Calculus Third Editon By Strauss, Bradley and Smith sec6.4

Calculus (3rd Edition)

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Unformatted text preview: F-1{1 + :2} a: (”51311: = 12 _ 1 + man-.311 = W = «or—w . = (1%=IF+%+[§€T4]R _, Page 269 l1 .— r=sinfl+cuafl 'r2=rainfl'+rcnafl L‘ - The curve 1' = lin’gis traced out-fq'tHI-hé. I] 55 <2r. We fin. the-w: length banning sing sing? + ms: 55 = M: =4 1113- IL 5 = I Eva-(5)3111 E1“! (5]: + [(5).]: d9 9 I}: = I aruzsfsin a .19 fl . = 503(th all 1r": = 25' - 5.4, Arc [myth ind Surface Am “92]? + (2“? d3. 9 a + 4 as = gm +413f3|: W: —3 ai—wL gun—“.- I] Page 2?? It appears the arc length is appmjmmly 1_33_ 29'- {(3}: 5(312:l+;{— 11:- ]|=.:2~m 1+Lf’{z)]! =1+ 391— %-I—IE?-= {:2 + fir: 1* S= 21 £145 + Influx“ + 4—511: = arr Bar“ + I31+ fir+ far—3F: : 2W 112—35?" ]I:=1151151r T 3 :‘h + if [11]} }'~’ d: :1? + £5111: = 21r[;:"+;1ur]11 = 23M! +%1n 3] Numerical approximation is 12M amfhhfifl-h4fi :11: in + (1’33 .. 139—2”): dy = (1+:yfl3}2 _ %+ (iy-afll)! d‘y = 1111*” + if“)? 11 yarn — 2.13 ”gnaw-ht” b. .92 2x —211' Hr—|p lull—q + 39' d! 3. About Lin: y—Im'n. distance is —x: 3=2f—J1‘[_U(%F5I3_ MHXW" 19—21(3)” I] = —211' (:9 ”3 -%-#”W® alt—h... __ —21r(3% 1013 3 '1 _ Hawajafll II I h. .1 I Ii 3 R I: :11 M i: Chipm- 6. Additrbm.‘ Applications of the Intggral' I in. About the n-axis, diatanm in 3.1: I Frying” + iii—2”") in = 211' (Na-1» if“) 119. =2flgqu + 3 m4}3)|1=__fl_9r R: 3.5343 c. About. the lint: y- _ -1, distance is y +' 1 S= 21f[11+ Him”l +1-11-2’31d11 D 2 Wings 1,213+ = Eflg’yflfil +3 3 ”5:3 flh‘fih all. 1 £1.13 +iy—2fal d” §M¢ffl 1 _.1 fins”! =% :1 12.0156 31.1111: -111; a;=‘{1+(—§=dx=§m d3 = 21m 115‘ S: gm'l‘ir dz =W£13113 =31r‘f1l} RS 29.3E38 1fl+tdr is = 211‘: d5 = ‘2‘11'3‘1'1 + a: [it 3 S=21rJl1lfl+md= Lelu=l+x -4w(=c+ 11mm: — 21 3 232, _ —“11—"- 32. .1; = :1“ «is; 111 = fl = 1-5- 25 43.59 33. «£11 =[%r"1'fl — first“) (is: its: {1+(1I—1f2 __ %:l!=:l=dz = 12-112 + 1,112; .11 _1T=(1:—1,i2+%11;2] at: = 5:11:11” + cm} .1: 1 Station 6.4, An: Length Ind Surface Area .—. 1'5: 3: 1 gig — 1%] as 27.12 —(4— 31‘1” dz . 35. Set: the solution to Problem 35. I. p- M]. 11 = 2.1m 1 _ g; .1333} 3”: “113d; {1 _ £2153 }34f'2(2= —1f3 d3) = -a<2)t§}nu ~ :“r'anmlg =21 57. y = tans, y'zmaz. = 111 + 506:]: (11:. 1 5 = 2wItanxV1+ seclr d: u a: 3.6322 by calculatar 3 ‘33. L=I1f1+(y']1 .1. u f *1 1' (£11 01+ {1'11}! 0.0 3.7 2.8328 [1.3 3.9 4.0202 0.5 4.1 4.2202 [1.9 4.1 4.2202 1.2 4.2 4.31.74 1.5 4.4 4.5122 1.3 4.5 4.2074 2.1 4.1] 5.0011] 2.4 5.2 5.2953 2.7 5.5 5.5902 3.0 0.0 5.0825 Page 2.?3 L as 113.3323 + 29.0252] + + 2115.511112] + B.fl528]{0.3} s. 14.11543 39. s: T212111 + {1'55}: .1. "r. r'rz.) =.a.fl+lr'(=n= 0.0 5.7 0.0000 0.3 3.9 1.2079 0.5 4.1 2.5321 0.9 4.1 2.7052 1.2 J1.2 5.1309 1.5 4.4 6.7553 1.8 4.5 3.4?33 2.1 4.9 10.5021 2.4 5.2 12.70 87 2.? 5.5 15.05 35 3.0 5.0 13.2434 5' s: 3qu + 40.2019) + 2(2.5:121} + +4(15.0935) Ham-110.3) :2: 141.3974 — ' _ 1" _ 40. y-fny _-E'.d.afi 1+fid‘z d5 = 2n:- da = 211% r]———h—M d1: {1 S: 21(fijffii + PI: 11: n _ um 41. Let the an: of the graph be aulndivided into small elements, each approximated by .213. Each of 111m in the hypotenuse of a right triangle With legs 22.: and fly. 11111111.I 01”}2 +151 2 = .111 11%): .1: .13 = 2n.’1+(:—::)2 .1... Thus, 5: £12“: 1 + dim, A: fl-iy-l-Ofl- 1:! E; 1' . .21 5: 2a- ally-loo l5:31:11'1 «reg—3:): Ark = 2x I: 1+6?!) dz Page 2.74 42. = 2n- fn Ill l+[f'[:]]2rf: By the MVT there exists an a": such that . {(It— i—fl‘fl fin 1' in") = W = 1?. its -._- ”qfirijlg-I-(Ayfla: 1%): filth = all + [rain]? at a. Let Pr— 1 and Pt be the points with coordinates [it]: _ 1, n_ 1] and {ab Ft)? respectively. When the line segment P*_1P1_ in revolved about an: Nixie, it generates the i'ruatnm of a cane with radii r1 = n_ 1 and r2 = 3* and slant height I? = £1- In the text, We showed that such a fruatum has surface area 'IrIIr'1 + r2}! = r{y*_1 + ydli b. Assume rlgri. Then r1=§r1+%r:¢_:§r1+l£n2 and r=1r+1r >1r+ r 2 52 E? -I‘t '52 aothat ‘ 5 tin-H‘s} S r A similar argument applies if r: 2 r1. Since 1-1 = y*_1 and “I- — y” weaee that {flew 1 + “1' Is a number between 3"]:— and pi. and by applying the intermediate value theorem, i‘ii’r—I + 5'9 = 11¢le for some nnmber ck between :k- 1 and 1.}. Thus, lit—1 + ilk: 2flzck} e. The slant height t", of the line segment P*_ 113* is {re — rh_1]i + {F1- — lift—ll2 = {an}? + [Hal - fiat—I}: l‘IiVT, we have fife: : Ifitn1l_ I,“ if) for some number a: between :*_1 and It. Time, fin) — “It—Ill —.f (”t lire — 171—1) Chapter 6, Additional Appflrafinns efthe integral = f 1:;th and Er: fifirtla + [Ital — .l'itr—lla = f1 + ”1::le ‘33:: It [allows that the surface area generated by revolving the line segment Pi_1P,‘ about the .II-axia ia fist = mitt—1 + Filer = renew: + Lr'ImI‘H” be where efi is the number formed in part b. By adding all such areas associated with the paetitloin P'1 we obtain an approxi- mation for the total surfaee area. of retrain tinn Send by taking the limit as II I’ll—III!I we are let the define S by =“gilerHE i': Earn-ah! 1+[r [=3] 1]“ be 11. :E_1(cl.£:*—oz&=:rauci—Ixmd n-woonallPIImfl. = linlw ): azmkhh + [Haj]1t a: = 2:! m) 1 + If uni a 1H. (r-RJ=+‘y'=r3er1r=1lr=-(e~ R3 r: = a? _ (z — efl'mmrz — R] ' =02 _ (a — lam-ma — m d3: 21:1l1+{y']2 tit : 2r: 1 +[Flea—iffm] d: _ P-(s-maux— R)“ bill: will? I" :21: d; ;;f2 — (I: — R! fir 19:2er 7' '26: R— :2 — (a- R} Letu=.r- R;du=d1;z=u+fi;if e=R+rthenu=nandifz=Rmn thena= —r: = 211'] 732.13%“ r r ._ d -21r1:[ rgfuz-l-RI [Eu—n] -r -r = 21r[-—{:-3 it)”: + ruin—‘1“. j" This corresponds to rotating the upper semi- circle about the y—axis. For the full circle, 3: 432m. 45. y = 0:1“ + He'll-“1 g“ = 21.09““ + 2(1 — unit; ‘2' 1+ (9')? =1+[2nc?“‘1+ 2(1 r renal—1"]? = 1 Hawaii?" ' ‘3 — anon — non 6.5 +4“ — #027394"! = mum-n +t 1- + an — n]3D1:2u_h] Since Snfin — HOD =% = [snafu-1 + 201 — nine ““11 Thus. the arc length is t L: Inna?“ + 2n» — noel-2'] a 2' = [219'- — new-“i1“ = CHEW 02"] _ DH!“ -n] “in-Ht]! 4t. 9' = ”2 — :2; - :3 1+(y)“=1+ '3'— 1“ = :4 +1+___.(r2+ 2)? a 33 g :t ‘3' :5 _ z” 2 _ :3 3 _ 9 2 For the general fomulo, if y = Ct“ + 3:1 _ " 3 1 + (9')” =1+[mr:='*—1 + (2 — one! H]! [now—1 — {n - ewe-“F 4?. We are given Section 6.5. Physical Applications: Work, Liquid Force. and Cell-timid: Hat} =_ full + [IT-2)]: dt = ]n[aec .r + tan 2) Page 2375 0 [or all a: in [1}. 1]. Differentiathtg both sides of this equation! we obtain 2 r/——- + we) a = m—Wtr: re == m 1+ U‘tr)!” = weir [)"(zfllI .—- seat:2 ’5 — 1 = tang: .Iffliz} = :l:t.u.n : fit] = :Flnlcos :H— G Since fill] = l] [the curve 9 7: fix} pnaees through the origin]. 0 = fl and flat] = :hlnlcos :rL Physical Applications: Work, Liquid Force, and Gmtmsids, Page: 495401 The work W done on 3 object moving a distance d in a straight line against a constant farce Fis W = Pd. If the force varies with .r for e 5 :5 t, the work is given by the Integral ‘ W= {life} d: I Fluid pressure in force per unit area exerted by a fluid. The fluid force on a unit haritontni plate of area. A at depth h is phtl. where p in the densit}r of the fluid. For a. plate submerged vertically from depth : = a to a: = t, the fluid force is given by the integrai F = ighlflflzj d1: I where Mr] is the depth at sand 12(3) is the corresponding length of a typical horieontal approximating strip. Dentinr p is a ratio of mass per unit length or per unit area1 or per unit volume. if the density varies in terms or a variable, find the element ol’ mass and then sum up (integrate). Let f and g be continuous and satisfy 1(a)? fix} on the interval [n. t], and consider a thin plate [lamina] of uniform denuit;r p that covers the region R between the graphs of y : {[3] and y = 9(a) on the interval [n, t]. Then the centroid of R is the point (1,?) such that ...
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