10 Vector Integral Calculus


 Ethelbert Taylor
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1 Vector Integrl lculus Vector integrl clculus extends integrls s known from clculus to integrls over curves ("line integrls"), surfces ("surfce integrls") nd solids ("volume integrls"). These integrls hve sic engineering pplictions in solid mechnics, fluid flow nd het prolems. These different kinds of integrls cn e trnsformed into one nother. This is done to simplify evlutions or to gin useful generl formuls, for instnce, in potentil theory (see Sec..8).. Line integrls The concept of line integrl is simple nd nturl generliztion of definite integrl, f ( xdx ), (..) known from clculus. In eqn (..) the integrnd f ( x ) is integrted from x long the x  xis to x. In line integrl given function (lso clled the integrnd) is integrted long curve in spce (or in the plne). Hence, curve integrl would e etter nme, ut line integrl is stndrd. Represent the curve y prmetric representtion (see Sec. 9.5), r() t x() t i y() t j z() t k, ( t ). (..) The curve is clled the pth of integrtion, A: r ( ) its initil point, nd B : r ( ) its terminl point. urve is now oriented. The direction from A to B, in which t increses, is clled the positive direction on nd cn e mrked y n rrow (Fig. 7). The points A nd B my coincide (Fig. 7) in which cse the curve is clled closed pth. The curve is clled smooth curve if it hs t ech point unique tngent whose direction vries continuously long curve. This mens tht r() t in eqn (..) is differentile nd the derivtive r ( t) dr/ dt is continuous nd different from the zero vector t every point of curve. Generl ssumption: In this chpter nd ll other chpters every pth of integrtion of line integrl is ssumed to e piecewise smooth, i.e., consists of finitely mny smooth curves... Definition nd evlution of line integrls A line integrl of vector function Fr ( ) over curve : r( t) (s in eqn (..)) is defined y, Fr () dr= Fr (()) t r () t dt, (..3) June 5,./6
2 where r ( t) dr ( t) / dt. In terms of components, with dr( t) dx i dy j dz k s in Sec. 9.5 nd () d()/ dt, eqn (..3) ecomes, Fr () dr= ( F dx + F dy + F dz), 3 ( Fx Fy Fz 3 ) dt. = + + (..3*) If the pth of integrtion is eqn (..3) is closed curve, then insted of we lso write. Note tht the integrnd Fr (()) t r () t in eqn (..3) is sclr nd indeed F r / r is the tngentil component of F. The integrl in eqn (..3) on the righthnd side is definite integrl of function of t tken over the intervl t on the t xis in the positive direction (the direction of incresing t ). This definite integrl exists for continuous F nd piecewise smooth, ecuse this mkes F r piecewise continuous. Line integrls eqn (..3) rise nturlly in mechnics, where they give the work done y force F in displcement long (see exmples elow). Hence, the line integrl eqn (..3) my lso e clled the work integrl. Other forms of the line integrl will e discussed lter in this section. Exmple : Evlution of line integrl in the plne Find the vlue of the line integrl eqn (..3) when Fr ( ) y i xy j nd is the circulr rc in Fig. 8 from A to B. Solution: Represent the circulr rc y r( t) costi sin tj, where t / x() t cos, t y() t sint nd. Then Fr ( ()) t y() t i x() t y() t j sinti costsintj. By differentition, r ( t) sinti costj, so tht y eqn (..3) [use eqn () in Appendix 3., set cos t u in the second term], / Fr ( ) dr= (sin ti costsin tj) (sin ti+ cos tj) dt, / (sin t cos tsin t) dt, =  = =  = 4 3 / ( cos tdt ) u( du).45. June 5,./6
3 Exmple : Line integrl in spce The evlution of line integrls in spce is prcticlly the sme s it is in the plne. To illustrte this, find the vlue of eqn (..3) when Fr () zi xj yk nd is the helix (Fig. 9), r() t costi sintj 3 tk, t. (..4) Solution: From eqn (..4), we hve x() t cos, t y() t sin t, z() t 3t. Thus, ( ( )) ( ) (3 cos sin ) ( sin cos 3 ) Frt r t ti tj tk ti tj k. Hence, eqn (..3) gives, ( ) dr= ( 3tsin t+ cos t+ 3sin t) dt= 7 =.99. Fr Simple generl properties of the line integrl eqn (..3) follow directly from corresponding properties of the definite integrl in clculus, nmely, kf dr= k F dr, ( k constnt) (..5) d ( F+ G) dr= F dr+ G r, (..5) F dr= F dr+ F dr, (see Fig. ) (..5c) where in eqn (..5c) the pth is sudivided into two rcs nd tht hve the sme orienttion s (Fig. ). In eqn (..5) the orienttion of is the sme in ll three integrls. If the sense of integrtion long is reversed, the vlue of the integrl is multiplied y. However, we note the following independence if the sense is preserved. June 5,.3/6
4 Theorem : Directionpreserving prmetric trnsformtions Any representtions of tht give the sme positive direction on lso yield the sme vlue of the line integrl eqn (..3). Proof: A proof follows y the chin rule, where r () t is the given representtion, t (*) t with positive derivtive dt / dt * is the trnsformtion, with * t* * corresponding to t, in eqn (..3), nd we write r() t r ( ( t*)) r*(*) t. Then dt ( dt / dt*) dt * nd, * dr dt Fr (*) dr* = Fr (((*))) t dt*, * dt dt * dr = Fr (()) t dt= Fr () r. dt d.. Motivtion of the line integrl eqn (..3): Work done y force The work W done e constnt force F in the displcement long stright segment d is W F d; see Ex. in Sec. 9.. This suggests tht we define the work W done y vrile force F in the displcement long curve : r( t) s the limit of sums of work done in displcements long smll chords of. It will e shown tht this definition mounts to defining W y the line integrl eqn (..3). For this we choose points t( ) t... tn( ). Then the work Wm done y Fr ( ( t m )) in the stright displcement from ( ) to r ( ) is, r t m tm W Fr (( t )) [( rt ) r( t )] Fr (( t )) r ( t ) t, m m m m m m m where tm tm t m. The sum of these n works is Wn W... W n. if we choose points nd consider W for every n ritrrily ut so tht the gretest pproches zero s n, n then the limit of Wn s n is the line integrl eqn (..3). This integrl exists ecuse of our generl ssumption tht F is continuous nd is piecewise smooth; this mkes r ( t) continuous, except t finitely mny points where my hve corners or cusps. Exmple 3: Work done y vrile force If F in Exmple is force, the work done y F in the displcement long the qurtercircle is.45, mesured in suitle units, sy Newtonmeters ( nt m, Nm, lso clled joules, revition J). Similrly in Exmple. t m Exmple 4: Work done equls the gin in kinetic energy Let F e force, so tht eqn (..3) is work. Let t e time, so tht velocity. Then we cn write eqn (..3) s, dr/ dt v, W = F dr= F(()) r t v() t dt By Newton's second lw ( force mss ccelertion ), F mr () t mv () t,. (..6) June 5,.4/6
5 where m is the mss of the ody displced. Sustitution into eqn (..5) gives, æv vö m W = mv vdt= m ç dt= v çè ø t=. t= On the right, m v / is the kinetic energy. Hence the work done equls the gin in kinetic energy. This is sic lw in mechnics...3 Other forms of line integrls The line integrls, Fdx,, Fdy 3 re specil cses of eqn (..3) when F F i or F j or Fdz, (..7) F3 k, respectively. Furthermore, without tking dot product s in eqn (..3) we cn otin line integrl whose vlue is vector rther thn sclr, nmely, Fr () dt = Fr (()) t dt = [ F (()) rt i+ F (()) rt j+ F (()) rt k] dt. (..8) Another form of line integrl is, 3 f () r dt = f (()) r t dt, (..8*) with s in eqn (..). The evlution is similr to tht efore. Exmple 5: A line integrl of the form eqn (..8) Integrte Fr () xyi yzj zk long the helix in Exmple. Solution: Fr (()) t cossin t ti 3sin t tj 3 tk, integrted with respect to t from to gives,..4 Pth dependence 3 Frt dt ti t t t t ( ( )) = [ cos + (3sin  3 cos ) j+ k] = [i 6 j+ 6 k]. Pth dependence of line integrls is prcticlly nd theoreticlly so importnt tht we formulte it s theorem. A whole section (Sec..) will e devoted to conditions under which pth dependence does not occur. Theorem : Pth dependence The line integrl eqn (..3) generlly depends not only on F nd on the endpoints A nd B of the pth, ut lso on the pth itself long which the integrl is tken. Proof: Almost ny exmple will show this. Tke the stright segment : r( t) ti tj k nd the prol : r ( t) ti t j k with t (Fig. ) nd integrte June 5,.5/6
6 4 F i xy j k. Then Fr ( ()) t r () t t, Fr ( ()) t r () t t, so tht integrtion gives / 3 nd /5, respectively. June 5,.6/6
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