5.2. LINE INTEGRALS 265. Let us quickly review the kind of integrals we have studied so far before we introduce a new one.


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1 5.2. LINE INTEGRALS Line Integrls Introduction Let us quickly review the kind of integrls we hve studied so fr before we introduce new one. 1. Definite integrl. Given continuous relvlued function f, b f (x) dx represents the re below the grph of f, between x nd x b, ssuming tht f (x) between x nd x b. 2. The definite integrl cn lso be used to compute the length of curve. If curve is given by its position vector r (t) x (t), y (t) in 2D or r (t) x (t), y (t), z (t) in 3D for t b, then the length L of the curve is given by L b r (u) du The rc length function ws defined to be s (t) t r (u) du so tht, using the fundmentl theorem of lculus, we hve ds dt r (t) 3. Double integrls. Given relvlued function f of two vribles, D f (x, y) da represents the volume of the solid bove the region D nd below the grph of z f (x, y). 4. The double integrl cn lso be used to find the re of region by the formul re of D da In this section, we study n integrl similr to the one in exmple 1, except tht insted of integrting over n intervl, we integrte long curve Line Integrls Along Plne urves Let us consider the following problem: Suppose tht we hve plne curve given by its position vector D r (t) x (t), y (t) for t b. (5.1)
2 266 HAPTER 5. VETOR ALULUS Let us ssume is smooth curve ( r is continuous nd r (t) ). Suppose further tht we hve continuous function z f (x, y), we will ssume for now f (x, y). onsider the surfce S given by x (t), y (t), z. This surfce will intersect the grph of z f (x, y) in curve. We wish to find the surfce re of S between the curves nd. To help you visulize S, think of curtin hnging. Except tht the curtin in not hnging long stright ril, but curved one. Furthermore, the ril is not necessrily horizontl, it hs whtever shpe f hs. This is shown in figure Imgine we wnt to find the re of the curtin.we will first pproximte the re using technique similr to the one used when defining the definite integrl. We will outline the steps. 1. Split [, b] into n subintervls [t i 1, t i ] of equl length. Let x i x (t i ), y i y (t i ). 2. The corresponding points P i (x i, y i ) divide into n subrcs of length s 1, s 2,..., s n. 3. In ech subrc, pick point P i (x i, y i ) (this corresponds to point t i in [t i 1, t i ]. 4. Drw the rectngle with bse [P i 1, P i ] nd height f (x i, y i ). The re of this rectngle is f (x i, y i ) s i. 5. The re of S cn be pproximted by f (x i, y i ) s i. i1 6. The lrger n is, the better the pproximtion.
3 5.2. LINE INTEGRALS 267 This llows us to define: Definition 388 With the nottion bove, the re of S, denoted A (S) is defined to be A (S) lim f (x i, yi ) s i n i1 Definition 389 If f is ny continuous function (not just positive one), defined on smooth curve given in eqution 5.1, then the line integrl of f long is defined by if this limit exists. f (x, y) ds lim n i1 f (x i, yi ) s i (5.2) You will note tht we re integrting with respect to rc length. Remembering tht ds dt r (t), it follows tht ds r (t) dt nd therefore, the line integrl cn be evluted s follows: Theorem 39 If f is ny continuous function (not just positive one), defined on smooth curve given in eqution 5.1, then the line integrl of f long cn be computed by the following formul f (x, y) ds b b f (x (t), y (t)) r (t) dt (5.3) (dx ) 2 ( ) 2 dy f (x (t), y (t)) + dt dt dt Remrk 391 We used nd b for the limits of integrtion becuse they re the limits of the vrible t. Remrk 392 Note tht the line integrl is with respect to rc length. However, to compute it, we use the prmetriztion of the curve, whtever it is. We rewrite everything in terms of the prmeter used for the curve. Exmple 393 Evlute ( 2 + x 2 y ) ds where is the upper hlf of the unit circle x 2 + y 2 1. First, we must write in prmetric form. The upper hlf of the unit circle is
4 268 HAPTER 5. VETOR ALULUS Figure 5.5: Piecewise smooth curve x (t) cos y, y (t) sin t, t π. Then f (x, y) ds b π π π f (x (t), y (t)) r (t) dt ( 2 + cos 2 t sin t ) sin 2 t + cos 2 tdt ( 2 + cos 2 t sin t ) dt 2dt + π [ ] 2t cos3 t π 3 2π ( 3 2 π + 1 ) 3 cos 2 t sin tdt Remrk 394 In the bove theorem, the given formul to find f (x, y) ds requires tht be smooth curve. However, It is still possible to compute line integrl when the curve is not smooth curve, s long s it is piecewise smooth, tht is mde of smooth pieces, s the one shown in figure 5.5. In this cse f (x, y) ds f (x, y) ds + 1 f (x, y) ds + 2 f (x, y) ds + 3 f (x, y) ds 4
5 5.2. LINE INTEGRALS 269 Exmple 395 Evlute 2xds where 1 2, 1 being the rc of the prbol y x 2 between (, ) nd (1, 1) nd 2 being the verticl line from (1, 1) to (1, 2). 2xds 2xds + 1 2xds 2 We evlute ech integrl seprtely. 1. To evlute 1 2xds,we need to prmetrize 1. y x 2 cn be prmetrized by x t, y t 2, t 1. Thus 1 2xds t 1 + 4t 2 dt 2. To evlute 2 2xds,we need to prmetrize 2. A verticl line between the given points cn be prmetrized by x 1, y t, 1 t 2. Thus 2 2xds dt 3. Therefore 2xds Two other integrls cn be obtined using similr technique. When we form the sum, we cn use x i x i x i 1 or y i y i y i 1 insted of s i. The integrls we obtin re f (x, y) dx lim f (x, y) dy lim n i1 n i1 f (x i, yi ) x i (5.4) f (x i, yi ) y i (5.5) These re clled line integrls of f long with respect to x nd y. If x x (t), then dx x (t) dt. Similrly, dy y (t) dt. So, these integrls cn be computed s follows: b f (x, y) dx f (x (t), y (t)) x (t) dt (5.6) f (x, y) dy b f (x (t), y (t)) y (t) dt
6 27 HAPTER 5. VETOR ALULUS Remrk 396 It often hppens tht these integrls pper together s in P (x, y) dx+ Q (x, y) dy. In this cse, we will write P (x, y) dx + Q (x, y) dy P (x, y) dx + Q (x, y) dy Remrk 397 The line integrl in eqution 5.3 is clled the line integrl of f long with respect to rc length. The line integrls in eqution 5.6 re clled line integrls of f long with respect to x nd y. Remrk 398 As you hve noticed, to evlute line integrl, one hs to first prmetrize the curve over which we re integrting. Here re some pointers on how to do it. 1. ircle of rdius r: ounterclockwise: x r cos t, y r sin t with t 2π. lockwise: x r cos t, y r sin t with t 2π. 2. A curve given by function y f (x): x t, y f (t). For exmple, y x 2 cn be prmetrized by x t, y t Verticl line through (, b): x, y t. 4. Horizontl line through (, b): x t, y b. 5. Line segment between r x, y, z nd r 1 x 1, y 1, z 1 : Vector form: r (t) (1 t) r + t r 1, t 1. In coordinte form: x (t) (1 t) x +tx 1, y (t) (1 t) y +ty 1, z (t) (1 t) z + tz 1. Note tht it is similr in 2D, simply drop the z coordinte. Exmple 399 Evlute y2 dx + xdy in ech cse below: 1. is the line segment from ( 5, 3) to (, 2). 2. is the rc of the prbol x 4 y 2 between ( 5, 3) nd (, 2). Solution of 1 cn be prmetrized by x 5 (1 t) + t 5t 5 nd y 3 (1 t) + 2t 5t 3 with t 1. Therefore y 2 dx + xdy ( 1 (5t 3) 2 5dt + ( 25t 2 25t + 4 ) dt ) 5t 5 5dt
7 5.2. LINE INTEGRALS 271 Solution of 2 cn be prmetrized by y t nd x 4 t 2 3 t 2. Therefore y 2 dx + xdy t 2 ( 2t) dt + ( 4 t 2) dt ( 2t 3 t ) dt with Remrk 4 In these two computtions, we were evluting the sme integrl, between the sme points, long different pths. We got different nswers. This indictes tht the integrl depends on the pth chosen. We will elborte on this in the next section Line Integrls Along Spce urves We cn define similr integrl if is spce curve given by. Definition 41 Let be smooth curve given by x x (t), y y (t) nd z z (t), t b. The line integrl of f (x, y, z) long is defined to be f (x, y, z) ds lim n i1 f (x i, yi, zi ) s i Theorem 42 The bove line integrl is evluted using the formul f (x, y, z) ds b b f (x (t), y (t), z (t)) r (t) dt (5.7) (dx ) 2 ( ) 2 ( ) 2 dy dz f (x (t), y (t), z(t) + + dt dt dt dt As before, we cn define line integrls with respect to x, y, nd z. They re evluted s follows: f (x, y, z) dx f (x, y, z) dy f (x, y, z) dy b b b f (x (t), y (t), z (t)) x (t) dt (5.8) f (x (t), y (t), z (t)) y (t) dt f (x (t), y (t), z (t)) z (t) dt Exmple 43 Evlute y sin zds where is the helix given by x cos t,
8 272 HAPTER 5. VETOR ALULUS y sin t, nd z t, t 2π. y sin zds 2π 2 2 2π sin 2 t sin 2 t + cos 2 t + 1dt 2π 2π sin 2 tdt 1 (1 cos 2t) dt 2 Summry 44 Let us recpitulte wht we lerned bout line integrls of function long curve. 1. If is smooth curve in the plne, then to compute f (x, y) ds, we do the following: () Find smooth prmetriztion of, sy r (t) x (t), y (t), t b. (b) The integrl is evluted by the formul f (x, y) ds b f (x (t), y (t)) r (t) dt 2. If is smooth curve in spce, then to compute f (x, y, z) ds, we do the following: () Find smooth prmetriztion of, sy r (t) x (t), y (t), z (t), t b. (b) The integrl is evluted by the formul f (x, y, z) ds b f (x (t), y (t), z (t)) r (t) dt 3. If is mde by joining finite number of smooth curves end to end in other words n then n f (x, y) ds 1 f (x, y) ds+ 2 f (x, y) ds n f (x, y) ds. 4. If is plne curve nd f (x, y) then the geometric mening of f (x, y) ds is the re of the curtin with bse below the grph of z f (x, y). 5. Line integrls re lso used in physics. An importnt mening is the following. The mss of thin wire lying long smooth curve is f (x, y, z) ds where f (x, y, z) is the density of the wire t (x, y, z). This formul llows us to find the mss of thin wire for which the density is not constnt long the wire.
9 5.2. LINE INTEGRALS Line Integrls of Vector Fields In the previous section, we lerned the mening of nd how to compute line integrls of sclr functions. We now turn to line integrls of vector field. We motivte this type of integrl with n ppliction. In physics, work is defined s force cting upon n object to cuse displcement. There re three key words in this definition  force, displcement, nd cuse. In order for force to qulify s hving done work on n object, there must be displcement nd the force must cuse the displcement. There re severl good exmples of work which cn be observed in everydy life  horse pulling plow through the fields, fther pushing grocery crt down the isle of grocery store, freshmn lifting bckpck full of books upon her shoulder, weight lifter lifting brbell bove her hed, n Olympin lunching the shotput, etc. In ech cse described here there is force exerted upon n object to cuse tht object to be displced. We first remind the reder of some formuls giving the work done by force in simple cses. You my recll lerning in physics clss tht the work W done by vrible force f (x) moving n object from to b long the xxis is given by W b f (x) dx (5.9) In this cse, we hve vrible force, long stright line, the xxis. Another simple cse is when we hve constnt force F which moves n object between two points P nd Q in spce. The work done in this cse is W F P Q (5.1) These two cses were firly simple. In the first one, the motion ws long the xxis. In the second, though the motion ws in spce, the force ws constnt. In this section, we wish to compute the work done by force in more generl setting. The force will be vrible force, the object will be moving long ny smooth curve. This cn hppen when, for exmple, n object in moving long curve, in vector force field. At ech point long the curve, the force pplied to n object will be given by vector field. In other words, suppose we hve vector field F (x, y, z) P (x, y, z), Q (x, y, z), R (x, y, z). We wish to compute the work done by F in moving prticles long smooth curve. We will use technique similr to the technique used in the previous section when we defined line integrls. We will lso use the sme nottion. I will not repet the detils here. is divided into subrcs P i 1 P i of length s i. In the i th subrc, we select point Pi (x i, y i, z i ). This point corresponds to vlue of t we cll t i. If s i prticle moves from P i 1 to P i in the direction of T (t i ), the unit tngent vector t Pi. Thus the work done by F in moving the prticle from P i 1 to P i cn be pproximted by W i F (x i, y i, z i ) s i T (t i )
10 274 HAPTER 5. VETOR ALULUS So, the work done by F moving the prticle long cn be pproximted by: W F (x i, yi, zi ) T (t i ) s i i1 As n becomes lrger, this pproximtion becomes better. So, we cn define: Definition 45 The work W done by force field F (x, y, z) cting on moving prticle long smooth curve cn be given by the limit of the bove sum s n. This is precisely the line integrl we defined in the previous section. Thus, W F (x, y, z) T (x, y, z) ds F T ds You will recll tht if the curve is given by position vector r (t) then T (t) r (t) r (t) Also, Therefore ds r (t) dt T ds r (t) r (t) r (t) dt r (t) dt It follows tht W b F ( r (t)) r (t) dt This integrl is often bbrevited s F d r. Keep in mind tht F (r (t)) F (x (t), y (t), z (t)), lso d r r (t) dt. Definition 46 Let F be continuous vector field defined on smooth curve given by position vector r (t), t b. The line integrl of F over is F d r b F ( r (t)) r (t) dt (5.11) F T ds Remrk 47 Keep in mind tht F ( r (t)) F (x (t), y (t), z (t)), lso d r r (t) dt. In prticulr, we must use the prmetrized form of.
11 5.2. LINE INTEGRALS 275 Remrk 48 The bove formul is vlid in both 2D nd 3D. Remrk 49 There re six diff erent wys to write the integrl corresponding to the work of vector field F (x, y, z) P (x, y, z), Q (x, y, z), R (x, y, z) over curve given by r (t) x (t), y (t), z (t) for t b. They re shown below. Keep in mind they re different wys of writing the sme thing. F T ds, the definition. F d r, clled the compct differentil form. b F ( r (t)) r (t) dt, since d r r (t) dt. b F (x (t), y (t), z (t)) x (t), y (t), z (t) dt, using the components of r (t). b [P (x (t), y (t), z (t)) x (t) + Q (x (t), y (t), z (t)) y (t) + R (x (t), y (t), z (t)) z (t)] dt, using the components of F. P dx + Qdy + Rdz, the most common form. Exmple 41 Find the work done by F y x 2, z y 2, x z 2 over the curve r (t) t, t 2, t 3 from (,, ) to (1, 1, 1). First, let us note tht the two points correspond to t nd t 1. To evlute the integrl, we proceed s for line integrls of sclr functions. We write everything in terms of t. Now, r (t) 1, 2t, 3t 2 Also So Hence F (x, y, z) y x 2, z y 2, x z 2 t 2 t 2, t 3 t 4, t t 6, t 3 t 4, t t 6 F (r (t)) r (t), t 3 t 4, t t 6 1, 2t, 3t 2 Work 2t 4 2t 5 + 3t 3 3t F T ds F (r (t)) r (t) dt ( 2t 4 2t 5 + 3t 3 3t 8) dt
12 276 HAPTER 5. VETOR ALULUS Exmple 411 An object of mss m moves long the curve given by the position vector r (t) αt 2, sin βt, cos βt, t 1, α nd β re constnts. Find the totl force cting on the object nd the work done by this force. You will recll tht Newton s second lw of motion sys tht F m (t) m r (t) And therefore, the work W will be given by W 1 m r (t) r (t) dt from eqution Now, r (t) 2αt, β cos βt, β sin βt nd So tht r (t) 2α, β 2 sin βt, β 2 cos βt r (t) r (t) 4α 2 t It follows tht W 1 2α 2 m 4mα 2 tdt Exmple 412 An object cted on by force F (x, y) x 3, y moves long the prbol y 3x 2 from (, ) to (1, 3). lculte the work done by F. Since the curve is not prmetrized, we must do so first. For this prbol, we cn use r (t) x (t), y (t) where x t, y 3t 2, t 1. Now, Also F ( r (t)) F (x (t), y (t)) x 3 (t), y (t) t 3, 3t 2 r (t) 1, 6t So tht It follows tht F ( r (t)) r (t) t t 3 19t 3 W 1 19t 3 dt 19 4
13 5.2. LINE INTEGRALS 277 Remrk 413 (independence of the prmetriztion) This remrk pplies to ll the integrls studied in this section, tht is line integrls of sclr functions long plne or spce curves s well s line integrls of vector fields. To evlute these integrls, one must hve prmetriztion of the curve involved. Since the sme curve cn be prmetrized different wys nturl question is to know if the results depends on the prmetriztion. It cn be proven tht it does not. Given smooth curve nd ny prmetriztion for it, line integrl long this curve will hve unique nswer, which does not depends on the prmetriztion chosen Assignment 1. Evlute yds where is the curve given by the position vector r (t) t 2, t, t Evlute xy4 ds where is the right hlf of the circle x 2 + y Evlute xydx+(x y) dy where consists of line segments from (, ) to (2, ) nd from (2, ) to (3, 2). 4. Evlute (xy + y + z) ds long the curve r (t) 2t, t, 2 2t, t Evlute xeyz ds where is the line segment from (,, ) to (1, 2, 3). 6. Evlute x2 y zdz where is the curve given by the position vector r (t) t 3, t, t 2, t Evlute 1 2 ( x + y z 2 ) ds where 1 is given by r 1 (t) t, t 2,, t 1 nd 2 is given by r 2 (t) 1, 1, t, t Find the mss of the wire tht lies long the curve r (t), t 2 1, 2t, t 1 if the density of the wire is given by f (x, y, z) 3t Find the work done by F (x, y, z) 3y, 2x, 4z over ech pth below: () 1 given by r (t) t, t, t, t 1. (b) 2 given by r (t) t, t 2, t 4, t Find the work done by F (x, y, z) 3x 2 3x, 3z, 1 over ech pth below: () 1 given by r (t) t, t, t, t 1. (b) 2 given by r (t) t, t 2, t 4, t Find the work done by F (x, y, z) 2y, 3x, x + y over the curve given by r (t) cos t, sin t, t, t 2π 6
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