4. F = i + sin xj. (page 811) 1. F = xi + xj. 5. F = e x i + e x j. 2. F = xi + yj. 3. F = yi + xj. 6. F = (x 2 y) = 2xi j.

Transcription

1 ETION 5 PAGE 8 R A ADAM: ALULU HAPTER 5 ection 5 pge 8 VETOR FIELD Vector nd clr Fields F i + j The field lines stisf d d, ie, d d The field lines re +, stright lines prllel to 4 F i + sin j The field lines stisf d d sin Thus d sin The field lines re the curves d cos + Fig 54 F i + j The field lines stisf d Fig 5 d Thus ln ln + ln, or The field lines re stright hlf-lines emnting from the origin 5 F e i + e j The field lines stisf d e d e Thus d d e The field lines re the curves e + Fig 5 F i + j The field lines stisf d d Thus d d The field lines re the rectngulr hperbols nd their smptotes given b Fig 55 6 F i j The field lines stisf d d The re the curves ln + Fig 5 Fig 56 57

2 INTRUTOR OLUTION MANUAL ETION 5 PAGE 8 7 F ln + The field lines stisf d lines nd i + j + d Thus the re rdil Fig 57 8 F cos i cos j The field lines stisf d d, tht is, cos cos cos d+ cos d Thus the re the curves sin + sin v This implies tht sin + The stremlines re the spirls in which the surfces sin intersect the clinders + i + j + + The stremlines stisf d nd d d Thus nd The stremlines re horiontl hlf-lines emnting from the -is v i + j + k The field lines stisf d d d, or, equivlentl, d/ d/ nd d d Thus the field lines hve equtions, +, nd re therefore prbols 4 v e i + j + k The field lines stisf d d d, so the re given b,ln ln / or, equivlentl, e / 5 v, i j The field lines stisf d/ d/, so the re given b ln / + ln, or e / 6 v, i + + j The field lines stisf Fig 58 9 v,, i j k The stremlines stisf d d d Thus +, + The stremlines re stright lines prllel to i j k v,, i + j k The stremlines stisf d d d Thus +, The stremlines re stright hlflines emnting from the -is nd perpendiculr to the vector i + k v,, i j + k The stremlines stisf d d d Thus d+ d, so + Therefore, d d d d + d d + v + dv d + v Let v d d v + dv d + v Thus dv/d /, nd so v ln + The field lines hve equtions ln + 7 F ˆr + r ˆθ The field lines stisf dr, so the re the spirls r θ + 8 F ˆr + θ ˆθ The field lines stisf dr r/θ,or dr/r /θ, so the re the spirls r θ 9 F ˆr + θ ˆθ The field lines stisf dr/ r/θ,or dr/r /θ, so the re the spirls r θ F r ˆr ˆθ The field lines stisf dr/r r, or dr/r, so the re the spirls /r θ +, or r /θ + 57

3 ETION 5 PAGE 8 R A ADAM: ALULU ection 5 pge 89 onservtive Fields F i j + k, F, F, F We hve F F, F F, F F Therefore, F m be conservtive If F φ, then,, Evidentl φ,, + is potentil for F Thus F is conservtive on R F i + j + k, F, F, F We hve F F, F F, F F Therefore, F m be conservtive If F φ, then,, Therefore, φ,, d +, +,, φ,, + Thus φ,, + is potentil for F, nd F is conservtive on R i j F +, F +, F We hve + F +, F + Thus F cnnot be conservtive 4 F i + j +, F +, F F + F We hve + Therefore, F m be conservtive If F φ, then Therefore, φ, +, d ln + Thus we cn choose, nd c c φ, ln + is sclr potentil for F, nd F is conservtive everwhere on R ecept t the origin 5 F i + + j k, F, F +, F We hve F F, F F, F F Therefore, F m be conservtive If F φ, then, +, Therefore, φ,, d +, + +, + φ,, Thus φ,, + is sclr potentil for F, nd F is conservtive on R 57

4 INTRUTOR OLUTION MANUAL ETION 5 PAGE 89 6 F e + + i + j + k F e + +, F e + +, F e + + We hve F e + + F, F + e + +, F + e + + F Thus F cnnot be conservtive 7 φr r r r r r r r r r r r r r r r 4 8 ince similr formuls hold for the other first prtils of φ, wehve F φ [ ] r r 4 i + j + k r r r r 4 This is the vector field whose sclr potentil is φ ln r r ln r r r r r i + j + k r r r 9 F i + j + k, F, F, F + We hve F F, F F, F F Therefore, F m be conservtive in R ecept on the plne where it is not defined If F φ, then Therefore, φ,, d +,, + φ,, Thus φ,, + is potentil for F, nd F is conservtive on R ecept on the plne The equipotentil surfces hve equtions +, or + Thus the equipotentil surfces re circulr prboloids The field lines of F stisf d d d + From the first eqution, d d,so A for n rbitrr constnt A Therefore d d + so + A d d Hence d + A, + A + B, or + + B, where B is second rbitrr constnt The field lines of F re the ellipses in which the verticl plnes contining the -is intersect the ellipsoids + + B These ellipses re orthogonl to ll the equipotentil surfces of F F i + j + k G + k, where G is the vector field F of Eercise 9 ince G is conservtive ecept on the plne, so is F, which hs sclr potentil,, + φ,, , 57

5 ETION 5 PAGE 89 R A ADAM: ALULU since + is potentil for G nd is potentil for the vector k or The equipotentil surfces of F re φ,,, + + which re spheres tngent to the -plne hving centres on the -is The field lines of F stisf d d d + As in Eercise 9, the first eqution hs solutions A, representing verticl plnes contining the -is The remining equtions cn then be written in the form d d + A This first order DE is of homogeneous tpe see ection 9, nd cn be solved b chnge of dependent vrible: v We hve v + dv d d d v + A v dv d v + A v v + + A v v v dv v + + A d ln v + + A ln + ln B The sclr potentil for the two-source sstem is φ,, φr m r lk m r + lk Hence the velocit field is given b vr φr mr lk mr + lk + r lk r + lk mi + j + lk mi + j + + lk [ + + l + ] / [ + + l ] / Observe tht v if nd onl if, nd v if nd onl if Also l v,, m l + + l + l k, which is if nd onl if Thus v onl t the origin At points in the -plne we hve v,, mi + j + + l / The velocit is rdill w from the origin in the -plne, s is pproprite b smmetr The speed t,, is v,, m l / ms s + l gs, / where s + For mimum gs we set v + + A B + + A B + + B These re spheres centred on the -is nd pssing through the origin The field lines re the intersections of the plnes A with these spheres, so the re verticl circles pssing through the origin nd hving centres in the -plne The technique used to find these circles ecludes those circles with centres on the -is, but the re lso field lines of F Note: In two dimensions, circles pssing through the origin nd hving centres on the -is intersect perpendiculrl circles pssing through the origin nd hving centres on the -is Thus the nture of the field lines of F cn be determined geometricll from the nture of the equipotentil surfces s + l / g s m ss + l / s s + l ml s s + l 5/ Thus, the speed in the -plne is gretest t points of the circle + l / The sclr potentil for the source-sink sstem is φ,, φr r + r k Thus, the velocit field is v φ r r r k r k i + j + k i + j + k + + / + + / 574

6 INTRUTOR OLUTION MANUAL ETION 5 PAGE 89 For verticl velocit we require + + / + + /, nd similr eqution for Both equtions will be stisfied t ll points of the -is, nd lso wherever + + / + + / / K K K, where K / / / This ltter eqution represents sphere,, since K K > The velocit is verticl t ll points on, s well s t ll points on the -is ince the source t the origin is twice s strong s the sink t,,, onl hlf the fluid it emits will be sucked into the sink B smmetr, this hlf will the hlf emitted into the hlf-spce > The rest of the fluid emitted t the origin will flow outwrd to infinit There is one point where v This point which is esil clculted to be,, + lies inside tremlines emerging from the origin prllel to the -plne led to this point tremlines emerging into > cross nd pproch the sink tremlines emerging into < flow to infinit ome of these cross twice, some others re tngent to, some do not intersect nwhere so the velocit t n point,, is 4 For v, m v unit vector in direction i + j + mi + j + mi + j +,wehve v m + v, so v m be conservtive, ecept t, We hve d φ, m + m ln + + m + m + + d d Thus we m tke, nd obtin φ, m ln + m ln r, s sclr potentil for the velocit field v of line source of strength of m 5 The two-dimensionl dipole of strength µ hs potentil Fig 5 φ, [ m lim ln + l mlµ ln + µ lim l ppl l Hôpitl s Rule l + l µ lim l µ + µ r l ln + + l ] l ln + + l l + l + + l Fluid emitted b intervl in time intervl [, t] occupies, t time t, clinder of rdius r, where πr Z vol of clinder πmt Thus r mt, nd r dr m The surfce of this dt clinder is moving w from the -is t rte dr dt m r m +, Now µ r r µ r 4 r r µ r r 4 µ r 4 Thus µ F φ + i + j 575

7 ETION 5 PAGE 89 R A ADAM: ALULU 6 The equipotentil curves for the two-dimensionl dipole hve equtions or µ µ + + µ µ 4 These equipotentils re circles tngent to the -is t the origin 7 All circles tngent to the -is t the origin intersect ll circles tngent to the -is t the origin t right ngles, so the must be the stremlines of the two-dimensionl dipole As n lterntive derivtion of this fct, the stremlines must stisf d d, or, equivlentl, d d This homogeneous DE cn be solved s ws tht in Eercise b chnge in dependent vrible Let v Then v + dv d d d v v dv d v v v + v v v dv v + d lnv + ln + ln v These stremlines re circles tngent to the -is t the origin 8 The velocit field for point source of strength mdt t,, t is m i + j + tk v t,, / + + t Hence we hve v t,, dt m i + j + tk + + t / dt dt mi + j / + + t Let t + tn θ dt + sec θ mi + j π/ + cos θ π/ mi + j +, which is the velocit field of line source of strength m long the -is The definition of strength of point source in -spce ws mde to ensure tht the velocit field of source of strength hd speed t distnce from the source This corresponds to fluid being emitted from the source t volume rte of 4π imilrl, the definition of strength of line source gurnteed tht source of strength gives rise to fluid speed of t unit distnce from the line source This corresponds to fluid emission t volume rte π per unit length long the line Thus, the integrl of -dimensionl source gives twice the volume rte of -dimensionl source, per unit length long the line The potentil of point source mdt t,, t is m φ,, + + t This potentil cnnot be integrted to give the potentil for line source long the -is becuse the integrl m dt + + t does not converge, in the usul sense in which convergence of improper integrls ws defined 9 ince r cos θ nd r sin θ, wehve Also, r θ cos θ + sin θ r sin θ + r cos θ i + j ˆr cos θi + sin θj r i + j ˆθ sin θi + cos θj r 576

8 INTRUTOR OLUTION MANUAL ETION 5 PAGE 84 Therefore, r ˆr + r θ ˆθ cos θ + sin θ cos θ i + sin θ + cos θ sin θ j + sin θ sin θ cos θ i + cos θ sin θ + cos θ i + j φ j If F F r r,θˆr + F θ r,θˆθ is conservtive, then F φ for some sclr field φr,θ, nd b Eercise 9, r F r, r θ F θ For the equlit of the mied second prtil derivtives of φ, we require tht F r θ tht is, F r θ r F θ r r rf θ F θ + r F θ r, F θ If F r sinθˆr + r cosθˆθ φr,θ, then we must hve r r sinθ, r θ r cosθ Both of these equtions re stisfied b φrθ r sinθ+, so F is conservtive nd this φ is potentil for it If F r cos θ ˆr + αr β sin θ ˆθ φr,θ, then we must hve r r cos θ, r θ αr β sin θ From the first eqution φr,θ r cos θ + θ The second eqution then gives θ r sin θ θ αr β+ sin θ This eqution cn be solved for function θ independent of r onl if α / nd β In this cse, θ constnt F is conservtive if α nd β hve these vlues, nd potentil for it is φ r cos θ + ection 5 Line Integrls pge 84 : r cos t sin ti + sin tj + cos tk, t π/ ince r cos t sin t + sin 4 t + cos t for ll t, must lie on the sphere of rdius centred t the origin We hve Thus ds cos t sin t + 4 sin t cos t + sin tdt cos t + sin t + sin tdt + sin tdt π/ ds cos t + sin tdt Let u sin t du cos tdt + u du Let u tn φ du sec φ dφ π/4 sec φ dφ [ sec φ tn φ + ln sec φ + tn φ + ln + ] π/4 : t cos t, t sin t, t, t π We hve Thus ds cos t t sin t + sin t + t cos t + dt + t dt ds π t + t dt Let u + t +4π u/ +4π u / du du t dt + 4π / / 577

9 ETION 5 PAGE 84 R A ADAM: ALULU Wire: r ti + t j + t k, t v i + 6tj + 6t k v + 4t + 4t 4 + t If the wire hs densit δt + t g/unit length, then its mss is m + t + t dt t t t t4 8g 4 The wire of Emple lies in the first octnt on the surfces nd, nd, therefore, lso on the surfce,or +, circulr clinder ince it goes from,, to,, it cn be prmetried r cos ti + sin tj + cos k, t π/ v sin ti + cos tj cos t sin tk v + sin t cos t ince the wire hs densit δ sin t cos t sint, its mss is m 4 π/ cos t sint dt v dv Let v cost dv sint dt v dv, which is the sme integrl obtined in Emple, nd hs vlue π + /8 5 : r e t cos ti + e t sin tj + tk, t π ds e t cos t sin t + e t sin t + cos t + dt + e t dt The moment of inerti of bout the -is is I δ + ds δ π +e 4π e t + e t dt Let u + e t du 4e t dt δ udu 4 δ +e4π 6 u/ δ [ + e 4π / /] 6 6 is the sme curve s in Eercise 5 We hve e ds π e t + e t dt tπ sec θ t Let e t tn θ e t dt sec θ [ ] tπ sec θ tn θ + ln sec θ + tn θ t e t + e t + ln e t + + e t eπ + e 4π + e π ln + + e 4π + 7 The line of intersection of the plnes + nd + + from,, to,, cn be prmetried r ti + tj tk, t Thus ds 4 dt nd ds 4 9t dt 4 8 The curve of intersection of + nd cn be prmetried r cos ti + cos tj + sin tk, t π Thus ds sin t + 4 sin t cos t + cos tdt + sin tdt We hve + 4 ds π π π + 4 cos t sin t + sin tdt + sin t dt + π π cos 4t dt π 578

10 INTRUTOR OLUTION MANUAL ETION 5 PAGE 84 9 r cos ti + sin tj + tk, t π v sin ti + cos tj + k, v If the densit is δ t, then m π M M M π π π tdt π t cos tdt t sin tdt π t dt 8π We hve omitted the detils of the evlution of these integrls The centre of mss is, π, 4π Here the wire of Eercise 9 etends onl from t to t π: m π tdt π M M M The centre of mss is π π π t cos tdt t sin tdt π t dt π 4 π, π, π r e t i + tj + e t k, t v e t i + j e t k v e t + + e t e t + e t + ds e t + e t e t + e t dt m M M e t + e t + e t + e t dt e + e e e e t + e t dt e e e t e t + e t dt e + te t + e t dt e e M e t e t + e t dt e e e + e The centroid is e, e +, e e e The first octnt prt of the curve +,, cn be prmetried r cos ti + sin tj + cos tk, t π/ We hve ds + sin tdt,so π/ ds cos t + sin tdt Let sin t tn θ cos tdt sec θ tπ/ sec θ t [ sec θ tn θ + ln sec θ + tn θ [ sin t + sin t + ln sin t + [ + ln + ] ] tπ/ t + sin t ] π/ 4 On, wehve Thus cn be prmetried Hence r ti + tj + t t k, t ds We hve ds + + t t t dt dt t t t t dt t t 5 The prbol +, +, cn be prmetried in terms of t since + + t t t + t Thus ds t + + t dt + t dt, nd ds + t + / t dt + / dt + t tn t π π 579

11 ETION 5 PAGE 84 R A ADAM: ALULU 6 :,, from,, to, 4, 6 Prmetrie b ection 54 pge 8 Line Integrls of Vector Fields r ti + t j + t 4 k, t ince ds + 4t + 6t 6 dt,wehve ds t 7 + 4t + 6t 6 dt 7 Heli: cos t, b sin t, ct < < b ds sin t + b cos t + c dt c + b b sin tdt b + c k sin tdt k b b + c One complete revolution of the heli corresponds to t π, nd hs length π L b + c k sin tdt π/ 4 b + c k sin tdt 4 b + c Ek 4 b + c E b b + c units The length of the prt of the heli from t to t T <π/is T L b + c k sin tdt b + c Ek, T b + c E b b + c, T units 8 The stright line L with eqution A + B,, lies t distnce D / A + B from the origin o does the line L with eqution D ince + depends onl on distnce from the origin, we hve, b smmetr, ds L + ds L + d + D D tn D π D π D π A + B F i j : r ti + t j, t F dr [t t t] dt F cos i j sin : sin from, to π, F dr sin π,, t dt 4 F i + j k : r ti + tj + tk, t F dr t + t t dt t 4 F i j + k : r ti + t j + t k, t F dr [t t t + tt ] dt 5t dt 5t F i + j + k : curve from,, to,, ince F is conservtive, it doesn t mtter wht curve,, F dr,, 6 F i + j + k + + is given polgonl pth from,, to,, but n other piecewise smooth pth from the first point to the second would do s well +,, F dr +,, 7 F + i + j + k + + The work done b F in moving n object from,, to,, is +,, W F dr +,, units 58

12 INTRUTOR OLUTION MANUAL ETION 54 PAGE 8 8 is mde up of four segments s shown in the figure On,, d, nd goes from to On,, d, nd goes from to On,, d, nd goes from to On 4,, d, nd goes from to Thus d + d d + d d d + d d d + d 4 Finll, therefore, d + d Fig 548, 9 Observe tht if φ e + sin +, then φ e + sin + i + e + sin + + cos + j + e + cos + k Thus, for n piecewise smooth pth from,, to, π 4, π 4,wehve e + sin + d + e + sin + + cos + d + e + cos + d φ dr φ,,,π/4,π/4,, e +π/4 F + i + j + b + k is conservtive if F F F F F F b If nd b, then F φ where φ + d + +, + F, d d + F + + Thus φ is potentil for F F A ln i + B j + + k is conservtive if F F F F F F A B If A nd B, then F φ where φ ln + If is the stright line t +,, t +, t, from,, to,,, then ln d+ d+ d φ dr d+ d,, ln + [t + + t + ] dt,, t 4ln+ + t 4ln F cos + i + sin 4j + + k sin The curve : sin t, t, t, t, goes from,, to π/,, The work done b F in moving prticle long is W F dr sin + π/,, 4 +,, + 4π π 58

13 ETION 54 PAGE 8 R A ADAM: ALULU For ln +,, from to, we hve [ sinπ e d ] + π cosπ e d e d sinπ e dr e d sinπ e,,ln +,, + d 9 7 consists of two prts: On,, d, nd goes from to On, cos t, sin t, t goes from to π d d+ d π + cos tdt π, d d+ d + π cos t dt π 4 {, : >, } is simpl connected domin b {, :, } is not domin It hs empt interior c {, :, > } is domin but is not connected There is no pth in from, to, d {,, : > } is domin but is not connected There is no pth in from,, to,, e {,, : + > } is connected domin but is not simpl connected The circle +, lies in, but cnnot be shrunk through to point since it surrounds the clinder + which is outside f {,, : + + > } is simpl connected domin even though it hs bll-shped hole in it 5 is the curve r cos ti + sin tj, t π d d π π cos tcos tdt π sin t sin t dt π Fig is mde up of four segments s shown in the figure On,, d, nd goes from to On,, d, nd goes from to On,, d, nd goes from to On 4,, d, nd goes from to d + d d d +, 6 is the curve r cos ti + b sin tj, t π d d π π cos tbcos tdt πb b sin t sin t dt πb 4 Fig

14 INTRUTOR OLUTION MANUAL ETION 54 PAGE 8 9 is mde up of three segments s shown in the figure On,, d, nd goes from to On, bt, t, nd t goes from to On,, d, nd goes from b to d + d + + t bdt + b bt dt + b d c f D 4 Fig 54 g b Fig 549 fg + f g + g + f f g + f g i + g f + f g Thus, since goes from P to Q, k f g + f g j onjecture: If D is domin in R whose boundr is closed, non-self-intersecting curve, oriented counterclockwise, then d re of D, d re of D f g dr + g f dr fg dr fg f QgQ f PgP Q P Proof for domin D tht is -simple nd -simple: ince D is -simple, it cn be specified b the inequlities c d, f g Let consist of the four prts shown in the figure On nd, d On, g, where goes from c to d On, f, where goes from d to c Thus d d c + The proof tht c g f uses the fct tht D is -simple g d + + f d d d re of D d re of D is similr, nd : cos t, sin t, t π d d π + π π Fig 54 cos t + sin t cos t + sin dt t Fig 54b 4 58

15 ETION 54 PAGE 8 R A ADAM: ALULU b ee the figure hs four prts On,, d, goes from to On,, d, goes from to On,, d, goes from to On 4,, d, goes from to d d π + [ d π + + d + d ] + + d + dt π + t π tn t π π 4 + π 4 is not conservtive on n domin in R tht contins the origin in its interior ee Emple 5 However, the integrl will be for n closed curve tht does not contin the origin in its interior An emple is the curve in Eercise c 4 If is closed, piecewise smooth curve in R hving eqution r rt, t b, nd if does not pss through the origin, then the polr ngle function θ θ t, t θt cn be defined so s to vr continuousl on Therefore, θ, tb t π w, where w is the number of times winds round the origin in counterclockwise direction For emple, w equls, nd respectivel, for the curves in prts, b nd c of Eercise ince 4 θ θ i + θ j i + j +, Fig 54 c ee the figure hs four prts On,, d, goes from to On, cos t, sin t, t goes from to π On,, d, goes from to On 4, cos t, sin t, t goes from π to d d π + [ π 4 cos t + 4 sin t + π 4 cos t + 4 sin t dt cos t + sin ] t + + π cos t + sin t dt π π π Although + + for ll,,, Theorem does not impl tht d d + is ero for ll closed curves in R The set consisting of points in R ecept the origin is not simpl connected, nd the vector field i + j F + we hve d d π + θ dr π tb θ, π w ection 55 pge 84 t urfces nd urfce Integrls The polr curve r gθ is prmetried b gθ cos θ, gθ sin θ Hence its rc length element is ds d + d g θ cos θ gθ sin θ + g θ sin θ + gθ cos θ gθ + g θ 584

16 INTRUTOR OLUTION MANUAL ETION 55 PAGE 84 The re element on the verticl clinder r gθ is d ds d gθ + g θ d The re element d is bounded b the curves in which the coordinte plnes t θ nd θ + nd the coordinte cones t φ nd φ + dφ intersect the sphere R ee the figure The element is rectngulr with sides dφ nd sin φ Thus sin φ d sin φ dφ φ d dφ imilrl,, so the surfce re element on the sphere cn be written d The required re is dd d d 4 R π/ π/ π/ 4 sin θ d d 4 u / du 4 cos θ rdr 4 r Let u 4 r du r dr 8 cos θ 6 π/ θ sin θ 8 π sq units θ 4 Fig 55 The plne A + B + D hs norml n Ai + Bj + k, nd so n re element on it is given b d n A d d + B + d d n k Hence the re of tht prt of the plne ling inside the elliptic clinder is given b + b + b πb A + B + A + B + d d sq units 4 One-qurter of the required re is shown in the figure It lies bove the semicirculr disk R bounded b +, or, in terms of polr coordintes, r sin θ On the sphere + + 4,wehve, or r sin θ Fig d F,, F,, d d d F,, F,, dd 6 The clinder + intersects the sphere on the prbolic clinder + 4 B Eercise 5, the re element on + is d i + j dd + dd + + dd dd 585

17 ETION 55 PAGE 84 R A ADAM: ALULU The re of the prt of the clinder inside the sphere is 4 times the prt shown in Figure 5 in the tet, tht is, 4 times the double integrl of d over the region, 4, or d / 4 d d 4 / 6 sq units d + ds d + 7 On the surfce with eqution /wehve / nd / Thus d + d d If R is the first qudrnt prt of the disk +, then the required surfce integrl is d R + d d + d d 4 d Let u u du du d π 4 π 8 8 The norml to the cone + mkes 45 ngle with the verticl, so d d d is surfce re element for the cone Both nppes hlves of the cone pss through the interior of the clinder +, so the re of tht prt of the cone inside the clinder is π squre units, since the clinder hs circulr cross-section of rdius 9 One-qurter of the required re lies in the first octnt ee the figure In polr coordintes, the rtesin eqution + becomes r sin θ The rc length element on this curve is ds r + dr Fig 559 One-eighth of the required re lies in the first octnt, bove the tringle T with vertices,,,,, nd,, ee the figure The surfce + hs norml n i + k, son re element on it cn be written d n n k d d dd d d The re of the prt of tht clinder ling inside the clinder + is T + dd 8 d 8 sq units d + d Thus d + ds r 4 sin θ on the clinder The re of tht prt of the clinder ling between the nppes of the cone is T 4 π/ 4 sin θ 6 sq units,, Fig

18 INTRUTOR OLUTION MANUAL ETION 55 PAGE 84 Let the sphere be + + R, nd the clinder be + R Let nd be the prts of the sphere nd the clinder, respectivel, ling between the plnes nd b, where R b R Evidentl, the re of is π Rb squre units An re element on the sphere is given in terms of sphericl coordintes b + b + b R R d R sin φ dφ On we hve R cos φ, so lies between φ cos b/r nd φ cos /R Thus the re of is Fig 55 π R π R cos φ cos /R cos b/r cos /R cos b/r sin φ dφ Observe tht nd hve the sme re π Rb sq units R A norml to is n j + k, nd the re element on is d n bdd d d n j b Let R be the region of the first qudrnt of the -plne bounded b + b,,, nd Let R be the qurter-disk in the first qudrnt of the plne bounded b +,, nd Then Fig 55 b R We wnt to find A, the re of tht prt of the clinder + inside the clinder + b, nd A, the re of tht prt of + b inside + We hve A 8 re of, A 8 re of, where nd re the prts of these surfces ling in the first octnt, s shown in the figure A norml to is n i + k, nd the re element on is d n dd dd n i A d 8 R π/ b d π/ b sin tdt b sin tdt 8b 8bE sq units b A 8 8b 8b 8b d 8b R 8bE sin /b b d sin /b b, sin b d b Let sin t d cos tdt d b Let b sin t d b cos tdt b sin tdt b sin tdt sq units d d 587

19 ETION 55 PAGE 84 R A ADAM: ALULU The intersection of the plne + nd the cone + hs projection onto the -plne the elliptic disk E bounded b Note tht E hs re A π nd centroid, If is the prt of the plne ling inside the cone, then the re element on is Thus d + d d d d d dd Aȳ π E 4 ontinuing the bove solution, the cone + hs re element d d d d d d d If is the prt of the cone ling below the plne +, then d dd Aȳ 6π E 5 If is the prt of in the first octnt nd inside tht is, below, then hs projection E onto the -plne bounded b,or 4 +, n ellipse ince hs re element d + 4 d d,wehve d E / / d d d d 6 4 d Let u 6 4 u / du 96 du 64 d 6 The surfce hs re element d + + d d + + d d + d d If its densit is k, the mss of the specified prt of the surfce is m k k 5 d 5 5 d k + d + d + d 5k units 7 The surfce is given b e u cos v, e u sin v, u, for u, v π ince, u,v eu sin v e u cos v eu cos v, u,v e u cos v e u sin v eu sin v, u,v eu cos v e u sin v e u sin v e u cos v eu the re element on is d e u cos v + e u sin v + e 4u du dv e u + e u du dv If the chrge densit on is + e u, then the totl chrge is + e u d π e u + e u du dv π e u + eu π e + e 4 8 The upper hlf of the spheroid + + c hs circulr disk of rdius s projection onto the -plne ince + c c, 588

20 INTRUTOR OLUTION MANUAL ETION 55 PAGE 84 nd, similrl, c, the re element on the spheroid is d + c4 + 4 d d + c + d d 4 + c r r rdr in polr coordintes Thus the re of the spheroid is 4π 4π 4πc 4 + c r r rdr Let u r udu rdr 4 + c u du π c c u du c c u du For the cse of prolte spheroid < < c, let k c c Then 4πc 4πc k πc k k u du Let ku sin v kdu cos v dv sin k cos v dv v + sin v cos v sin k πc c c sin + π sq units c 9 We continue from the formul for the surfce re of spheroid developed prt w through the solution bove For the cse of n oblte spheroid < c <, let k c c Then 4πc 4πc k + k u du Let ku tn v kdu sec v dv tn k sec v dv πc tn k sec v tn v + lnsec v + tn v k [ ] πc c c c + ln c + c c π + πc c ln + c c sq units u cos v, u sin v, bv, u, v π This surfce is spirl helicl rmp of rdius nd height πb, wound round the -is It s like circulr stircse with rmp insted of stirs We hve, u,v cos v u sin v sin v u cos v u, u,v sin v u cos v b b sin v, u,v b cos v u sin v b cos v d 4 u + b sin v + b cos v du dv u + b du dv The re of the rmp is A π π u + b du u + b du u πb sec θ u dv Let u b tn θ du b sec θ πb u sec θ tn θ + ln sec θ + tn θ u πb u u + b u + b + ln u + b b π + b + πb + ln + b sq units b 589

21 ETION 55 PAGE 84 R A ADAM: ALULU πb Fig 55 The distnce from the origin to the plne P with eqution A + B + D, D is δ D A + B + If P is the plne δ, then, since the integrnd depends onl on distnce from the origin, we hve P P d + + / d + + / π π δ π u δ rdr r + δ / Let u r + δ du r dr du u / π δ π A + B + D Thus M B smmetr, ȳ, A so the centroid of tht prt of the surfce of the sphere + + ling in the first octnt is,, The cone h + hs norml n i j + k h i + j + k, + so its surfce re element is h d + d d + h d d The mss of the conicl shell is m σ d σ + h π πσ + h + The moment bout is M σ + + h d d + h πσh + h πσh + h r rdr Thus h B smmetr, ȳ The centre of mss is on the is of the cone, one-third of the w from the bse towrds the verte h h h + Use sphericl coordintes The re of the eighth-sphere is A 8 4π π sq units The moment bout is M d π/ π π/ π/ sin φ cos φ sin φ dφ dφ π 4 Fig 55 4 B smmetr, the force of ttrction of the hemisphere shown in the figure on the mss m t the origin is verticl The verticl component of the force eerted b re element d sin φ dφ t the position with sphericl coordintes,φ,θ is kmσ d df cos φ kmσ sin φ cos φ dφ 59

22 INTRUTOR OLUTION MANUAL ETION 55 PAGE 84 Thus, the totl force on m is F kmσ π π/ sin φ cos φ dφ πkmσ units m h,,b ψ D d d φ Fig 555 m Fig is the clindricl surfce +, h, with rel densit σ Its mss is m πhσ ince ll surfce elements re t distnce from the -is, the rdius of grtion of the clindricl surfce bout the - is is D Therefore the moment of inerti bout tht is is 5 The surfce element d d t the point with clindricl coordintes,θ, ttrcts mss m t point,, b with force whose verticl component see the figure is kmσ d kmσb d df cos ψ D kmσ b d + b / The totl force eerted b the clindricl surfce on the mss m is π h F h πkmσ h πkmσ sin tdt h πkmσ cos t D kmσ b d + b / Let b tn t d sec tdt tn tsec tdt sec t h πkmσ + b πkmσ + b h + b I m D m πσ h 7 is the sphericl shell, + +, with rel densit σ Its mss is 4πσ Its moment of inerti bout the -is is I σ + d π π σ sin φ sin φ dφ π πσ 4 sin φ cos φdφ Let u cos φ du sin φ dφ πσ 4 u du 8πσ4 The rdius of grtion is D I /m 8 The surfce re element for conicl surfce, h +, hving bse rdius nd height h, ws determined in the solution to Eercise to be d + h d d 59

23 ETION 55 PAGE 84 R A ADAM: ALULU The mss of, which hs rel densit σ, ws lso determined in tht eercise: m πσ + h The moment of inerti of bout the -is is I σ + d σ + h πσ + h π r rdr 4 4 πσ + h The rdius of grtion is D I /m 9 B Eercise 7, the moment of inerti of sphericl shell of rdius bout its dimeter is I m Following the rgument given in Emple 4b of ection 57, the kinetic energ of the sphere, rolling with speed v down plne inclined t ngle α bove the horiontl nd therefore rotting with ngulr speed v/ is KE mv + I mv mv v m The potentil energ is PE mgh, so, b conservtion of totl energ, 5 6 mv + mgh constnt Differentiting with respect to time t, we get 5 dv m v 6 dt + mgdh 5 dv mv + mgv sin α dt dt Thus the sphere rolls with ccelertion dv dt g sin α 5 ection 56 Oriented urfces nd Flu Integrls pge 848 F i + j The surfce of the tetrhedron hs four fces: On,, ˆN i, F ˆN On,, ˆN j, F ˆN, d d d On,, ˆN k, F ˆN i + j + k On 4, + + 6, ˆN, F ˆN +, 4 4 d d 4 d ˆN j d d We hve F ˆN d F ˆN d 6 F ˆN d d d 6 d F ˆN d d + d d d The flu of F out of the tetrhedron is F ˆN d Fig 56 On the sphere with eqution + + we hve ˆN i + j + k 59

24 INTRUTOR OLUTION MANUAL ETION 56 PAGE 848 If F i + j + k, then F ˆN on Thus the flu of F out of is F ˆN d 4π 4π ˆN + F i + j + k The bo hs si fces F ˆN on the three fces,, nd On the fce, wehve ˆN i, sof ˆN Thus the flu of F out of tht fce is re of the fce bc Fig 564 ˆN B smmetr, the flu of F out of the fces b nd c re lso ech bc Thus the totl flu of F out of the bo is bc c Fig 56 4 F i + k Let be the conicl surfce nd be the bse disk The flu of F outwrd through the surfce of the cone is F ˆN On : ˆN Thus + b i + j + + k, d d d F ˆN d d d + + π π π π π r dr On : ˆN k nd, so F ˆN Thus, the totl flu of F out of the cone is π/ 5 The prt of ling bove b < lies inside the verticl clinder + b For, the upwrd vector surfce element is ˆN d i + j + k d d Thus the flu of F i + j + k upwrd through is F ˆN d [ + + ] d d + b π b r + r dr b π + 4 b π b b 6 For the upwrd surfce element is ˆN d i + j + k d d The flu of F i + j + k upwrd through, the prt of inside + is F ˆN d + + d d + π cos θ r dr + + π π π 4 4 π 7 The prt of 4 ling bove + hs projection onto the -plne the disk D bounded b + 4, or Note tht D hs re 4π nd centroid, For 4, the downwrd vector surfce element is ˆN d i j k d d 59

25 ETION 56 PAGE 848 R A ADAM: ALULU Thus the flu of F i + j + k downwrd through is F ˆN d d d D use the smmetr of D bout the -is da 4π 4π D 8 The upwrd vector surfce element on the top hlf of + + is i + j + k i + j ˆN d d d + k d d The flu of F k upwrd through the first octnt prt of the sphere is π/ F ˆN d r rdr π4 8 9 The upwrd vector surfce element on is i + 4j + k ˆN d d d If E is the elliptic disk bounded b +, then the flu of F i + j through the required surfce is F ˆN d + 4 d d Let u, v E d d du dv 4 u + v du dv now use polrs 4 u +v π r dr π : r u vi + uv j + v k, u, v, hs upwrd surfce element ˆN d r u r du dv v uvi + v j u i + uvj + v k du dv v 4 i 6uv j + u v k du dv The flu of F i + j + k upwrd through is F ˆN d du 6u v 5 6u v 5 + u v 5 dv u du 6 : r u cos vi + u sin vj + uk, u, v π, hs upwrd surfce element ˆN d r u r du dv v u cos vi u sin vj + uk du dv The flu of F i + j + k upwrd through is F ˆN d du π u u du u cos v u sin v + u dv π dv 4π : r e u cos vi + e u sin vj + uk, u, v π, hs upwrd surfce element ˆN d r u r du dv v e u cos vi e u sin vj + e u k du dv The flu of F i j + + k upwrd through is F ˆN d du π e 4u du ue u sin v cos v + ue u sin v cos v + e 4u dv π dv π e4 4 F mr mi + j + k r + + / B smmetr, the flu of F out of the cube,, is 6 times the flu out of the top fce,, where ˆN k nd d d d The totl flu is Fig 56 R 594

26 INTRUTOR OLUTION MANUAL ETION 56 PAGE 848 d d 6m + + / rdr 48m r + / 48m 4m R R s shown in the figure π/4 sec θ rdr r + / Let u r + du r dr +sec θ du u / + sec θ π/4 π/4 48m π 48m 4 π 48m 4 π/4 π/4 cos θ cos θ + cos θ sin θ Let sin v sin θ cos v dv cos θ π π/6 cos v dv 48m 4 cos v π 48m 4 π 4πm 6 4 The flu of F mr out of the cube,, r is equl to three times the totl flu out of the pir of opposite fces nd, which hve outwrd normls k nd k respectivel This ltter flu is mi mi, where I k d d + + k / Let + k tn u d + k sec udu d + k cos udu d + k sin u d + k + + k J k J k, where d J kn n + k + n + k Let n + k tn v d n + k sec v dv sec v dv n ] [n + k tn v + k sec v cos v dv n n + k sin v + k cos v cos v dv n k + n sin Let w n sin v v dw n cos v dv dw k + w w k tn k Thus n sin v tn k k n k tn k + n + k tn k I k k n k 4 + n + k tn [ tn 4 k 8 + k tn + tn k + k ] n k + n + k k 5 + k The contribution to the totl flu from the pir of surfces nd of the cube is mi mi [ m tn tn + tn 6 tn 4 + tn 6 tn ] Using the identities we clculte tn tn, nd tn π tn, tn tn 4 π + 4 tn tn 6 tn π 6 tn 6 Thus the net flu out of the pir of opposite fces is B smmetr this holds for ech pir, nd the totl flu out of the cube is You were wrned this would be difficult clcultion! 595

27 ETION 56 PAGE 848 R A ADAM: ALULU 5 The flu of the plne vector field F cross the piecewise smooth curve, in the direction of the unit norml ˆN to the curve, is F n ds The flu of F i + j outwrd cross the circle + is i + j F ds π π b the boundr of the squre, is 4 i + j 6 F + i + j i d 4 d 8 The flu of F inwrd cross the circle of Eercise 7 is i + j ds ds π π i + j b The flu of F inwrd cross the boundr of the squre of Eercise 7b is four times the flu inwrd cross the edge, Thus it is 4 i + j d + i d tn π The flu out of the other two pirs of opposite fces is lso Thus the totl flu of F out of the bo is b If is sphere of rdius we cn choose the origin so tht hs eqution + +, nd so its outwrd norml is ˆN i + j + k Thus the flu out of is F + F + F ds, since the sphere is smmetric bout the origin Review Eercises 5 pge 848 : t, e t, e t, t v + 4e t + 4e 4t + e t ds + e t e t dt e t + et e e cn be prmetried t, t, t + 4t, t Thus d+ d+ d [4t + t + + 8t] dt t + dt 48 7 The flu of ˆN cross is ˆN ˆN d d re of 8 Let F F i + F j + F k be constnt vector field If R is rectngulr bo, we cn choose the origin nd coordinte es in such w tht the bo is, b, c On the fces nd we hve ˆN ind ˆN i respectivel ince F is constnt, the totl flu out of the bo through these two fces is F F dd b c The cone + hs re element d + + d d d d If is the prt of the cone in the region which itself lies between nd, then d d d + 4 d

28 INTRUTOR OLUTION MANUAL REVIEW EXERIE 5 PAGE The plne + + hs re element d d d If is the prt of the plne in the first octnt, then the projection of on the -plne is the tringle, Thus d 6 d d d Let u 6 du d u u du For, the upwrd vector surfce element is ˆN d i j + k d d The flu of F i j upwrd through, the prt of stisfing nd is F ˆN d d d + d d 6 The plne hs downwrd vector surfce element i j k ˆN d d d If is the prt of the plne in the first octnt, then the projection of on the -plne is the tringle, 6 Thus i + j + k ˆN d d d r sin ti + cos tj + btk, t 6π r j, r6π j + 6πbk The force F mgk mg is conservtive, so the work done b F s the bed moves from r6π to r is W t t6π F dr mg 6πb 6πmgb 8 b v cos ti sin tj + bk, v + b A force of constnt mgnitude R opposing the motion of the bed is in the direction of v, soitis F R v v R + b v ince dr v dt, the work done ginst the resistive force is W 6π R + b v dt 6π R + b F dr cn be determined using onl the endpoints of, provided F + i b j + b + c k is conservtive, tht is, if + F F + F F b + b F F b + c Thus we need, b, nd c With these vlues, F + + Thus F dr + +,,,,, 9 F /i + j + k The field lines stisf d d d Thus d/ d/ nd the field lines re given b +, ln + The field line psses through,, provided nd In this cse the field line lso psses through e, e,, nd the segment from,, to e, e, cn be prmetried rt e t i + e t j + tk, t Then F dr e t + e t + dt e t + t e F + e + i + e + + j k e + + Thus F is conservtive 597

29 REVIEW EXERIE 5 PAGE 848 R A ADAM: ALULU b G + e + i + e + + j k F + j + k : r te t i + tj + tk, t r,,, r,, Thus G dr F dr + j + k dr,, e + + +,, t t + dt e + t e ince the field lines of F re, nd so stisf d+ d, or d d, thus F λ, i j ince F, if,,, λ, ±/ +, nd i j F, ± + ince F, i j/, we need the plus sign Thus F, i j +, which is continuous everwhere ecept t, The first octnt prt of the clinder + 6 hs outwrd vector surfce element ˆN d j + k d d j + k d d 6 The flu of i j k outwrd through the specified surfce is 5 4 F ˆN d d d d 9 d hllenging Problems 5 pge 849 Given: + cos v cos u, + cos v sin u, sin v for u π, v π The clindricl coordinte r stisfies r + + cos v r + cos v r + This eqution represents the surfce of torus, obtined b rotting bout the -is the circle of rdius in the -plne centred t,, ince v π implies tht, the given surfce is onl the top hlf of the toroidl surfce B smmetr, nd ȳ A ring-shped strip on the surfce t ngulr position v with width dv hs rdius +cos v, nd so its surfce re is d π + cos v dv The re of the whole given surfce is π π + cos v dv 4π The strip hs moment d π + cos v sin v dv bout, so the moment of the whole surfce bout is π M π + cos v sin v dv π cos v 4 π cosv 8π Thus 8π 4π The centroid is,, /π π This is trick question Observe tht the given prmetrition ru,v stisfies ru + π,v ru, v Therefore the surfce is trced out twice s u goes from to π It is Möbius bnd ee Figure 58 in the tet If is the prt of the surfce corresponding to u π, nd is the prt corresponding to π u π, then nd coincide s point sets, but their normls re oppositel oriented: ˆN ˆN t corresponding points on the two surfces Hence F ˆN d F ˆN d, 598

30 INTRUTOR OLUTION MANUAL HALLENGING PROBLEM 5 PAGE 849 for n smooth vector field, nd F ˆN d F ˆN d + F ˆN d We hve mde the chnge of vrible t cos φ to get the lst integrl This integrl cn be evluted b using nother substitution Let u bt + b Thus,, b m t + b u, dt udu b b, b t u + b b ψ b cos φ D When t nd t wehveu + b nd u b respectivel Therefore cos φ φ d b F πkmσ u + b +b bu πkmσ +b b + b u πkmσ b b u b +b u b udu b du Fig -5 The mss element σ d t position [,φ,θ] on the sphere is t distnce D + b b cos φ from the mss m locted t,, b, nd thus it ttrcts m with force of mgnitude df kmσ d/d B smmetr, the horiontl components of df coresponding to mss elements on opposite sides of the sphere ie, t [,φ,θ] nd [,φ,θ + π] cncel, but the verticl components kmσ d dfcos ψ D b cos φ D reinforce The totl force on the mss m is the sum of ll such verticl components ince d sin φ dφ, it is F kmσ π πkmσ π b cos φsin φ dφ + b b cos φ / b tdt bt + b / There re now two cses to consider If the mss m is outside the sphere, so tht b > nd b b, then F πkmσ b +b b b +b+ 4πkmσ b However, if m is inside the sphere, so tht b < nd b b, then F πkmσ b + b + b b + b 599