15_3.pdf - TheregionRismore R(r,0 |2 gr 5 5,050 5 21r Thus...

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Unformatted text preview: . TheregionRismore easilydesaibedbypolarooordinates: R: {(r,0) |2 gr 5 5,050 5 21r}. Thus ”A f(z,y)dA=f02"f: f(rcoso,rsin0)rdrd0. . TheregionRismoreeasilydescfibedbyrectangularcoordinates: R: {(z,y) | —l 525 l, —:v 53/5 1}. Thus ffn “2,10“ = IL I; flat, 1;) dz; d2. . TheregionRismoreeasilydescfibedbypolarcoordinates: R: {(r,0) | 0 5r 5 1,1r5 0 5211’}. Thus ffR f(x,y)dA = ff" fol f(rcos 0,rsin0)rdrd0. . TheregionRismoreeasilydescribedbypolarcoordinates: R: {(r,0) | 0 51'53, — Thus HR f(x y)dA= f?" f: f(rc039, rsin0)rdrd0. . 'I'heintegl'alf37/4f12 rdrdOrepresentstheareaoftheregion R: {(130) | l firs 2,1/430531/4},thetopqlm‘terporfionofa ring (annulus). jaw/4f: rdrdO: (f3;/4d0)(f12 Mir)I = [911714 [1 ”L— — 3-1- 7. ThehalfdischanbedescxibedinpolarcoordinatesasD: {(r,0) | 0 955,0 5 05 1r}. Then ffD xzydA =fo"f: (roosfi)2(rsin0)rdrd0= (foxcoszosinada)( :r4dr) —% 0063913 [V]: = -% —1 — 1) - 625 = $ 9. {£2 8in(z2 + ya)“: IW/z 1'13 sin(r2) rdrd0= I[WT/'4! d0 11-3 rsin(r2) dr— _ [error/2 [_% “(72”: = (1) [_;(mg_ml)] = 11m -0089) 11. [be e’-= -v dA: ff” 2 e’-' ”we: ff” d0 fo’re-r’dr 1r/2 Oe 1r/2 = [grif/2 He—r’]: = «Hm—4 — co) = ga — e—4) //m= // — // :2 +y2 <4 (c—1)2+y251 c>0,y>0 y>0 = 3'” or’oosodrdo— f’V’o 2“" r2 casedrda _ —g'/’ 1(8coso)do— f"/2 1(8cos4 o)do 1—823[cos Osin0+ a(0+sin0c030)]::/2 —%[o+%<1)1=¥ 21. 22+y+z=4 4: z:4—2z—y,sothevolumeofthesolidis V =ffl,+v,<1(4—2z-y)dA=f:'£(4-2rc030—rsin0)rdrd0 = fj'fo‘ [4r—r2 (2cosa+sin€)] «mm = [02"[2r’ — §r3(2coso+sino)]:; do = 02"[2—%(2<:080+si110)]do=[20—§(2si110—coa30)];";7r _ 2 25. Theconez = Va."+112intersectsthesphereatz+112 +22 = lwhenzz +y2+ (\/:c’+y2) v: // (m_\/m)dA=/0"[M(m_r)rdrdo z2+y2 51/2 I/fi _ 2 3/2 _ l a] = _ r ) 31' o 27r( 26. The two paraboloids intersect when6 —:1:2 — y2 = 2:02 + 2112 or 22 +312 = 2. For 2:2 +112 5 2, the paraboloid 2:6—z2—y2isabovez=2a:2+2yzso V: // [(6—zz—y2)—(222+2y2)]dA= // [6—3022+y2)]dA=A2"Afi(6—3r2)rdrd0 32+y252 314-1352 = [0" d0 fofl(6r — 3r3) d,» = [a]:r [3r’ — 3-1-4] 3’5 = 21; (e _ 3) = (a,r 28. (a) Hue the region in the zy-plane is the annular region 1'? S 22 + y2 5 1'3 and the desired volume is twice that above the zy-plane Huce 2w 2r V: 2 ff fir: —x’—y2dA= 2/ /: fir: —r2rdrd0= 2/ 0:9ng —r2rdr r1<sz+y3<r2 = 2(21r) [—%(r§ — rafm]r2 = 47"(1'2 — r0372 7'1 (b) Act'oss-secflonalcutisshowninthefigm'e. Sor§ = Gk)2 +r'f or fih’ =r§ —r§. Thusthevolumeintermsofhis V = gem?" = g-hs. Aa/W (22+y)dzdy =/:/o‘a(2rcosa+rsin0)rdrdo _W =fo*(2coso+sino)da 1:1'2dr = [2sin0—<2050]:,r §r3]: =[(o+1)—(o—1)1-%<a‘°‘—0)= 31. TheregionDofintegrationisshowninfllefigme. Inpolarcoordinatesflle line: = flyiso =1r/6,so 1/2 1—,,2 /s 1 [o / zyzdmdy=f / (rcos0)(rsin0)2rdrd0 fig 0 o = o"/‘sin29coeodo f; r‘dr = Gain 0 3" #13 =[§<%> —o] [%—o1=1—;.. 31. AsinExercise 15.2.61, 1'... = fiffo f(a:,y)dA. HereD = {(r,0) | a. S r S b,0 g 0 S 21r}, so A(D) = 1rb2 — 1m2 = «(b2 — a2) and 1 2n _ J——2(b a) _ =—1r(b2 a2) ["10 [r]:— 1r(b2—a2)(21r)(b a): (b+ a)(b— a.) a—+b 4— :2 39.]12 [J— wydydz+/f 2[:I.-ydyd.'i:+/: ./o xydyda: 1/\/_ 1—32 1r/4 f/4 =/ f: r 3cm08in9drd0= / [r—cosesin91mzd0 r—l 15 15 sin’o "’4 1_5 T o /4sin90089d9— TI: 2 :|o= —1—6 40. (3) IL,“ e'('2+”2)dA = L217: re"2 drd0 = 21r [—%e_’2]: = 1r(l — e42) for each a. Then lim 1r(l — e43) = 1r 8—0” since e‘“2 —r 0 as a —> 00. Hence 11’; ff; (3—way?) dA = 1r. (b) Us. e-‘°’+"” M = £21; mt = (I; dz) (I; at) f“ “ch“- Then, from(a), 1r = ffn, —(z2 + y2)dA, so w-alingoffse'('2+’2)dA=1im(f:e")(ffaefl’ady)=(f:°°°e"2dx)(f:°°°e'”2dy). a—‘W To evaluate lim (fie -=’ dz) ( “ r9” dy), we ate using the fact that these integrals are bounded. This is true since 2 _ 2 2 on [—1, 1], 0 < e" S 1 while on (—00, —1), 0 < e" S e' and on (1, oo), 0 < e" < e". Hence 0 S ff; 6“, dz 5 f3; 6' dm + fit dz + L“ e" dz = 2(::-1 + 1). (c) Since ([3; e-""2 dx) (fax e‘"2 dy) = 7r and y canbe replaced by x, (If; e", (1:17)2 = 1r implies that f°° e"2da:=:E\/1—r.Bute‘”, 20fmaflx,sof:°we"2dz=fi. —00 (d) Lettingt = fie, f_°°°° e-'=2 da: = f_°°°° 715 (ta-‘92) dt, so that w? = 715 f; e-*’/2 dt or f; 6"” dt = J27. 41. (a) Weintegiatebypartswithu=zanddv=ze"2dz.Thendu=dzandv=—%e f’te-“de: limfgxze“2dx=lim t—too t—boo t—too = the (—%te‘d‘:)+% ;°e-=’de=o+; o°°e e-= dz [bylHOSpital’sRule] =3 f°°°° e" [since e“"2 isaneven function] = w? [by Exercise 40(c)] (b)Letu=fi.Thmu2=z => dz=2udu => fo°°z e="de 11m};:':e“‘t:l:t,-=limit?”I ue‘“ 2udu=2f°° uze‘“2du=2(%~/7r) lbYPa“(a)]=%~/7'- t—om t—OOO ...
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