1 13. Line Integls Line integls e simil to single integl, f ( x) dx ws e unde f ( x ) if ( ) 0 Insted of integting ove n intevl [, ] (, ) f xy ds f x., we integte ove cuve, (in the xy-plne). **Figue - get this imge in you hed, it s get visul! f xy,, long cuve, is the lue cutin, o fence ceted fom the xy-plne up to The line integl of ( ) whee it hits the sufce, f (, ) xy, pictued in ed on top of the fence. The line integl is equl to the e of tht lue cutin. If f ( x) 0, then f ( xy, ) dsepesents the positive e of one side of the cutin. Scl Line Integls e used to epesent totl mss nd totl chge nd to find electic potentils. We ll use them in this section to solve wok done y foce field on pticle moving long pth nd to find mss nd cente of mss of thin wies in the shpe of cuves (cuve ). Figue - How cn we find the e of one of these lue veticl ectngles? s suc length Δ s - chnge in suc length So we just multiply f ( xy, ) Δ s (height) (length) This is 1 piece. We need to sum up ll the pieces which is Riemnn n Sum, (, ) i= 1 f x y Δs. As usul, when we tke the nume of i i i suintevls to ppoch then the Riemnn Sum tuns into the exct integl: Line Integl of f long cuve is This is the ig pictue ut we need to f ( xy, ) ds know how to ctully evlute these. * f (, ) xy is sufce. * is cuve (so mye they should e clled cuve integls). We ll need to define cuve. *ds is in suc length, so ou s s un fom wht to wht on ou limits of integtion? This is going to e ou new ttchment (like ddθ in pol, dzddθ in cylindicl, ρ sinφ dρ dθ dφ in spheicl) nd gin it s ecuse of how these things e mesued. We ll find ou new ttchment ds. Lowe cse s. This will llow us to evlute the line integl.
2 Mentioned in this textook ck in Section 10.3 on the fist pge of the lesson (ut elly intoduced ck in lculus II, section 6.4 in the Stewt lculus ook): The Length of cuve with pmetic equtions x = f ( t), y= g( t), t, is the limit of lengths of inscied polygons nd, fo the cse whee f nd g e continuous, we ived t the fomul dx dy L= + dt dt dt This is wht we ttch on fo ds. This is how suc lengths e mesued. So, when is defined pmeticlly (x nd y s functions of t): dx dy ds = + dt dt dt,, f ( xy) ds = f( xt () yt ()) + dt MEMORIZE THIS! dx dy dt dt ***Notice thee s no hee, t The new integl is in tems of t, dt. You will need to descie/wite cuve pmeticlly (in tems of t). When setting up line integls this is usully the most difficult pt!!!!!!!! You ve got to come up with pmetic epesenttion of cuve; cuve. You will e using t s you pmete. When you see the lowe cse s in ds, think of suc nd don t foget to ttch plce of the ds. dx dy + dt dt dt in A moe commonly used compct fom ises if we use vecto nottion to descie cuve : t = x t, y t, t ( ) ( ) ( ) f xt yt f t ( ( ), ( )) ( ( )) dx dy + x t + y t t dt dt *** (, ) = ( ( )) c () () () f xy ds f t t dt 1443 ds Notice if you function = 1 If ll heights = 1, it s (c length) 1 = c length. 1 t dt which is the c length fomul ck in Section 10.3, p. 708: L= t dt
3 Let s tlk out pmeteizing cuves. If cuve is given s y is function of x, so y g( x) = (like y = x ) The ook will choose x fo the pmete, x x, y g( x) consistent with how we ve pmeteized cuves: x t, y g( t) = = ut I will lwys use t s the pmete to sty = =. Exmple: If y = x, then I ll pmetize the cuve s x = t, y = t o t () = t, t Then we e ck to ( ) ( ( ) ( )) dx dy f xy, ds = f xt, yt + dt dt dt The sme ide cn e used if cuve is given x s function of y, like ou fist exmple. Pmeteize the following cuves : ) is the c of the cuve y x = e fom (1, 0) to ( e,1) ) consists of the top hlf of the cicle ( 1,0)to (,3) x + y = 1 fom (1, 0) to ( 1, 0) nd the line segment fom c) is the line segment fom ( 1,5,0) to (1,6,4) d) : the cicle x + y = 4 oiented counte-clockwise
4 Exmple 1: Evlute xy ds, whee is the line segment fom ( ) ( ) So this is the line integl of sufce ( ) 0,1 to 3,4. ds efes to sucs f xy, = xy, long cuve, with espect to c length. Visulize Figue. If f ( x) 0, ou nswe will e the e of the cutin unde the sufce f ( xy, ) = xy, long cuve, the line segment fom ( 0,1 ) to ( 3,4 ). f t t dt ( ()) o ( (), ()) dx dy f xt yt + dt dt dt Anothe ppliction: IF the function epesents line density t point (x,y) of thin wie in the shpe of cuve, THEN the nswe epesents the totl mss of the wie. Exmple : Evlute 3 x y ds if is x y =, 0 x. If f ( x) 0, ou nswe will e the e of the cutin unde the sufce f ( x, y) 3 x =, long cuve, y x y =, 0 x. IF insted the function epesents line density t point (x,y) of thin wie in the shpe of cuve, THEN the nswe epesents the totl mss of the wie.
5 Just like odiny single integls, if f ( xy, ) is positive function f (, ) xy dsepesents the e of one side of the cutin (whose se is nd whose height ove the point ( x, y ) is (, ) If the sufce, f (, ) f xy. xy, dips elow the xy-plne, it gives us the signed e of the cutin (e ove e elow) f xy, ds f xy, ove line cuve (lying in the xy-plne). ( ) - The line integl of function ( ) Let s tke function f nd cuve nd shift them up dimension. f ( xyz,, ) ds - The line integl of function f (,, ) It just dds z onto the fomul on the left:,,,, xyz ove cuve which my lie nywhee in spce. f ( xyz) ds = f( xt () yt () zt ()) + + dt o ( ()) dx dy dz dt dt dt sme s it ws efoe f t t dt Exmple 3: Evlute the line integl (with espect to x, y, nd z), z dx + x dy + y dz, whee is the given cuve x t y t z t 3 =, =, = fom ( ) ( ) 0, 0, 0 to 1,1,1. Notice the line integl is NOT with espect to c length (ds), ut insted with espect to x, y, nd z. Thee is no ds, ut dx, dy, nd dz insted. Becuse it is not mesued with sucs, we do not ttch dx dy dz + + o t. dt dt dt Insted, you ll need to clculte dx, dy, nd dz nd expess eveything in tems of pmete t.
6 If is given in septe pts, you need to set up two septe integls. 1 (, ) = (, ) + (, ) f xy ds f xy ds f xy ds 1 Thee my e moe thn two. These line integls we e looking t ight now depend on the pth they tke, so they must e delt with septely. They e clled Dependent of Pth. 1 3 You d get diffeent e if you took the line integl using cuve 3 insted of 1 +. Think of the e unde the cutin Exmple 4: Evlute the line integl, ( + ) x y z ds whee is the pt of the pol c fom ( 0,0 ) to ( 1,1 ) in the xy-plne nd the line segment fom ( ) ( ) 1,1, 0 to 1,1,1. y= x in the 1 st qudnt Notice f ( xyz,, ) = x+ y z, not f ( xy, ), so it s not the e of the cutin. Notice cuve hs two pts nd lso notice we e with espect to c length ds. Wht could ou nswe men? If x + y z is the mss long cuve, then ou nswe epesents the totl mss. If x + y z is the chge density long, then ou nswe epesents the totl chge.
7 Insted of integting ove 1 Do you think f ( x, y) ds= +? 3 1 +, wht if you just mke ee-line fom ( ) ( ) 0,0,0 to 1,1,1 nd cll it 3? 3 1 Even though the cuves hve the sme endpoints, the vlues won t necessily e equl. The vlue of line integl depends not just on the endpoints of the cuve, ut lso on the pth it tkes. c ( + ) x y z ds whee is the pt of the pol 0,0 to 1,1 in the xyplne nd the line segment fom ( ) ( ) 1,1, 0 to 1,1,1. y = x in the 1 st qudnt fom ( ) ( ) Insted let s sy is the line segment fom ( 0,0,0 ) to ( 1,1,1 ). : t = ttt,,, 0 t 1 3 3() (hee is the pmeteiztion of the line segment) 3 = 1,1,1 = ( t) () t 1 ( ) ( ) } ds, f x y ds = t+ t t dt = Not the sme. 6 0 When this hppens we sy the integl is dependent of pth (fom stt to finish it depends on the pth it tkes to get thee). When scl function f tuns into vecto function F we will sy F is not consevtive vecto field. Lte, we ll integte functions tht e independent of pth. We ll get the sme nswe no mtte wht pth we tke fom stt to finish. We ll look fo quick nd esy wy (lte on). When scl function f tuns into vecto function F we will sy F is consevtive vecto field. So when the integl is dependent of pth you hve to do it in pieces. We ll e lening how to simplify this pocess lso (insted of hving to do it in unch of pieces).
8 So, in summy so f: Line integl of f (, ) xy ove line cuve lying in the xy-plne. - Ae of the side of cutin. - If f ( xy, ) epesents line density t point (x,y) of thin wie in the shpe of cuve, then the nswe epesents the totl mss of the wie. Line Integls f (, ) xy ds= pmeticlly defined dx dy f ( xt (), yt ()) + dt dt dt vecto nottion o f () t t dt ( ) () 13, ( t) x( t), y( t) = Splitting into pts if piece-wise 1 Line integl of f (,, ) xyz ove cuve which my lie nywhee in spce.,,,, f ( xyz) ds = f( xt () yt () zt ()) + + dt o ( ()) dx dy dz dt dt dt f t t dt Oienttion, Figue 8. Theoeticl Ide. Genelly, given pmetiztion x = xt (), y= yt (), t detemines n oienttion of cuve, with the positive diection coesponding to incesing vlues of the pmete t. If the line integl epesents the e of the side of the cutin, is it going to mtte which wy we tvese cuve? No. When we integte with espect to c length, ds, the vlue of the integl does not chnge when we evese the oienttion of the cuve (it doesn t mtte if you go fom to o to ). The e will e the sme vlue. (, ) = (, ) f xy ds f xy ds Ac length is just length; it s lwys positive. So this is tue of ds, ut not tue of dx o dy. When you chnge fom ds to dt you ve picked THE diection of the pth y you choice of t s.
9 ***Think of this s new section completely. Hee s n ppliction of line integls (don t eplce this with plin line integls). Appliction 1 ente of Mss If ρ ( x, y) epesents line density t point (, ) mss of the wie, m x y of thin wie shped like cuve, then the totl (, ) c ente of mss ( x, y) 1 x= x ρ ( x, y) ds m 1 y = y ρ ( x, y) ds m m= ρ x y ds Know these. Use you gphing clculto to clculte them quickly. Don t foget to ttch t dt fo ds. Let s ty one: Exmple 5: A wie tkes the shpe of the semicicle x + y = 1, y 0, nd it is thicke ne its se thn ne the top. Find the cente of mss of the wie if the line density t ny point is popotionl to its distnce fom the line y = 1. Agin, this is just n ppliction of line integls. Don t eplce this with plin line integls.
10 Appliction Wok Line Integls of Vecto Fields We ll use Vecto Line Integls to find the wok done y foce field, F, when moving pticle long pth. Don t confuse this with plin line integls of function f. Bck in section: 6.5 W = f ( x) dx Wok done y vile foce f ( ) 9.3 W = F D to long the x-xis. x in moving pticle fom Wok done y constnt vecto foce F (such s W = F D cosθ electicl o gvittionl) in moving n oject fom point P to point Q, whee D = uuu PQ, the displcement vecto. Hee F cts on the oject nd we must wok ginst the foce field to move the oject. F θ P D Q Now, we e going to compute the wok done y foce field, F, in moving pticle long smooth cuve,. The fomul you e going to use is: W = F ( () t ) t dt Explntion of the fomul: F xyz o = PQR,, foce field in 3-dim.(like gvittionl o electic foce fields) t = x t y t z t (defines cuve in spce) (,, ) F () (), (), () T = ( x, y, z) o T t is the unit tngent vecto, T() t = t ( ) () t diffeent points on cuve.. Think of the tngent vecto t = () t Divide the cuve into unch of sucs of length s ( ds) Δ. Δ s If Δ s is vey smll, then s the pticle moves long the cuve in the ppoximte diection of the unit tngent vecto T. t = T Δ s (W = F D so now F TΔs )
11 Wok done y the foce field F in moving the pticle long cuve is ppoximtely F Δ st o F TΔs. F T This dot poduct computes wok t ech point on cuve. Totl wok done in moving the pticle long is the sum of Δ s ll F ( xyz,, ) T( xyz,, ) Δs As lwys, s the nume of eqully spced sucs, n, the wok done y foce field F is the limit of Riemnn sums which ecomes the integl. Hee s visuliztion in -dimensions: uve is in lck, vecto field F is in lue nd t cetin points on cuve in geen, nd the unit tngent vecto T is in ed Pticle is moving west. Pticle is moving est. Hee most of the dot poducts F T e Hee most of the dot poducts F T e negtive ecuse the ngles etween positive ecuse the ngles etween the vectos e otuse so the line the vectos e otuse so the line integl will e negtive. integl will e positive. The vecto field is moe often in the opposite The vecto field is moe often in the sme diection s the pth the oject tkes. diection s the pth the oject tkes. W= F x y z T x y z ds = F T ds (,, ) (,, ) Scl-educes ck to egul lines Wok is the line integl with espect to c length of the tngentil component of the foce. Wow!
12 W= F x y z T x y z ds = F T ds But this fomul (,, ) (,, ) doesn t mtch the fomul I gve you W = F ( () t ) t dt so let s do little moe wok. If cuve is given y the vecto eqution ( t) x( t), y( t), z( t) ( t) T() t = (sect 10.4) () t =, then unit Tngent vecto F xyz T xyz ds s We cn ewite (,, ) (,, ) ( t) W = F( () t ) t dt = F( () t ) t dt 1443 t 1443 F 1443 ds T evited W = F d Wok done y foce field, F, in moving pticle long smooth cuve,. This occus in othe es of physics lso like clculting the flux coss plne cuve defined s the integl of the noml component of vecto field, the thn the tngentil component Definition: Let F e continuous vecto field defined on smooth cuve, given y vecto function ( t), t. Then the line integl of F long is: u F d = F t t dt = F T ds 1443 ( ()) d = WORK Appliction F t ( ( )) mens F ( xt ( ), yt ( ), zt ( )) This is the wok done y foce field, F, in moving pticle long smooth cuve,. Notice, this is vey simil to ou plin line integl, f ( () t ) t dt, the wok ppliction one is just missing the solute vlue nd hs the dot poduct. Don t mix them up!!
13 Exmple 6: F, F x, y = y, x Find the wok done y the foce field ( ) unit cicle centeed t the oigin. W = F ( () t ) t dt Need: ( t), F( ( t) ), ( t), nd F( ( t) ) ( t), in moving n oject fom ( 1, 0 ) to ( 0,1 ) on the
14 Why did we get negtive wok? When we get negtive wok, the vecto field must siclly e in the opposite diection s the pth the oject took. Moe of the dot poducts F T e negtive ecuse the ngle etween them is otuse. (Figue 1, no vectos stting on the (lue) cuve,, point oughly in the sme diection s the cuve. In fct most vectos point in the opposite diection). The foce field impedes the movement of the pticle long the cuve. Hee is ou Exmple 6 done y hnd (o just show them on Mthemtic): Find the wok done y the foce field F, F( x, y) = y, x unit cicle centeed t the oigin. ( 0,1 ) ( ),, in moving n oject fom ( ) ( ) 1, 0 to 0,1 on the ( 1, 0 ) Domin Rnge Point Vecto F = y, x ( 1, 0 ) 0, 1, ( 0,1 ) 1, 0, 0.7, Negtive π ecuse ou vectos in the vecto field e pointing in the opposite diection s ou pth fom ( 1, 0 ) to ( 0,1) on. 1 Hee, it mttes which wy we go (the diection of the pth) with vecto field F in this wok ppliction. F d = F d + Tvese uve in opposite diections one nswe is positive nd the othe negtive. But when we used c length, ds, ( length is length, it s lwys positive) it didn t mtte which f xyds, = f xyds, we we tvesed cuve : ( ) ( ). +
15 I m going to sk this next question two diffeent wys so you cn see thei connection. It s just mtte of nottion! 1. Evlute + ( + ) = fom ( ) ( ) xy dx x y dy long the cuve y x 1,1 to, 4. F xy, = xy, x+ y long the cuve y x. Find the wok done y foce field ( ) W = F ( () t ) t dt = fom ( 1,1 ) to (, 4). Let s ty the fist one: 1. Evlute xy dx + ( x+ y) dy long the cuve y nything s we go so they cn see the connection to the second wy. = x fom ( 1,1 ) to (, 4). Do not simplify Notice these line integls of f long e with espect to x nd y insted of c length ds. Let s stt the second one until you see it is the sme s the fist one: F xy, = xy, x+ y long the cuve. Find the wok done y foce field ( ) y = x fom ( 1,1 ) to (, 4). W = F t t dt ( ( )) ( )
16 This is the connection etween line integls of scl fields nd line integls of vecto fields. The polem I sked two diffeent wys: xy dx + ( x+ y) dy could e witten s F d if F = xy, x+ y Textook Exmple 6 is: ydx+ zdy+ xdz which could e witten s F d if F = y, z, x scl vecto field Just think of tht d s the deivtive of ( t) x( t), y( t), z( t) = nd do the dot poduct. W = Pdx+ Qdy + Rdz = F d lin integl of line integl of scl fields vecto field F = = if F P( xyz,, ), Q( xyz,, ), R( xyz,, ), ( t) xt ( ), yt ( ), zt ( ) If they don t see it, hee s the long explntion: = = Suppose vecto field F on 3-dim. is F P( xyz,, ), Q( xyz,, ), R( xyz,, ), ( t) xt ( ), yt ( ), zt ( ) F d = F t t dt ( ()) F t = P x t y t z t Q x t y t z t R x t y t z t F t ( ( )) ( (), (), ()), ( (), (), ()), ( (), (), ()) ( ( )) = P, Q, R fo ese. dx dy dz t =,, dt dt dt Let s gee to cll it dx dy dz F( () t ) t = P + Q + R dt dt dt So F t t dt ( ()) dx dy dz = P + Q + R dt dt dt dt = P dx + Q dy + R dz Which is the line integl long with espect to x, y, nd z. W= F d = Pdx+ Qdy+ Rdz line integl of line integl of vecto field F scl fields so given this nottion, you know F = P, Q, R.