Section 4.3. Line Integrals
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1 The lculus of Functions of Severl Vribles Section 4.3 Line Integrls We will motivte the mthemticl concept of line integrl through n initil discussion of the physicl concept of work. Work If force of constnt mgnitude F is cting in the direction of motion of n object long line, nd the object moves distnce d long this line, then we cll the quntity F d the work done by the force on the object. More generlly, if the vector F represents constnt force cting on n object s it moves long displcement vector d, then F d d (4.3.1) is the mgnitude of F in the direction of motion (see Figure 4.3.1) nd we define ( F ) d d F d (4.3.2) d to be the work done by F on the object when it is displced by d. F F. u Figure Mgnitude of F in the direction of d is F u, where u d d d We now generlize the formultion of work in (4.3.2) to the sitution where n object P moves long some curve subject to force which depends continuously on position (but does not depend on time). Specificlly, we represent the force by continuous vector field, sy, F : R n R n, nd we suppose P moves long curve which hs smooth 1 opyright c by Dn Sloughter 21
2 2 Line Integrls Section 4.3 F (ϕ( t)) D ϕ( ) t P ϕ( ) t Figure Object P moving long curve subject to force F prmetriztion ϕ : I R n, where I [, b]. See Figure To pproximte the work done by F s P moves from ϕ() to ϕ(b) long, we first divide I into m equl subintervls of length t b m with endpoints t < t 1 < t 2 < < t m b. Now t time t k, k, 1,..., m 1, P is moving in the direction of Dϕ(t k ) t speed of Dϕ(t k ), nd so will move distnce of pproximtely Dϕ(t k ) t over the time intervl [t k, t k+1 ]. Thus we my pproximte the work done by F s P moves from ϕ(t k ) to ϕ(t k+1 ) by the work done by the force F (ϕ(t k )) in moving P long the displcement vector Dϕ(t k ) t, which is vector of length Dϕ(t k ) t in the direction of Dϕ(t k ). Tht is, if we let W k denote the work done by F s P moves from ϕ(t k ) to ϕ(t k 1 ), then W k F (ϕ(t k )) Dϕ(t k ) t. (4.3.3) If we let W denote the totl work done by F s P moves long, then we hve W m 1 k W k m 1 k F (ϕ(t k )) Dϕ(t k ) t. (4.3.4) As m increses, we should expect the pproximtion in (4.3.4) to pproch W. Moreover, since F (ϕ(t)) Dϕ(t) is continuous function of t nd the sum in (4.3.4) is left-hnd rule pproximtion for the definite integrl of F (ϕ(t)) Dϕ(t) over the intervl [, b], we should hve m 1 b W lim F (ϕ(t k )) Dϕ(t k ) t F (ϕ(t)) Dϕ(t)dt. (4.3.5) m k
3 Section 4.3 Line Integrls 3 Exmple Suppose n object moves long the curve prmetrized by ϕ(t) (t, t 2 ), 1 t 1, subject to the force F (x, y) (y, x). Then the work done by F s the object moves from ϕ( 1) ( 1, 1) to ϕ(1) (1, 1) is W t F (ϕ(t)) Dϕ(t)dt F (t, t 2 ) (1, 2t)dt (t 2, t) (1, 2t)dt 3t 2 dt ( ) t Exmple The function ψ(t) 2, t2 4, 2 t 2, is lso smooth prmetriztion of the curve in the previous exmple. Using the sme force function F, we hve 2 2 F (ψ(t)) Dψ(t)dt t ( t 2 4, t ) 2 2 3t 2 8 dt ( 1 2, t ) dt 2 This is the result we should expect: s long s the curve is trversed only once, the work done by force when n object moves long the curve should depend only on the curve nd not on ny prticulr prmetriztion of the curve. We need to verify the previous sttement in generl before we cn stte our definition of the line integrl. Note tht in these two exmples, ψ(t) ϕ ( t 2). In other words, ψ(t) ϕ(g(t)), where g(t) t 2 for 2 t 2. In generl, if ϕ(t), for t in n intervl [, b], nd ψ(t), for t in n intervl [c, d], re both smooth prmetriztions of curve such tht every point on corresponds to exctly one point in I nd exctly one point in J, then there exists differentible function g which mps J onto I such tht ψ(t) ϕ(g(t)). Defining such g is strightforwrd: given ny t in [c, d], find the unique vlue s in [, b] such tht ϕ(s) ψ(t) (such vlue s hs to exist since is the imge of both ψ nd ϕ). Then g(t) s. Proving tht g is differentible is not s esy, nd we will not provide proof here. However, ssuming tht g is differentible, it follows tht for ny continuous
4 4 Line Integrls Section 4.3 vector field F, Now if we let in (4.3.6), then d c F (ψ(t)) Dψ(t)dt d c d c d c u g(t), du g (t)dt, F (ψ(t)) Dψ(t)dt F (ϕ(g(t)) D(ϕ g)(t))dt if g(c) nd g(d) b (tht is, ϕ() ψ(c) nd ϕ(b) ψ(d), nd d c F (ψ(t)) Dψ(t)dt b F (ϕ(g(t)) Dϕ(g(t))g (t)dt. (4.3.6) b F (ϕ(u)) Dϕ(u)dt (4.3.7) F (ϕ(u)) Dϕ(u)dt F (ϕ(u)) Dϕ(u)dt (4.3.8) b if g(c) b nd g(d) (tht is, ϕ() ψ(d) nd ϕ(b) ψ(c). Note tht the second cse occurs only if ψ prmetrizes in the reverse direction of ϕ, in which cse we sy ψ is n orienttion reversing reprmetriztion of ϕ. In the first cse, tht is, when ϕ() ψ(c) nd ϕ(b) ψ(d), we sy ψ is n orienttion preserving reprmetriztion of ϕ. Our results in (4.3.7) nd (4.3.8) then correspond to the physicl notion tht the work done by force in moving n object long curve is the negtive of the work done by the force in moving the object long the curve in the opposite direction. From now on, when referring to curve, we will ssume some orienttion, or direction, hs been specified. We will then use to refer to the curve consisting of the sme set of points s, but with the reverse orienttion. Line integrls Now tht we know tht, except for direction, the vlue of the integrl involved in computing work does not depend on the prticulr prmetriztion of the curve, we my stte forml mthemticl definition. Definition Suppose is curve in R n with smooth prmetriztion ϕ : I R n, where I [, b] is n intervl in R. If F : R n R n is continuous vector field, then we define the line integrl of F long, denoted F ds, by F ds b F (ϕ(t)) Dϕ(t)dt. (4.3.9)
5 Section 4.3 Line Integrls 5 As consequence of our previous remrks, we hve the following result. Proposition Using the nottion of the definition, F ds depends only on the curve nd its orienttion, not on the prmetriztion ϕ. Moreover, F ds F ds. (4.3.1) Exmple Let be the unit circle centered t the origin in R 2, oriented in the counterclockwise direction, nd let ( F (x, y) y ) x 2 + y 2, x 1 x 2 + y 2 x 2 ( y, x). + y2 To compute the line integrl of F long, we first need to find smooth prmetriztion of. One such prmetriztion is for t 2π. Then F ds 2π 2π 2π 2π 2π. ϕ(t) (cos(t), sin(t)) F (cos(t), sin(t)) ( sin(t), cos(t))dt 1 cos 2 (t) + sin 2 ( sin(t), cos(t)) ( sin(t), cos(t))dt (t) (sin 2 (t) + cos 2 (t))dt dt Note tht ψ(t) (sin(t), cos(t)), t 2π, prmetrizes, from which we cn clculte 2π F ds F (sin(t), cos(t)) (cos(t), sin(t))dt 2π 2π 2π 2π, 1 sin 2 ( cos(t), sin(t)) (cos(t), sin(t))dt (t) + cos 2 (t) ( cos 2 (t) sin 2 (t))dt in greement with the previous proposition. dt
6 6 Line Integrls Section Figure Rectngle with counterclockwise orienttion A piecewise smooth curve is one which my be decomposed into finite number of curves, ech of which hs smooth prmetriztion. If is piecewise smooth curve composed of the union of the curves 1, 2,..., m, then we my extend the definition of the line integrl to by defining F ds F ds + 1 F ds F ds. m (4.3.11) The next exmple illustrtes this procedure. Exmple Let be the rectngle in R 2 with vertices t (, ), (2, ), (2, 1), nd (, 1), oriented in the counterclockwise direction, nd let F (x, y) (y 2, 2xy). If we let 1, 2, 3, nd 4 be the four sides of, s lbeled in Figure 4.3.3, then we my prmetrize 1 by α(t) (t, ), t 2, 2 by t 1, 3 by t 2, nd 4 by β(t) (2, t), γ(t) (2 t, 1), δ(t) (, 1 t), t 1. Then F ds F ds + 1 F ds + 2 F ds + 3 F ds 4 2 F (t, ) (1, )dt + 1 F (2, t) (, 1)dt + 2 F (2 t, 1) ( 1, )dt
7 Section 4.3 Line Integrls (, ) (1, )dt + + dt F (, 1 t) (, 1)dt 1 (t 2, 4t) (, 1)dt + ((1 t) 2, ) (, 1)dt 4tdt + 2 ( 1)dt + 1 dt 2 (1, 4 2t) ( 1, )dt 2t Note tht it would be slightly simpler to prmetrize 3 nd 4, using ϕ(t) (1, t), t 2, nd ψ(t) (t, ), t 1, respectively, thn to prmetrize 3 nd 4 directly. We would then evlute F ds F ds + F ds F ds F ds A note on nottion Suppose is smooth curve in R n, prmetrized by ϕ : I R n, where I [, b], nd let F : R n R n be continuous vector field. Our nottion for the line integrl of F long comes from letting s ϕ(t), from which we hve which we my write, symboliclly, s ds dt Dϕ(t), ds Dϕ(t)dt. Now suppose ϕ 1, ϕ 2,..., ϕ n nd F 1, F 2,..., F n re the component functions of ϕ nd F, respectively. If we let x 1 ϕ 1 (t), x 2 ϕ 2 (t),. x n ϕ n (t),
8 8 Line Integrls Section 4.3 then we my write b F ds F (ϕ(t)) Dϕ(t)dt b b b F (x 1 (t), x 2 (t),..., x n (t)) (ϕ 1(t), ϕ 2(t),..., ϕ n(t))dt (F 1 (x 1 (t), x 2 (t),..., x n (t))ϕ 1(t) + F 2 (x 1 (t), x 2 (t),..., x n (t))ϕ 2(t)) + + F n (x 1 (t), x 2 (t),..., x n (t))ϕ n(t))dt F 1 (x 1 (t), x 2 (t),..., x n (t))ϕ 1(t)dt b b F 2 (x 1 (t), x 2 (t),..., x n (t))ϕ 2(t)dt F n (x 1 (t), x 2 (t),..., x n (t))ϕ n(t)dt. (4.3.12) Suppressing the dependence on t, writing dx k for ϕ k (t)dt, k 1, 2,..., n, nd using only single integrl sign, we my rewrite (4.3.12) s F 1 (x 1, x 2,..., x n )dx 1 + F 2 (x 1, x 2,..., x n )dx F n (x 1, x 2,..., x n )dx n. (4.3.13) This is common, nd useful, nottion for line integrl. Exmple We will evlute ydx + xdy + z 2 dz, where is the prt of helix in R 3 with prmetric equtions x cos(t), y sin(t), z t, t 2π. Note tht this is equivlent to evluting F ds, where F : R 3 R 3 is the vector field F (x, y, z) (y, x, z 2 ). We hve 2π ydx + xdy + z 2 dz (sin(t)( sin(t)) + cos(t) cos(t) + t 2 )dt 2π 2π (cos 2 (t) sin 2 (t) + t 2 )dt (cos(2t) + t 2 )dt 1 2π 2 sin(2t) + 1 2π 3 t3 8π3 3.
9 Section 4.3 Line Integrls 9 Grdient fields Recll tht if f : R n R is 1, then f is continuous vector field on R n. Suppose ϕ : I R n, I [, b], is smooth prmetriztion of curve. Then, using the chin rule nd the Fundmentl Theorem of lculus, f ds b b f(ϕ(t)) Dϕ(t)dt d dt f(ϕ(t))dt f(ϕ(t)) b f(ϕ(b)) f(ϕ()). Theorem If f : R n R is 1 nd ϕ : I R n, I [, b], is smooth prmetriztion of curve, then f ds f(ϕ(b)) f(ϕ()). (4.3.14) Note tht (4.3.14) shows tht the vlue of line integrl of grdient vector field depends only on the strting nd ending points of the curve, not on which prticulr pth is tken between these two points. Moreover, (4.3.14) provides simple mens for evluting line integrl if the given vector field cn be identified s the grdient of sclr vlued function. Another interesting consequence is tht if the beginning nd ending points of re the sme, tht is, if v ϕ() ϕ(b), then f ds f(ϕ(b)) f(ϕ(b)) f(v) f(v). (4.3.15) We cll such curves closed curves. In words, the line integrl of grdient vector field is long ny closed curve. Exmple If F (x, y) (y, x), then F (x, y) f(x, y), where f(x, y) xy. Hence, for exmple, for ny smooth curve strting t ( 1, 1) nd ending t (1, 1) we hve F ds f(1, 1) f( 1, 1) Note tht this grees with the result in our first exmple bove, where ws the prt of the prbol y x 2 extending from ( 1, 1) to (1, 1). Exmple If f(x, y) xy 2, then f(x, y) (y 2, 2xy).
10 1 Line Integrls Section 4.3 If is the rectngle in R 2 with vertices t (, ), (2, ), (2, 1), nd (, 1), then, since is closed curve, y 2 dx + 2xydy, in greement with n erlier exmple. Similrly, if E is the unit circle in R 2 centered t the origin, then we know tht y 2 dx + 2xydy, with no need for further computtions. E In physics, force field F is sid to be conservtive if the work done by F in moving n object between ny two points depends only on the points, nd not on the pth used between the two points. In prticulr, we hve shown tht if F is the grdient of some sclr function f, then F is conservtive force field. Under certin conditions on the domin of F, the converse is true s well. Tht is, under certin conditions, if F is conservtive force field, then there exists sclr function f such tht F f. Problem 9 explores one such sitution in which this is true. The function f is then known s potentil function. Problems 1. For ech of the following, compute the line integrl F nd curve prmetrized by ϕ. () F (x, y) (xy, 3x), ϕ(t) (t 2, t), t 2 ( ) y (b) F (x, y) x 2 + y 2, x x 2 + y 2, ϕ(t) (cos(t), sin(t)), t 2π (c) F (x, y) (3x 2y, 4x 2 y), ϕ(t) (t 3, t 2 ), 2 t 2 (d) F (x, y, z) (xyz, 3xy 2, 4z), ϕ(t) (3t, t 2, 4t 3 ), t 4 F ds for the given vector field 2. Let be the circle of rdius 2 centered t the origin in R 2, with counterclockwise orienttion. Evlute the following line integrls. () 3xdx + 4ydy (b) 8xydx + 4x 2 dy 3. Let be the prt of helix in R 3 prmetrized by ϕ(t) (cos(2t), sin(2t), t), t 2π. Evlute the following line integrls. () 3xdx + 4ydy + zdz (b) yzdx + xzdy + xydz 4. Let be the rectngle in R 2 with vertices t ( 1, 1), (2, 1), (2, 3), nd ( 1, 3), with counterclockwise orienttion. Evlute the following line integrls. () x 2 ydx + (3y + x)dy (b) 2xydx + x 2 dy
11 Section 4.3 Line Integrls Let be the ellipse in R 2 with eqution x y2 9 1, with counterclockwise orienttion. Evlute F ds for F (x, y) (4y, 3x). 6. Let be the upper hlf of the circle of rdius 3 centered t the origin in R 2, with counterclockwise orienttion. Evlute the following line integrls. () 3ydx (b) 4xdy 7. Evlute x x 2 + y 2 dx + y x 2 + y 2 dy, where is ny curve which strts t (1, ) nd ends t (2, 3). 8. () Suppose F : R n R n is 1 vector field which is the grdient of sclr function f : R n R. If F k is the kth coordinte function of F, k 1, 2,..., n, show tht x j F i (x 1, x 2,..., x n ) for i 1, 2,..., n nd j 1, 2,..., n. (b) Show tht lthough x i F j (x 1, x 2,..., x n ) xdx + xydy for every circle in R 2 with center t the origin, nevertheless F (x, y) (x, xy) is not the grdient of ny sclr function f : R n R. (c) Let ( F (x, y) y ) x 2 + y 2, x x 2 + y 2 for ll (x, y) in the set S {(x, y) : (x, y) (, )}. Let F 1 nd F 2 be the coordinte functions of F. Show tht y F 1(x, y) x F 2(x, y) for ll (x, y) in S, even though F is not the grdient of ny sclr function. (Hint: For the lst prt, show tht F ds 2π, where is the unit circle centered t the origin.)
12 12 Line Integrls Section Suppose F : R 2 R 2 is continuous vector field with the property tht for ny curve, F ds depends only on the endpoints of. Tht is, if 1 nd 2 re ny two curves with the sme endpoints P nd Q, then () Show tht for ny closed curve. F ds F ds. 1 2 F ds (b) Let F 1 nd F 2 be the coordinte functions of F. Define f : R 2 R by f(x, y) F ds, where is ny curve which strts t (, ) nd ends t (x, y). Show tht y f(x, y) F 2(x, y). (Hint: In evluting f(x, y), consider the curve from (, ) to (x, y) which consists of the horizontl line from (, ) to (x, ) followed by the verticl line from (x, ) to (x, y).) (c) Show tht f F.
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