Line integrls, nd rc length of curves. Some intuitive explntions nd definitions. Version 2. 8//2006 Thnks Yhel for correcting some misprints. A new remrk hs lso een dded t the end of this document. Let e curve in R 2 or R 3. Suppose tht hs prmetric representtion rt xt î + ytĵ+zt k with t. Explntion of nottion: In this document I wnt to simultneously tret the cse of R 2 nd R 3, so I use red letters to write ll terms which hve to e present if we re deling with curve in R 3 ut should not e there if our curve is in R 2. If you do not hve colour printer, or if you do not like reding documents from computer screen, you my prefer to use n lterntive version of this document in which ll symols ppering in red here re enclosed in doule squre rckets there, i.e., in tht document we write xtî + ytĵ[[+zt k]] insted of xtî + ytĵ+zt k If nyone thinks it would e useful, I cn esily prepre two seprte versions of this document, one for curves in R 2 nd the other for curves in R 3. So our curve cn e written s the set of ll points of the form xt, yt, zt for ll t [, ]. We re mking the usul ssumption tht xt nd yt nd zt re continuous functions on [, ]. For our purposes here we hve to mke one more ssumption, nmely tht the mp t rt is one to one, or lmost one to one. More precisely we hve to ssume the following property: * For ny two different points s, t in [, ], we hve rs rt, with the following possile exception: we llow the possiility tht perhps r r. Exmple. The curve with the prmetric representtion rt cos t î + sin t ĵ for t stisfies the condition * if nd only if 2π. If 2π we hve the llowed exception tht r r. This curve is of course circulr rc, or full circle, depending on nd. Plese rememer tht for ny given curve, there re infinitely mny different wys to choose prmetric representtion for. Well there is one exception to this, the rther silly cse where consists of just one point. In this document we wnt to try to nswer three questions, nd then use the nswers to them to motivte the definitions of two new types of integrls. Here re the questions: Question. Wht is the length often clled rc length orekh keshet of? Question 2. Suppose we end piece of wire so tht it hs exctly the shpe of. Suppose tht this piece of wire hs vrile liner density i.e. mss per unit length. Wht is the mss of this piece of wire? Question 3. Suppose tht prticle moves long nd t ech point it is cted on y force field, for exmple grvittionl, electrosttic or mgnetic field. How much work is done ltogether y the force field to move the prticle ll the wy long? Here first is n exct nd rigorous wy for clculting the nswer to Question. But it my sometimes e quite difficult to perform this clcultion. Let e prtition of the intervl [, ]. In other words let e finite collection of points t j for j 0,,..., n which stisfy t 0 < t <... < t n. Let e the polygonl curve metsul which consists of n stright line intervls from rt j to rt j for j, 2,..., n. Let L e the length of. lerly we must hve L n rt j rt j. Now we define the length of to e the supremum of the set of ALL numers L where rnges over ALL prtitions t 0 < t <... < t n of [, ]. Note tht n cn e different integer for different prtitions nd we hve to llow it to e ritrrily lrge. We will now use this definition to clculte the length of one rther exotic curve. { 0, x 0 Let g : R R e the function gx x cos. It is esy to check tht this function is x, x 0 continuous, lso t x 0. Let e the grph of this function on the intervl [0, /π]. So we cn choose the prmetric representtion rt t î + gt ĵ with 0 t /π for this curve. Now, for ech n let us consider the specil prtition of [0, ] given y n + points t j where t 0 0 nd, for j, 2,..., n, we choose t j πn+ j. It is ech to check tht this gives 0 t 0 < t <... < t n /π. Now let us clculte, or t lest estimte L for this. We hve L rt j rt j t j t j 2 + gt j gt j 2 gt j gt j gt j gt j j2 t j cos t j cos t j j2 t j.
2 Since t j πn+ j for the relevnt vlues of j in this lst sum, we see tht L πn + j cos πn + j cosπn + 2 j πn + 2 j. j2 If we set k n + j, then s j rnges from 2 to n, the numer k rnges from n to. So we cn rewrite the previous inequlity s n n L πk cos πk cosπk + πk + k k+ πk πk + k k n k + k n πk πk + n πk + πk + πk n π k. k k k k From sic results out numericl series we know tht the series k k diverges. This mens tht we cn mke n π k k s ig s we wish y choosing n sufficiently lrge. Since we cn choose n s ig s we wish when we define the prtition s we did ove, this mens tht L cn e mde igger thn ny given numer. So the length of this curve cn only e +. Despite this perhps emrssing result, there re lots of other curves which hve finite rc length, nd, insted of hving to consider the vlue of L for these curves for ll sorts of prtitions, there is n integrl which gives us the vlue of tht rc length. To use this integrl we need to suppose tht our prmetric representtion for our curve hs one more property: ** The functions xt nd yt nd zt ll hve continuous derivtives t every point of [, ], one sidedly t nd. If this property is stisfied then we define the derivtive of the vector vlued function rt xtî + ytĵ+zt k y d dt rt x t î + y t ĵ +z t k for ech t [, ]. Theorem. Suppose tht the curve hs prmetric representtion rt xt î + yt ĵ+zt k with t which stisfies oth * nd **. Then the rc length of is finite nd equls d dt rt dt x t 2 + y t 2 +z t 2 dt. Let us now stte two more theorems which, under suitle conditions, provide the nswers to Questions 2 nd 3 respectively. Theorem 2. Suppose tht the curve hs prmetric representtion rt xtî + ytĵ+zt k with t which stisfies oth * nd **. Let ρx, y, z e function which is defined in set contining nd is continuous t every point of. Then the mss of piece of wire which hs the sme shpe s nd whose liner density t ech point x, y, z of is equl to ρx, y, z is equl to 2 ρxt, yt, zt d dt rt dt ρxt, yt, zt x t 2 + y t 2 +z t 2 dt. It is sometimes convenient to use the nottion ρ rt for ρxt, yt, zt. Theorem 3. Suppose tht the curve hs prmetric representtion rt xtî + ytĵ+zt k with t which stisfies oth * nd **. Let 3 F x, y, z uxt, yt, zt î + vxt, yt, ztĵ+wxt, yt, zt k e vector field whose component functions u nd v nd w re defined in set contining nd re continuous t every point of. This mens tht when certin prticle is t the point x, y, z on, the field exerts the force F x, y, z on the prticle. Then the mount of work done y the force field to move the prticle long the full length of is equl to 4 F xt, yt, zt d dt rtdt F xt, yt, zt x tî + y tĵ+z t k dt uxt, yt, ztx t + vxt, yt, zty tĵ+wxt, yt, ztz t k dt.
3 It is sometimes convenient to use the nottion F rt for F xt, yt, zt. We will not prove these theorems, ut we will give intuitive explntions of why they re true. In fct since they re very closely connected it will e convenient to explin ll three of them simultneously. Remrk: As we mentioned ove, there re infinitely mny different wys to find prmetric representtion for given curve. And if we hve found one representtion which stisfies * nd ** then we cn find infinitely mny other ones which lso stisfy these two properties. It turns out tht when we clculte the integrl ppering in Theorem for ny two different representtions stisfying * nd ** then, lthough we my get two integrls which look very different, when we clculte them we will get the sme vlue for oth of them. A similr remrk pplies to the integrls ppering in Theorems 2 nd 3. This independence on the choice of prmetric representtion cn e proved mthemticlly, ut it the proof is too hrd for this course. Here we will explin this independence intuitively, from geometricl nd physicl considertions. The length of, the mss of our wire in the form of nd the work done y force field to move prticle long, re ech geometricl or physicl quntities depending only on the form nd shpe of nd the function ρ nd the vector field F. So they hve nothing t ll to do with ny prticulr choice of prmetric representtion. In Theorem 3 it is not quite true to sy tht the nswer does not depend on the prmetric representtion. It does depend on the direction of the prmetric representtion. If we chnge the direction megm of the movement of the prticle long, then the work done y the field will e multiplied y. Perhps the preceding remrk will mke more sense if you red it gin lter, fter you hve red the explntions of Theorems, 2 nd 3. Let us first prove the ove three theorems exctly in n esy specil cse. Suppose tht x t nd y t nd z t re ech constnts on [, ], i.e. suppose tht for some constnts nd β nd γ we hve x t nd y t β nd z t γ for ll t [, ]. This mens tht xt x + t nd yt y + t β nd zt z + t γ for ll t [, ]. This mens tht is the stright line segment from the point x, y, z to the point x, y, z x+, y+ β, z + γ. This segment is prllel to the vector î + βĵ+γ k, nd, if we think of it s vector, it is the sme s the vector 5 r r î + βĵ+γ k. In prticulr, the length of is simply r r î + βĵ+γ k. In this esy specil cse we will lso ssume tht ρx, y, z is constnt numer c nd F x, y, z is constnt vector G. In tht cse it is cler tht the mss of the stright segment must equl c r r nd the work done y the constnt force G to move the prticle long the stright line segment from r to r must equl G r r. It is good nd esy introductory exercise to clculte the three integrls in Theorems, 2 nd 3 directly in this esy cse nd to see tht indeed their vlues re r r, c r r nd G r r respectively. But we will not use those clcultions explicitly in the intuitive explntions which we will give now for the generl cses of Theorems, 2 nd 3. The ide is tht we will use very fine prtition t 0 < t <... < t n of [, ]. Since ech intervl [t j, t j ] is very smll, the continuous functions x t nd y t nd z t re ll nerly constnt on [t j, t j ]. This mens tht the prt, which we will cll j, of the curve from rt j to rt j is lmost stright line segment joining these two points tips of vectors. Since j is lso very smll, the continuous function ρx, y, z nd the continous functions ux, y, z nd vx, y, z nd wx, y, z ppering in 3 re lso lmost constnts on j. So the vector F x, y, z is lmost constnt on j. Now we pply the esy specil cse to the lmost stright curve j { rt : t j t t j } t lest in n pproximte wy. Here we hve t j in plce of nd t j in plce of. In view of our remrks in the previous prgrph, we cn lso tke x t j : j in plce of, nd y t j : β j in plce of β, nd z t j : γ j in plce of γ. We tke ρ rt j : c j in plce of c nd F rt j u rt j î + v rt j ĵ+w rt j k : G j in plce of the constnt vector G. So the exct clcultions of the esy specil cse suggest tht: i The length of j is pproximtely rt j rt j t j t j j î + β j ĵ+γ j k t j t j x t j î + y t j ĵ+z t j k t j t j x t j 2 + y t j 2 +z t j 2 : L j. ii The mss of the prt of the wire which coincides with j is pproximtely c j rt j rt j ρ rt j rt j rt j ρxt j, yt j, zt j t j t j x t j 2 + y t j 2 +z t j 2 : M j.
4 iii The work done y the lmost constnt force field to move the prticle long the very short curve j is pproximtely equl to G j rt j rt j F rt j t j t j x t j î + y t j ĵ+z t j k : W j. It follows from these three pproximte clcultions tht the totl length of, totl mss of M nd totl mount of work to move long ll of re pproximtely equl to n L j, n M j nd n W j. We wnt to show tht these three sums re pproximtely equl, respectively, to the three integrls ppering in Theorems, 2 nd 3. To do this we will use the following rther simple ide three times: Suppose tht φ : [.] R is continuous function. Then, for the ove very fine prtition t 0 < t <... < t n we ssert tht φ is lmost constnt, lmost equl to φt j on the very short intervl [t j, t j ]. So t j t j φtdt is pproximtely equl to t j t j φt j dt which equls t j t j φt j. It follows tht tj tj 6 φtdt φtdt φt j dt t j t j φt j. t j t j To finish our explntion of Theorem, we pply 6 in the specil cse where φt is the continuous function φt x t 2 + y t 2 +z t 2. This gives x t 2 + y t 2 +z t 2 dt t j t j x t j 2 + y t j 2 +z t j 2 s required. To finish our explntion of Theorem 2, we pply 6 in the specil cse where φt is the continuous function φt ρxt, yt, zt x t 2 + y t 2 +z t 2. This gives ρxt, yt, zt x t 2 + y t 2 +z t 2 dt t j t j ρxt j, yt j, zt j x t j 2 + y t j 2 +z t j 2 M j s required. Finlly, for Theorem 3, we pply 6 in the specil cse where φt is the continuous function φt F rt x tî + y tĵ+z t k F xt, yt, zt x tî + y tĵ+z t k. This gives F xt, yt, zt x tî + y tĵ+z t k dt t j t j F xt j, yt j, zt j W j. x t j î + y t j ĵ+z t j k This completes our intuitive explntion of these three theorems. Now perhps you might like to go ck to red the remrk out independence of the choice of prmetric representtion one more time. Theorem is truly mthemticl theorem ecuse we hve mthemticl definition of the length of curve. Theorems 2 nd 3 should proly not relly e clled theorems. They del with physicl quntities, which we hve not defined precisely from mthemticl point of view. But the mthemticl conclusion from ech of them, if we elieve the physics ehind them, is tht the vlues of integrls ppering in 2 nd 4, just like the integrl ppering in do NOT depend on our prticulr choice of the prmetric representtion for. Now we re lmost redy to mke conceptul jump forwrd. Rememer tht the originl motivtion for introducing the integrl β fxdx ws to e le to hve tool for clculting nd defining the re under the grph of positive function f on some intervl [, β]. But fterwrds we discovered tht the integrl hd ll sorts of other interprettions nd pplictions. For exmple, if fx πrx 2 the integrl is the volume L j
5 of ody otined y rotting the region {x, y : x β, 0 y rx} out the x xis. Or if ft is our velocity t time t then β fxdx is the distnce or displcement/ h tk tht we hve trvelled from time t to time t β. Ultimtely we understnd the integrl β fxdx for ritrry integrle or continuous functions in wy which is detched from those prticulr motivting pplictions. Exctly nlogously to this we now wnt to understnd the formulæ ppering in 2 nd 4 in more generl wy: First let us suppose tht ρ is some, ny, continuous function on the set. It does not hve to e density. It cn lso e negtive. Then we introduce the new nottion ρx, y, zds or simply ρds for the integrl ρxt, yt, zt x t 2 + y t 2 +z t 2 dt in 2 nd we cll it the line integrl of first type of the function ρ long the curve. In similr wy we will now let F e ny kind of vector field whose components re ll continuous functions on. It does not hve to e force field. It could e velocity field, or simply mthemticl vector field without ny prticulr physicl interprettion. We introduce the new nottion F d r for the integrl F xt, yt, zt x tî + y tĵ+z t k dt in 4 nd we cll it the line integrl of second type of the vector field F long the curve. The prmetriztion rt nd the endpoint vlues nd for t do not pper nywhere in the symols ρds nd F d r. But this is s it should e ecuse plese excuse me for sying this so mny times the vlues of ρds nd F d r do not depend on our prticulr choices of rt, nd. They DO of course depend on ρ nd F nd our choice of the curve. And F d r lso depends on the direction long which we move or our prmetric representtion moves long. There is nother lterntive nottion which you will sometimes see for the line integrl F d r. If u, v nd w re the components of F, s in 3 people sometimes use the nottion 7 ux, y, zdx + vx, y, zdy+wx, y, zdz insted of F d r. But I must wrn you tht 7 is very dngerous nottion! I sy this ecuse I hve seen mny students try to use this nottion to hint t wy for clculting the vlue of the integrl. There re ll sorts of historicl resons for using the nottion 7 nd lso the nottion F d r. But they only men one thing for us, the following thing: When we re sked to clculte such n integrl, we first hve to find some suitle prmetric representtion for the curve, nd then sustitute it the formul 4 nd clculte the Hedv m integrl which results from this sustitution. Any other pproch, such s trying to give independent menings to the symols d, dx or dy or dz or d r which re prt of F d r or prt of 7 will lmost certinly led you to mke mistke. Exercise. onsider the curve defined exctly s in Exmple with 0 nd 00π. Use the formul to clculte the length of this curve. Your nswer will e completely ridiculous. This shows wht cn go wrong if we do not impose the condition *. Remrk 2. The nswers to Questions, 2 nd 3, nd the definitions of the two kinds of line integrl cn e generlized to the cse of certin curves whose prmetric representtions do not quite stisfy condition **. We cn consider curves which re finite union of curves, 2,..., N, ech with representtions stisfying **. The integrl on is the sum of the integrls on ech of the curves j, j, 2,..., N. For exmple the curve in R 2 which is the oundry of the region D {x, y : x 0, y 0, x 2 + y 2 } consists of three curves ech with representtions stisfying **, two stright line segments nd qurter circle. There is no wy to find prmetric representtion rt for ll of which hs continuous derivtives for ll three vlues of t corresponding to the points 0, 0,, 0 nd 0,, which re corners pinot of. But we otin the line integrl of some function or vector field on y clculting the corresponding line integrls on ech of the three ove mentioned curves nd then tking their sum.