15_2.pdf - 1 f f0(8z— 2y dydz[15[Szy— yzfl"=5_dz...

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Unformatted text preview: 1. f: f0 (8z— 2y) dydz: [15 [Szy— yzfl "=5 _dz: ff [8z(z)— (z)2 — 8z(0) + (0)2]dz =f157z2dz: gz3]f= ;(125—1)=fl 2:1; 9. IL, e'Vz dA : L3 f0” e'yz dzdy : f: [ze'yfl] 2:0 AsatypeIregioleiesbetweenthelmva-bomdaryy:Oandtheupper boundaryyzzforogzg l,soD:{(z,y)|0$z5 1,03y3z}.1fwe desuibeDasatypenmgioleiesbetweenflleleflbonmdaryz=yandfl1e fightbomdaryz=lf0r0£ysl,soD={(w,y)Iosyslwszsl}. ThusfszdA: foL'zdydz: folk-31:: :dz: flzzdz:1z3]o: 3(1—0):§ ”panama: mdzdv= 1.. %21:::dy=- 'Ihecurvesy:z—2orz:y+2andz=y2intersectwheny+2=y2 c) y’—y—2:0 a (y—2)(y+1)=0 4: y:—1,y:2,sothepointsof intersectionare(l,—l)and(4,2)_HwedescfibeDasatypeImgiomtheupper boundarycurveisy:fibutthelawerbumdarywrveconsistsoftwoparts, y:—J5for0$z$landy:z—2forlgzg4. ThusD={(z,y)|05zsl,-x/55ny/5}U{(z,y)|15$S£z-2Sny/E}and fnydA:£f_;— ydydz+f4ff 2ydydz. Ifwedescn'beDasatypengiomDisemflosedbythelefiboundary z:y2mdthedghtbomdaryz=y+2fm—lSy$2,soD:{(z,y)|—15y$2,y2Szgy+2}and ”D y (M = fig,” ydz dy. In either case, the resulting iterated integrals am not diflicult to evaluate but the region D is more simply described as atype ngion, giving one iteratedintegral ratherflmna sumoftwo, sowe evaluate the latter integral: fnydA= ffl ,I'zydzdy = f31[zy]:::§'2dy = fie/+2 —y’)ydy = f_21(y2 +211 — y3)dy =[%f+v’-%y‘]’_1=(§+4-4)-(-%+1-%)= AsatypeIregion,D={(z,y)|0$z$£zgyg4}and ffp v’e’”dA = 13‘ J: v’e'"dydz- Asatypenregion. D={(z,y) I 03y34,09sy}andff,,y2erA=fifgyzewdxdy. Evaluating f y2e" dy requires integration by pm whereas f yze'” dz does not, so the iterated integral corresponding to D as a type II region appears easier to evaluate. 2e" dA = j: flyzenady= [6‘ [ye'”]::: dy: f: (gt/e"2 —y) dy = Bet #21:] =(; e1‘—s)— (-—o)= 5e follzfo xeosydydz: f0 [rain/:1: ‘la—dx f0 (L'sina';2 dz = _;m2]; = _;(cos1—coso> = al—mn m v” M = If f.,_ ’3" 9’ dwdy= f1 [W’IZZIi’ dz: = [12 [(7 _ 3y) _ (y _ 1)] 1/2 d9 = 13(81): — 4y3) dy —_gy3_y4]f= ——16—§+l =/::2zy w:_:/gdx =f_2 [2xW-%(4-=t’) +2IW+%(4—”2)]d” [0r,notethat4z\/—74—a: isanoddflmction,sof324z\/—’4—z dz=0.] V= fo1 £55633 + 2y) dydx = In1 [3314 + 1121:? dz = fol [QM/5+ z) — (323 + 24)] dz = 13(333/2 + a: — 32:3 — 2‘) dz =[3-52 1:55/2+-22—:z4—%25 V: ff2f:,(1 +z’y2)dzdy zf—z [3+3 '13 y 2]:::2 dy= f. 2(4+ 6313”2 _ 3y8)dy =[4y+%”-—7’_;y°12=s+—-:—1:+s+%-%:=22—:‘ 39. Thesolidliesbelowtheplanez: 1 —a:—y orz+y+z= landabovetheregion D:{(a:,y)|0$zgl,05ygl—z} inthe zy-plane. The solidisatetrahedron. 43. Thetwosurfacesintersectinfllecircles?+312 = l,z=0andfl1eregionofintegrationisthediskD: 2:2-l-y2 S l. UsingaCAS, thevolurneis/I./ID(1-:t:2 —y 2)dA: /: /;(1-22-y2)dydx=1r u. The projection onto the my-plane of the intersection of the two smfacesisthe circle a," + y2 = 2y => x2+y2-2y=0 => x2+(y—l)2=1,sotheregionofintegmtionisgivenby-lga:gl, —\/1—:r:2 gyg 1+VI—z2.1nfl1isregion,2y2z2+y2 so,usingaCAS,fl1evolumeis 1+J—1—c1 1r v= /_ [41—— [2y—(22+y’)1dydz=§ Because the region of integration is D={(z,y)IOSzSy.OSySI}={(z,y)IZSyslfiszsl} we have I: I: fix, y) dandy = I!» may) M = I: I: my) dy dz. Because the region of integration is D={@wHOSyShalSwS2L=KawIészszosyshm we have £2Ah2f(x,y)dydz=fAf(x,y)dA=Ah2[:f(x,y)dzdy Because the region of integration is D={(x,y) larctanzgys §,03x51} ={WJHOSZSumm05yS%} we have /[glcflawdydx=//Df(:c,y)dA=All4i/o‘.nf(a:,y)dxdy 1 1r/2 y=sinx or // COB$V1+0062$d$dy 0 arcainy x = arcsiny \ zrflo "n” coszv1+cosgxdydz =1?” coszv1+cos’x[y]::inl dz _ [2 . L60=cos:c,du=—sinzdz, —j;' coszv1+coszz smzda; “=du/(_sinz) =ff-W—1+u’du= —%(1+u’)3”]: 61. The average value of a function f of two variables defined on a rectangle Rwas definedinSection 15.1 as fm = :63 If): f(a:,y)dA. Extendingfllisdefiniflon togeneralregions D, we have fm = fl “'1, f(:v,y)dA. HereD = {(2,31) IO 5 a: 3 1,0 5 y g 3x},soA(D) = %(1)(3) : rm = n13 m, f(z,y)dA = 3‘; f: 03' xydydx :31: [233/ 2 :22“: 313933“: §x4];=% 67. WecanwriteffD(2z+3y)dA=ffD2sz+ffD3ydA ffD2szrepmentsfl1evolumeoffl1esolidlyinglmderflle planez=2xandabovetherectangleD. Thissolidregionisauiangularcylinderwflhlengthbandwhosemss-sectionisa triangle withwidthaandheight 2a. (Seethefirstfigm’e) (0. b. 3b) Thus its volumeis % - a - 20 - b = azb. Similarly, If!) 3ydAmpresentsthevolmne ofatriangflarcylinderwithlengtha, triangular cross-secfionwithwidth b andheight 3b, and volume % - b - 3b - a = gab”. (See the secondfigume.) Thus ffD(2z+3y)dA = a2b+ gab2 ...
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