Line Integrals. Chapter Definition

hpter 2 Line Integrls 2.1 Definition When we re integrting function of one vrible, we integrte long n intervl on one of the xes. We now generlize this ide by integrting long ny curve in the xy-plne. It is remrkble tht this process hs close connection with double integrls. Suppose then nd B re points in R 2 nd suppose is curve whose end-points re nd B. hoose points (x 0, y 0 ), (x 1, y 1 ),..., (x n, y n ) B on in such wy tht (x k, y k ) lies between (x k 1, y k 1 ) nd (x k+1, y k+1 ). Now let x k x k x k 1 nd y k y k y k 1, k 1,..., n. hoose (ξ k, η k ) on lying between (x k, y k ) nd (x k 1, y k 1 ) (see Figure 2.1). B(x n, y n ) (x k, y k ) (x 1, y 1 ) (x k 1, y k 1 ) (ξ k, η k ) (x 0, y 0 ) Figure 2.1 Suppose P nd Q re continuous functions on. Form the sum S n { } P (ξk, η k ) x k + Q(ξ k, η k ) y k k1 nd let n in such wy tht mx{ x k + y k } 0. Then it my be shown tht lim S n exists nd is independent of the choice of (x k, y k ) nd (ξ k, η k ). 1

2 HPTER 2. LINE INTERLS Line Integrl We denote this limit by or B P (x, y) dx + Q(x, y) dy P (x, y) dx + Q(x, y) dy (where the curve is specified). This is known s the line integrl from to B long. The process outlined bove extends to curves in R n (for n > 2) in the nturl wy. In prticulr in R 3, let P 1, P 2 nd P 3 be continuous functions of x, y nd z nd let be given curve. Then P 1 (x, y, z) dx + P 2 (x, y, z) dy + P 3 (x, y, z) dz lim k1 { } P1 (ξ k, η k, ζ k ) x k + P 2 (ξ k, η k, ζ k ) y k + P 3 (ξ k, η k, ζ k ) z k where the vribles hve the obvious menings. Letting P P 1 i + P 2 j + P 3 k nd dr dx i + dy j + dz k, we my write this integrl concisely s P dr. Similrly, if s k is the rc length long from (x k 1, y k 1 ) to (x k, y k ), one my define P (x, y) ds s lim k1 where the limit is tken s mx s k 0. P (ξ k, η k ) s k Theorem 2.1.1 Suppose P nd Q re continuous on the curve. Then () P (x, y) dx + Q(x, y) dy P (x, y) dx + Q(x, y) dy (b) Let hve end-points (, b) nd (e, f). If (c, d) is nother point on, then (e,f) (,b) P dx + Q dy (c,d) (,b) P dx + Q dy + (e,f) (c,d) P dx + Q dy where ll the integrls re evluted long or the relevnt prt of it. (c) If hs end points nd B then B ( P dx + Q dy B ) P dx + Q dy.

2.1. DEFINITION 3 Proof. () P (x, y) dx + Q(x, y) dy lim k1 lim k1 P dx + { } P (ξk, η k ) x k + Q(ξ k, η k ) y k P (ξ k, η k ) x k + lim Q dy k1 Q(ξ k, η k ) y k (b) Divide up the curve by the points (, b) (x 0, y 0 ), (x 1, y 1 ),..., (x n, y n ) (e, f) nd let (x m, y m ) (c, d) where 0 < m < n. Then m P (ξ k, η k ) x k P (ξ k, η k ) x k + k1 k1 nd it follows on tking limits s x k 0 tht (e,f) (,b) P dx The sme result holds for the integrls of Q. (c) Using the result of (b) B nd the result follows. P dx + Q dy + B (c,d) (,b) P dx + P dx + Q dy P (ξ k, η k ) x k m+1 (e,f) (c,d) P dx. P dx + Q dy 0 The next theorem is the essentil result tht enbles us to evlute line integrls esily. Theorem 2.1.2 Suppose P nd Q re continuous nd is described by the eqution y f(x) where f is differentible. Then (c,d) (,b) P (x, y) dx + Q(x, y) dy c P (x, f(x)) dx + c Q(x, f(x))f (x) dx. Proof. where (ξ k, η k ) lies on. (c,d) (,b) It follows tht η k f(ξ k ). Hence (c,d) (,b) P (x, y) dx lim P (x, y) dx lim k1 k1 P (ξ k, η k ) x k P (ξ k, f(ξ k )) x k c P (x, f(x)) dx.

4 HPTER 2. LINE INTERLS Furthermore (c,d) (,b) Q(x, y) dy lim k1 lim k1 lim k1 lim k1 Q(ξ k, η k ) y k Q(ξ k, f(ξ k ))(y k y k 1 ) Q(ξ k, f(ξ k ))(f(x k ) f(x k 1 )) Q(ξ k, f(ξ k ))(x k x k 1 )f (ζ k ), where ζ k lies between x k nd x k 1 (using the Men Vlue Theorem). Since x k x k 1 x k, we obtin c lim Q(ξ k, f(ξ k ))f (ξ k ) x k Q(x, f(x))f (x) dx, s required. k1 Remrks. 1. If is given by the eqution x g(y) similr result holds. 2. It is now simple mtter to see tht if is described by the prmetric equtions x φ(t), y ψ(t) then t1 { P (x, y) dx + Q(x, y) dy P (φ(t), ψ(t))φ (t) + Q(φ(t), ψ(t))ψ (t) } dt, t 2 where the end-points of re given by (φ(t 1 ), ψ(t 1 )) nd (φ(t 2 ), ψ(t 2 )). Exmple 2.1.1 Solution Evlute (2,3) (3x 2 y + 2xy) dx + (2xy 2 3y) dy (1,1) firstly long the stright line joining the end-points nd then long the curve which consists of two stright line segments from (1, 1) to (1, 3) nd then from (1, 3) to (2, 3). In the first cse the curve hs eqution y 2x 1. Using Theorem 3.1.2 nd noting tht dy 2 dx, we obtin 2 1 [3x 2 (2x 1) + 2x(2x 1)] dx + [2x(2x 1) 2 3(2x 1)]2 dx 77/2. In the second cse we evlute (1,3) (1,1), then (2,3) (1,3) nd dd. In moving from (1, 1) to (1, 3), x 1 so tht dx 0. We hve 3 1 (2y 2 3y) dy 16/3.

2.1. DEFINITION 5 In going from (1, 3) to (2, 3), y 3 nd dy 0. We obtin 2 1 (9x 2 + 6x) dx 30, giving totl of 16/3 + 30 106/3. Note tht the vlue of the line integrl depends on the pth chosen. Exmple 2.1.2 Evlute (4,8) (1,1) (x 2 + 2y) dx + (3x y) dy long the curve with prmetric equtions x t 2 nd y t 3. Solution First we see tht t (1, 1), t 1 nd t (4, 8), t 2. Proceeding formlly, dx 2t dt nd dy 3t 2 dt, so tht the integrl becomes Exmple 2.1.3 2 1 (t 4 + 2t 3 )2t dt + (3t 2 t 3 )3t 2 dt 701/10. Solution Evlute (1,2,3) (0,0,0) xyz dx + (3y 2 + z) dy + 4x 2 z 3 dz long the pth which goes long stright line segments from (0, 0, 0) to (1, 0, 0), then from (1, 0, 0) to (1, 2, 0) nd thence to (1, 2, 3). For the first prt y z 0 nd so dy dz 0. We hve 1 0 0 dx 0. For the second prt x 1, z 0, so tht dx dz 0 nd we obtin 2 0 3y 2 dy 8. Finlly we hve x 1, y 2, so dx dy 0 nd the integrl is On dding we obtin totl of 89. 3 0 4z 3 dz 81.

6 HPTER 2. LINE INTERLS Exmple 2.1.4 Evlute (1,2) (0,0) (3x 2 + 2y 2 ) ds long the stright line joining (0, 0) nd (1, 2), where s denotes rc length. Solution Let the line be described by the prmetric equtions x t, y 2t. Reclling tht ds/dt [(dx/dt) 2 + (dy/dt) 2 ] 1/2 5, we obtin 5 1 0 (3t 2 + 8t 2 ) dt 11 5 3. In evluting line integrls it is of prticulr interest to consider the cse where the curve is closed, tht is to sy tht its strting point nd end point re the sme. Simple losed urve Definition 2.1.3 simple closed curve is curve x φ(t), y ψ(t), t [t 1, t 2 ] in which φ(t 1 ) φ(t 2 ), ψ(t 1 ) ψ(t 2 ) nd such tht if s, t (t 1, t 2 ) with s t then (φ(t), ψ(t)) (φ(s), ψ(s)). Thus simple closed curve is one tht returns to its strting point nd does intersect itself (Figure 2.2). If is closed curve, we my trverse it in one of two directions. Henceforth, unless stted otherwise, we shll lwys move round in n nti-clockwise direction, tht is to sy tht s we proceed long the curve the region bounded by the curve will lwys be on the left. simple closed curve non-simple closed curve Figure 2.2 If is simple closed curve, we shll denote the line integrl round it (in the nti-clockwise direction) by. Simply onnected Region Definition 2.1.4 region of the plne is simply connected if its boundry is simple closed curve; otherwise it is multiply connected.

2.2. REEN S THEOREM 7 simply connected multiply connected Figure 2.3 2.2 reen s Theorem We re now redy to resume our study of line integrls. The next result yields the fundmentl connection between line integrls nd double integrls. Theorem 2.2.1 (reen s Theorem in the Plne) Let P, Q, P/ nd Q/ x be continuous on simply connected region with boundry. Then ( Q P dx + Q dy x P ) dy dx. Proof. We shll prove the theorem only in the cse tht the region is such tht ech line prllel to the xes cuts in t most two points. Hving done tht, we shll explin briefly how vrious extensions my be mde. Let consist of two segments EB nd BD s shown in Figure 2.4, nd let them hve equtions y f(x) nd y g(x) respectively. D y g(x) B y f(x) E.. b Figure 2.4 Then P dy dx b g(x) f(x) b b EB EB P (x, y) P dy dx ] g(x) f(x) dx { P (x, f(x)) P (x, g(x)) } dx P dx P dx + DB BD P dx P dx P dx.

8 HPTER 2. LINE INTERLS Similrly Q dy dx Q dy. x This proof is redily to the cse where prt of the contour consists of stright line segment prllel to one of the xes. Suppose for exmple the sitution is s shown in Figure 2.5. F E g(x) B. D f(x). b Figure 2.5 s before, P dy dx But on F, x, so dx 0 nd hence s required. b F DB DB ( P (x, f(x)) P (x, g(x)) ) dx P dx P dx + F EB BEF P dx 0. Thus P dx P dx. P dy dx P dx + P dx + P dx P dx, DB BEF F DBEF The cse where horizontl or verticl line cuts the curve in infinitely mny isolted points is hrder to del with. Before we show how the proof my be extended to more generl regions, let us give two exmples. Exmple 2.2.1 Verify reen s Theorem in the plne for xy 2 dx + 2xy 3 dy, where is the boundry of the region bounded by the curves y x 2 nd y x.

2.2. REEN S THEOREM 9 y (1, 1) y x y x 2 x Figure 2.6 Solution long the section where y x 2 we hve dy 2x dx nd the integrl is 1 0 x 5 dx + 2x 7 2x dx 11/18. lso long the section where y x (trversed from (1, 1) to (0, 0)), we hve dy dx, which yields 0 nd the totl is 11/18 13/20 7/180. Now s expected. ( Q x P ) dy dx 1 (x 3 + 2x 4 ) dx 13/20, 1 x 0 1 0 1 0 7/180, (2y 3 2xy) dy dx x 2 [y 4 /2 xy 2] x dx x 2 (x 4 /2 x 3 x 8 /2 + x 5 ) dx Sometimes reens Theorem my be used to evlute very intrctble line integrls. Exmple 2.2.2 Solution Evlute where is the circle x 2 + y 2 1. (y + e x2 ) dx + (3x y 3 + 1) dy ny ttempt to evlute the line integrl directly is futile. However, using reens Theorem we hve (y + e x2 ) dx + (3x ( y 3 + 1) dy D x (3x y 3 + 1) ) 2 (y + ex ) dx dy (3 1) dx dy D 2 d 2π D

10 HPTER 2. LINE INTERLS (where D is the disk x 2 + y 2 1). Now let us return to generliztions of the proof. First suppose the region is s shown in Figure 2.7, where verticl line cuts the boundry in four plces.. Figure 2.7 Divide up into two sub-regions, 1 nd 2 s shown in Figure 2.8, which ech verticl line cuts the boundries of 1 nd 2 just once. E 1... 2 B Figure 2.8 D Now omitting the integrnds, we hve DB + BE DB DB BD 1 +. + + + B B + BE 2 B B + + BE BE by Theorem 3.1.1(c) by the specil cse lredy proved We cn extend reen s Theorem to multiply connected regions in the following wy. onsider the region shown in Figure 2.9. Mke cut E from the outer bounding curve 1 to the inner curve 2. Then, by wht we lredy know. EF KEBD

2.2. REEN S THEOREM 11 1 D E K 2 F Figure 2.9 B Now the line integrl is E + + +. EF KE E BD 1 2 Thus ( Q x Q ) dx dy (P dx + Q dy) (P dx + Q dy). 1 2 If you prefer, you cn think of this s (P dx + Q dy) + (P dx + Q dy) 1 2 where the integrl round the outside contour, 1 is trversed nti-clockwise s usul, but the integrl round the inner contour 2 is trversed clockwise. Exmple 2.2.3 Evlute Γ y dx x 2 + y 2 + x dy x 2 + y 2 where Γ is the curve consisting of the prbol y x 2 2 for 2 x 2 together with the line segment from ( 2, 2) to (2, 2) (see Figure 2.10). Γ (2, 2) y x 2 2 Figure 2.10 Solution We cnnot use reen s Theorem directly becuse the integrnd in not continuous t the origin. To void this problem we consider the region, which lies inside Γ but outside the circle of rdius 1, centre (0, 0). We hve ( Q (P dx + Q dy) (P dx + Q dy) x P ) dx dy (2.1) Γ

12 HPTER 2. LINE INTERLS where P y/(x 2 + y 2 ) nd Q x/(x 2 + y 2 ). It is esy to check tht the right-hnd side of (3.1) is zero nd so. Setting x cos t nd y sin t the right-hnd side reduces to Γ 2π 0 dt 2π. reen s Theorem is useful in proving some theoreticl results which mke subsequent work much esier. Theorem 2.2.2 Let P nd Q be continuous with continuous prtil derivtives over simply connected region. Then P dx + Q dy 0 for every simple closed curve contined in if nd only if in. Proof. First suppose tht P Q x P dx + Q dy 0. If the conclusion were flse there would exist point (x 0, y 0 ) t which P/ Q/ x. Let us suppose, without loss of generlity, tht P (x 0, y 0 ) Q x (x 0, y 0 ) > 0. Since P/ Q/ x is continuous there must exist subregion with boundry which contins (x 0, y 0 ) nd on which P/ Q/ x > 0. But then ( Q 0 P dx + Q dy x P ) dx dy < 0. Hence the conclusion must be true. onversely suppose in. Then clerly P dx + Q dy Q x P ( Q x P ) dx dy 0. Theorem 2.2.3 Let P nd Q be continuous with continuous prtil derivtives. Then, given ny two points nd B, P dx + Q dy is independent of the pth tken from to B if nd only if B Q x P.

2.2. REEN S THEOREM 13 Proof. Suppose P/ Q/ x. hoose two points nd B nd let 1 nd 2 be two pths from to B. To void the possibility tht 1 nd 2 fil to form simple closed curve, tke nother pth 3, from to B which hs no points in common with 1 or 2 except nd B. E 3 B D 2 1 Figure 2.11 Tking points D nd E on 1 nd 3 respectively, DBE forms closed curve. Hence by Theorem 3.2.2 P dx + Q dy 0. But then Hence Similrly 0 DBE DBE DB 2 +. BE DB EB 1 3 3 1 nd so 3. 1 onversely, suppose tht, given ny two points nd B, whose end-points re nd B. 2. 1 2 for ny curves 1, 2 Let be ny simple closed curve. hoose nd B on, nd lbel 1 nd 2 s shown in Figure 2.12. 2 E B D 1 By hypothesis, Then 1 2, tht is DB 0 Figure 2.12 P dx + Q dy DB EB P dx + Q dy. +, EB DB BE

14 HPTER 2. LINE INTERLS which is to sy 0. Since ws n rbitrry closed curve, it follows from 3.2.2 tht Q/ x P/. Let Then This result my be rephrsed s follows: F(x, y, z) P (x, y)i + Q(x, y)j. curl F Hence curl F 0 if nd only if P/ Q/ x. i j k x z P Q 0 P z j + Q x k P k Q z i ( Q x P ) k. This formultion my be used to extend Theorem 3.2.3 to R 3, where P dx + Q dy + R dz B is independent of the pth tken if nd only if curl F 0, where F P i + Qj + Rk. We shll show this lter. Next note tht if F f, for some smooth function f then, by by Theorem?? curl F curl (grd f) 0. Hence if F f then by the rgument bove the line integrl is independent of the pth. D(x 0, y 1 ) E(x 1, y 1 ) (x 0, y 0 ) B(x 1, y 0 ) Figure 2.13 onversely, suppose, the line integrl is independent of the pth. Let (x 0, y 0 ) be fixed nd E (x 1, y 1 ) be ny point. Set D (x 0, y 1 ) nd B (x 1, y 0 ) (Figure 2.13). Let By hypothesis this is equl to f(x 1, y 1 ) BE DE P dx + Q dy. P dx + Q dy.

2.2. REEN S THEOREM 15 Now DE D On D we note tht x x 0 nd so dx 0. Thus P dx + Q dy Similrly Hence D DE f(x 1, y 1 ) P dx + Q dy y1 +. DE y1 y 0 x1 x 0 y 0 Q(x 0, y) dy + Then, by the first fundmentl theorem of clculus, Q(x 0, y) dy. P (x, y 1 ) dx. x1 f x 1 P (x 1, y 1 ), since the first integrl does not depend on x 1. Similrly x1 f(x 1, y 1 ) P (x, y 0 ) dx + BE x 0 x 0 P (x, y 1 ) dx. y1 y 0 Q(x 1, y) dy nd so Dropping the subscripts, we hve f 1 Q(x 1, y 1 ). P (x, y) f x nd Q(x, y) f. Thus F P i + Qj f. Hence if the line integrl is independent of the pth, then F is the grdient of some function f. In summry we hve the following result. Theorem 2.2.4 Suppose F P i + Qj where P nd Q hve continuous prtil derivtives. Then the following re equivlent. () Q x P (b) F dr 0 for every simple closed curve (c) For ny fixed nd B, (d) curl F 0 B (e) F f for some function f. F dr is independent of the pth between nd B

16 HPTER 2. LINE INTERLS Exmple 2.2.4 Evlute (xy 2 4x 3 y) dx + (x 2 y x 4 + y 2 ) dy where is the curve shown in Figure 2.14, which hs the form of sine wve of mplitude 1. (1, 1) Solution Figure 2.14 lerly it would be very difficult even to find the eqution of. However Hence P xy 2 4x 3 y nd Q x 2 y x 4 + y 2. P x Q. This mens tht the integrl is independent of the pth nd we my s well proceed long the stright line y x. The integrl reduces to 1 0 (2x 3 5x 4 + x 2 ) dx 1/2 1 + 1/3 1/6. 2.3 pplictions of Line Integrls Work First body moves from point to nother point B under constnt force F cting in the direction of motion. It is well known then tht the work done by the force is Force distnce moved. More generlly, suppose the body moves in R 2 long stright line from point to point B under force F (Figure 2.15). The component of F perpendiculr to the line of motion mkes no contribution to the motion. Thus if we let d B, then the work done is W F d cos θ F d where θ is the ngle between F nd the direction of motion.

2.3. PPLITIONS OF LINE INTERLS 17 F θ... B Now note tht Figure 2.15 d d d e B d where e d/ d is unit vector in the direction of motion nd B is the distnce moved. Thus W F e B. (2.2) Now suppose the body moves on curve. Let D nd E be nerby points on the curve, nd s be the rc-length long between D nd E (Figure 2.16). F T s D Figure 2.15 Let F be the force cting on the body nd T be the unit tngent vector t. If s is smll the T point be pproximtely in the direction D. Then, by (3.2) the work done from to D is W F T s. Summing over ll such rcs D gives This leds us to the following definition. W F T s. Definition 2.3.1 Let F be given force cting on body nd be smooth curve with end point nd B. Suppose T be the unit tngent vector to. Then the work done by the force in moving the body long the curve is W F T ds. then If is given by r(t) x(t)i + y(t)j + z(t)k, T(t) r (t) r (t) t b

18 HPTER 2. LINE INTERLS nd Hence ds (dx dt W ) 2 + ( dy dt b b B ) 2 ( dz ) 2 + dt r (t) dt. dt F(r(t)) F(r(t)) r (t) dt F(r) dr. r (t) r (t) r (t) dt Before proceeding, we need version of the fundmentl theorem of clculus. Theorem 2.3.2 Let be smooth curve given by r(t) (x(t), y(t), z(t)), where t b nd f : R 3 R be differentible function with continuous prtil derivtives. Then f dr f(r(b)) f(r()). Proof. We hve f dr b b b b f(r(t)) r (t) dt ( f x i + f j + f ) z k ( f dx x dt + f dy dt + f z df(r(t)) dt dt f(r(b)) f(r()). ( dx dt i + dy dt j + dz ) dt k dt ) dt dz dt Newton s Lw of rvittion Newton s Lw of grvittion sttes tht if two bodies of msses M nd m re seprted by distnce r then the mgnitude of the grvittionl force between them is given by F mm r 2 where is the universl grvittionl constnt. Suppose the mss M lies t the origin. Let r (x, y, z), so tht r r. Then the grvittionl force on m is towrds the origin nd the unit vector in this direction is r r. Hence Next note tht if F F r r mm r 3 r. f(x, y, z) mm x2 + y 2 + z 2

2.3. PPLITIONS OF LINE INTERLS 19 then f F(x, y, z) which is to sy tht F is grdient field. It follows from Theorem 3.3.2 tht the work done in moving the mss m from (x 1, y 1, z 1 ) to b (x 2, y 2, z 2 ) is mm b mm. Note tht no work is done if m returns to its strting plce. onservtion of Energy Suppose body of mss m moves under the influence of force F long pth given by r(t) where t [, b]. Then, by Newton s second lw Hence Thus W F(t) mr (t). b m m 2 m 2 F dr b F(r(t)) r (t) dt b b r (t) r (t) dt d dt [r (t) r (t)] dt d dt r (t) 2 dt m 2 r (b) 2 m 2 r () 2. W m ( v(b) 2 v() 2) 2 K(B) K() (2.3) where K is the kinetic energy t velocity v. Thus (3.3) shows tht work done chnge in kinetic energy. Now suppose F is conservtive field, so tht F f for some f. Let P (x, y, z) f(x, y, z). Then P is clled the potentil energy. Thus we hve F P. Hence W F dr P dr It follows from (3.3) nd (3.4) tht [P (r(b)) P (r())] (by Theorem 3.3.2) P () P (B). (2.4) K() + P () K(B) + P (B). Thus the totl energy remins the sme. We hve shown the following importnt principle.

20 HPTER 2. LINE INTERLS In conservtive system the sum of the kinetic nd potentil energies does not chnge.