smooth curves C1, C2, ., Cn, where, as illustrated in Figure 4, the initial point of Ci 1 is the terminal point of Ci. A piecewise-smooth curve Figure 4 Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals (11 of 20) Then we define the integral of f along C as the sum of the integrals of f along each of the smooth pieces of C: C f ( x, y ) ds C1 f ( x, y ) ds C2 f ( x, y ) ds Cn f ( x, y ) ds Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals (12 of 20) Any physical interpretation of a line integral f ( x, y ) ds depends on the physical C interpretation of the function f. Suppose that ρ(x, y) represents the linear density at a point (x, y) of a thin wire shaped like a curve C. Then the mass of the part of the wire from Pi 1 to Pi in Figure 1 is approximately ( xi , y i ) si and so the total mass of the wire is approximately ( ) xi , y i si . Figure 1 Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals (13 of 20) By taking more and more points on the curve, we obtain the mass m of the wire as the limiting value of these approximations: n m lim n ( i 1 ) xi , y i si C ( x, y ) ds 2 [For example, if f ( x, y ) 2 x y represents the density of a semicircular wire, then the integral in Example 1 would represent the mass of the wire.] Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals (14 of 20) The center of mass of the wire with density function ρ is located at the point ( x , y ) , where 4 1 x m C x ( x, y ) ds 1 y m C y ( x, y ) ds Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals (15 of 20) Two other line integrals are obtained by replacing Δsi by either Δxi xi xi 1 or Δyi yi yi 1 in Definition 2. They are called the line integrals of f along C with respect to x and y: 5 6 f ( x, y ) dx f ( x, y ) dx C C n lim n lim n ( ) ( ) i 1 n i 1 f xi , y i xi f xi , y i xi Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals (16 of 20) When we want to distinguish the original line integral f ( x, y ) ds from those in C Equations 5 and 6, we call it the line integral with respect to arc length. The following formulas say that line integrals with respect to x and y can also be evaluated by expressing everything in terms of t: x x(t), y y(t), dx x (t )dt , dy y (t )dt . 7 C f ( x, y ) dx C f ( x, y ) dy a f ( x (t ), y (t )) x′ (t ) dt b a f ( x (t ), y (t )) y ′ (t ) dt b Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals (17 of 20) It frequently happens that line integrals with respect to x and y occur together. When this happens, it’s customary to abbreviate by writing C P ( x, y ) dx C Q ( x, y ) dy C P ( x, y ) dx Q ( x, y ) dy When we are setting up a line integral, sometimes the most difficult thing is to think of a parametric representation for a curve whose geometric description is given. Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals (18 of 20) In particular, we often need to parametrize a line segment, so it’s useful to remember that a vector representation of the line segment that starts at r0 and ends at r1 is given by 8 r ( t ) (1 t ) r0 t r1 0 t 1 Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals (19 of 20) In general, a given parametrization x x(t), y y(t), a t b, determines an orientation of a curve C, with the positive direction corresponding to increasing values of the parameter t. (See Figure 8, where the initial point A corresponds to the parameter value a and the terminal point B corresponds to t b.) Figure 8 Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals (20 of 20) If C denotes the curve consisting of the same points as C but with the opposite orientation (from initial point B to terminal point A in Figure 8), then we have C f ( x, y ) dx C f ( x, y ) dx C f ( x, y ) dy C f ( x, y ) dy But if we integrate with respect to arc length, the value of the line integral does not change when we reverse the orientation of the curve: C f ( x, y ) ds C f ( x, y ) ds This is because Δsi is always positive, whereas Δxi and Δyi change sign when we reverse the orientation of C. Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals in Space Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals in Space (1 of 3) We now suppose that C is a smooth space curve given by the parametric equations x x (t ) y y (t ) z z (t ) a t b or by a vector equation r(t) x(t) i y(t) j z(t) k. If f is a function of three variables that is continuous on some region containing C, then we define the line integral of f along C (with respect to arc length) in a manner similar to that for plane curves: C f ( x, y , z ) ds lim n n ( i 1 ) f xi , y i , zi si Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals in Space (2 of 3) We evaluate it using a formula similar to Formula 3: 9 C f ( x, y, z ) ds a f ( x (t ), y (t ), z (t )) b 2 2 2 dx dy dz dt dt dt dt Observe that the integrals in both Formulas 3 and 9 can be written in the more compact vector notation a f (r (t ) ) r (t ) dt b For the special case f(x, y, z) 1, we get b C ds a r (t ) dt L where L is the length of the curve C. Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals in Space (3 of 3) Line integrals along C with respect to x, y, and z can also be defined. For example, C f ( x, y , z ) dz n lim n ( i 1 ) f xi , y i , zi zi a f ( x (t ), y (t ), z (t )) z′ (t ) dt b Therefore, as with line integrals in the plane, we evaluate integrals of the form 10 C P ( x, y, z ) dx Q ( x, y, z ) dy R ( x, y , z ) dz by expressing everything (x, y, z, dx, dy, dz) in terms of the parameter t. Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Example 5 Evaluate C y sin z ds, where C is the circular helix given by the equations x cos t, y sin t, z t, 0 t 2π. (See Figure 9.) Figure 9 Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Example 5 – Solution Formula 9 gives C y sin z ds 2 0 2 0 ( sin t ) sin t sin2 t 2 0 2 2 2 dx dy dz dt dt dt dt sin2 t cos2 t 1 dt 1 (1 cos 2t ) dt 2 2 2 1 t sin 2t 2 2 0 2 2 Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals of Vector Fields Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals of Vector Fields (1 of 9) We know that the work done by a variable force f(x) in moving a particle from a to b along the x-axis is W b a f ( x ) dx. Then we have found that the work done by a constant force F in moving an object from a point P to another point Q in space is W F · D, where D PQ is the displacement vector. 3 . Now suppose that F P i Q j R k is a continuous force field on (A force field on 2 could be regarded as a special case where R 0 and P and Q depend only on x and y.) We wish to compute the work done by this force in moving a particle along a smooth curve C. Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals of Vector Fields (2 of 9) We divide C into subarcs Pi 1 Pi with lengths Δsi by dividing the parameter interval [a, b] into subintervals of equal width. (See Figure 1 for the twodimensional case or Figure 11 for the three-dimensional case.) Figure 1 Figure 11 Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals of Vector Fields (3 of 9) Choose a point Pi ( xi , y i , zi ) on the ith subarc corresponding to the parameter value ti . If Δsi is small, then as the particle moves from Pi 1 to Pi along the curve, it proceeds approximately in the direction of T ti , the unit tangent vector at Pi . ( ) Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals of Vector Fields (4 of 9) Thus the work done by the force F in moving the particle from Pi 1 to Pi is approximately ( ) ( ) ( ) ( ) F xi , y i , zi si T t i F xi , y i , zi T t i si and the total work done in moving the particle along C is approximately n 11 ( i 1 ) ( ) F x , y , z T x , y , z s i i i i i i i where T(x, y, z) is the unit tangent vector at the point (x, y, z) on C. Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals of Vector Fields (5 of 9) Intuitively, we see that these approximations ought to become better as n becomes larger. Therefore we define the work W done by the force field F as the limit of the Riemann sums in (11), namely, 12 W C F ( x, y, z ) T ( x, y, z ) ds C F T ds Equation 12 says that work is the line integral with respect to arc length of the tangential component of the force. Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals of Vector Fields (6 of 9) If the curve C is given by the vector equation r(t) x(t) i y(t) j z(t) k, r (t ) then T ( t ) , so using Equation 9 we can rewrite Equation 12 in the form r (t ) W r (t ) F (r (t ) ) r (t ) dt r (t ) b a This integral is often abbreviated as physics as well. F (r (t ) ) r (t ) dt b a c F dr and occurs in other areas of Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals of Vector Fields (7 of 9) Therefore we make the following definition for the line integral of any continuous vector field. 13 Definition Let F be a continuous vector field defined on a smooth curve C given by a vector function r(t), a t b. Then the line integral of F along C is C F dr a F (r (t ) ) r (t ) dt C F T ds b When using Definition 13, remember that F(r(t)) is just an abbreviation for the vector field F(x(t), y(t), z(t)), so we evaluate F(r(t)) simply by putting x x(t), y y(t), and z z(t) in the expression for F(x, y, z). Notice also that we can formally write dr r ( t ) dt . Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Example 7 Find the work done by the force field F ( x, y ) x 2i xy j in moving a particle along the quarter-circle r ( t ) cos t i sin t j, 0 t 2 . Solution: Since x cos t and y sin t, we have F (r (t ) ) cos2 t i cos t sin t j and r (t ) sin t i cos t j Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Example 7 – Solution Therefore the work done is C F dr 0 F (r (t ) ) r (t ) dt 2 2cos t sin t ) dt 0 ( 2 2 2 cos3 t 2 3 0 2 3 Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals of Vector Fields (8 of 9) Finally, we note the connection between line integrals of vector fields and line integrals of scalar fields. Suppose the vector field F on 3 is given in component form by the equation F P i Q j R k. We use Definition 13 to compute its line integral along C: C F dr a F (r (t ) ) r (t ) dt b a ( P i Q j R k ) ( x (t ) i y (t ) j z (t ) k ) dt b ( x ( t ) , y ( t ) , z ( t ) ) x ( t ) Q ( x ( t ) , y ( t ) , z (t ) ) y (t ) dt a R ( x (t ) , y (t ) , z (t ) ) z (t ) b P Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Line Integrals of Vector Fields (9 of 9) But this last integral is precisely the line integral in (10). Therefore we have C F dr C P dx Q dy R dz where F P i Q j R k For example, the integral y dx z dy x dz could be expressed as F dr C where C F(x, y, z) y i z j x k Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Stewart, Calculus: Early Transcendentals, 8th Edition. 2016 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly .
Why Vector processors Basic Vector Architecture Vector Execution time Vector load - store units and Vector memory systems Vector length - VLR Vector stride Enhancing Vector performance Measuring Vector performance SSE Instruction set and Applications A case study - Intel Larrabee vector processor
Vector Calculus 16.1 Vector Fields This chapter is concerned with applying calculus in the context of vector fields. A two-dimensional vector field is a function f that maps each point (x,y) in R2 to a two-dimensional vector hu,vi, and similarly a three-dimensional vector field maps (x,y,z) to hu,v,wi.
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