Instrutor: Longfei Li Mth 43 Leture Notes 16. Line Integrls The invention of line integrls is motivted by solving problems in fluid flow, fores, eletriity nd mgnetism. Line Integrls of Funtion We n integrte funtion over ny urve, where is defined prmetrilly s x x(t), y y(t), t b We define the line integrl in the following wy: if we divide the prmeter intervl [, b] into n subintervls of equl wih: t < t 1 < t < < t n 1 < t n b, nd denote x i x(t i ), y i y(t i ), then the orresponding points P i (x i, y i ) divides the urve into n subrs s shown below. We then hoose ny smple points Pi (x i, y i ) in the ith subr, then we define the Riemnn sum: f(x i, yi ) S i, where S i is the length of the i th subr. Then we n formlly give the definition for the line integrls. Definition: If f is defined on smooth urve given prmetrilly s x x(t), y y(t), t b, then the line integrl of f long is f(x, y)ds lim f(x i, yi ) S i Rell: The r length funtion S(t) is S(t) t r (t) t (dx ) + So by Fundmentl Theorem of lulus: (dx ) ds r (t) + 1
Therefore, we my evlute the line integrl of f long the urve by evluting the following integrtion: b (dx ) f(x, y)ds f(x(t), y(t)) + Exmple: Evlute ( + x y) ds, where is the upper hlf of the unit irle x + y 1. Firstly, we need prmetri eqution of the urve. Sine is the upper hlf of unit irle, we my use the following prmetri equtions: So ( + x y) ds π x os(t), y sin(t), t π ( + os (t) sin(t) ) ( sin(t)) + os (t) π + 3 Property of line integrls: If is pieewise -smooth urve; tht is, is union of finite number of smooth urves 1,,..., n, then f(x, y) ds f(x, y) ds + 1 f(x, y) ds + + f(x, y) ds n Exmple: Evlute x ds, where onsists of the r 1 of the prbol y x from (, ) to (1, 1) followed by the vertil line segment from (1, 1) to (1, ). x ds x ds + 1 x ds Find prmetri equtions for 1 : x t, y t, t 1. Find prmetri equtions for : x 1, y t, 1 t.
So x ds 1 Line integrls of f long w.r.t x nd y: t 1 + 4t + f(x, y) dx lim f(x, y) dy lim 1 + 1 5 5 1 6 f(x i, yi ) x i f(x i, yi ) y i For distintion, we ll f(x, y) ds the line integrl of f long w.r.t r length. If line integrls w.r.t x nd y our together, then we write, for short, P (x, y) dx + Q(x, y) dy P (x, y) dx + Q(x, y) dy Line integrls in spe: If is smooth urve in spe given by prmetri equtions x x(t), y y(t), z z(t), t b (or by vetor eqution r(t) x(t)i + y(t)j + z(t)k), then the line integrl of f(x, y, z) long urve is defined s f(x, y, z) ds lim f(x i, yi, zi ) S i Similrly, we n evlute the line integrls in spe s b (dx ) f(x, y, z) ds f(x(t), y(t), z(t)) + Or using vetor nottion: f(x, y, z) ds b f(r(t)) r (t) + ( ) dz Exmple: Evlute y sin z ds, where is the helix given by x os t, y sin t, z t, t π. 3
π y sin z ds sin t sin t sin t + os t + 1 π sin t π Line Integrls of Vetor Fields We hve lerned tht the work done by onstnt fore F in moving n objet from point P to Q in spe is W F D, where D is the displement vetor. Now suppose F P i + Qj + Rk is ontinuous fore in R 3. We wnt to find the work done by this fore in moving prtile long smooth urve given by r(t), t b. We divide into subrs with length S i, then hoose smple point Pi (x i, y i, z i ) on the ith sub r orresponding to t t i. Thus the work done in moving the prtile from P i to P i+1 is pproximtely, W i F(x i, y i, z i ) T(x i, y i, z i ) S i Then we n find the work done by the fore F in moving the prtile long the urve by W F(x, y, z) T(x, y, z)ds lim Rell: ds r (t) nd T(t) r r (t), then we hve W F(x, y, z) T(x, y, z)ds b F(r(t)) F(x i, yi, zi ) T(x i, yi, zi ) S i r (t) r (t) r (t) b F(r(t)) r (t) b F dr Definition: Let F be ontinuos vetor field defined on smoothed urve given by r(t), t b. Then the line integrl of F long is b F dr b F(r(t)) r (t) Remrk: F(r(t)) is n bbrevition for F(x(t), y(t), z(t)). F Tds Exmple: Find the work done by fore F(x, y) xi xyj in moving prtile long the qurter-irle r(t) os ti + sin tj, t π. W F dr π π 3 F(r(t)) r (t) os t sin t Property: F dr F dr 4
here is with reversed orienttion. Suppose F P i + Qj + Rk, then Suggested problems: 16.: 1 11, 15, 17 1, 9. F dr b b < P, Q, R > < x (t), y (t), z (t) > [P x (t) + Qy (t) + Rz (t)] P dx + Qdy + Rdz 5