hw-16-sec-16.4-solns

# hw-16-sec-16.4-solns - SECTION 16.4 DOUBLE INTEGRALS IN...

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Unformatted text preview: SECTION 16.4 DOUBLE INTEGRALS IN POLAR COORDINATES ET SECTION 15.4 483 61. Since t /1 — m2 — y2 2 0, we can interpret ffD \/1 ~ 1‘2 — 3/2 (1A as the volume ofthe solid that lies below the graph of z : x/1 ~ m2 , y2 and above the region D in the try—plane. z : \/1 — 1'2 w y2 is equivalent to 3:2 + y2 + a2 2 1, z 2 0 which meets the any—plane in the circle 2:2 + y2 : 1, the boundary of D. Thus, the solid is an upper hemisphere of radius 1 which has volume %I§7r(1)3I = gm 62. To ﬁnd the equations of the boundary curves, we require that the z-values of the two surfaces be the same. In Maple, we use the command solve (4 —xA2 —y”2=1—X—y, y) ; and in Mathematica, we use Solve [4—x‘2—y”2==1—x-y,y] . We ﬁnd that the curves have 1::\/13+417-4x2 2 e uations : . To ﬁnd the two oints of intersection Cl 3; P of these curves, we use the CAS to solve 13 + 430 — 42:2 : 0, ﬁnding that 1:: m. So, using the CAS to evaluate the integral, the volume of intersection is 33 : 2 (1+x/ﬁ)/2 (1+ 13+417412)/2 2 497r V = 4 ~ — 2 — 1 — — d d = — /(1~\/iZ)/2 /(1—\/13+4Iv4z2>/2 R I y) ( a: 34)] y m 8 16.4 Double Integrals in Polar Coordinates ET 15.4 @ The region R is more easily described by polar coordinates: R : {(r’, 6) I 0 g r g 4, 0 S 0 g 3T" I». Thus ff}? f(a:,y) dA : (Jew/2f: f(r cos 0,Tsin6) rdrdO. @The region R is more easily described by rectangular coordinates: R : {(x, y) I H1 3 a: g 1, 0 S y S 1 — 2:2}. _ \$2 Thus ffR f(a:,y)dA : fil 01 f(;r,y)dyda:. @‘he region R is more easily described by rectangular coordinates: R = {(90, y) I —1 S :v S 1, 0 g y g %z + %} Thus ffR f(:c,y)dA = f_11 0W1)” f(m,y)dy dsc. .The region R is more easily described by polar coordinates: R = {(r, 0) I 3 S r S 6, — 325 g 0 S g}. Thus fat,y dA: j/2 6f7‘c056,'r'sin6I rdrdg. R 7r/‘2 3 @l' he integral ff" I47 7* dr d6 represents the area of the region R = {(r,0) I 4 S r g 7, 7r 3 0 g 27r}, the lower halfofa ring. ff” ff rdr d6 2 (ffr d6) (ff rdr) = [0]?"[%r217=7r-%(49v16)= — 7r 4 484 CHAPTER 16 MULTIPLE INTEGRALS ETCHAPTER 15 6. The integral [0 /2 04¢ch rdr d9 represents the area of the region R: {(739) I 0 S r S 4cosB,0 g 0 S 7r/2}. Sincer : 4cosH 4:) r2:4rc050 4: x2+y2:4w 4: (m~2)2+y2:4,Risthe portion in the ﬁrst quadrant of a circle of radius 2 with center (2,0). [5/2 f“% drd0= WEN]: 4C°\$9d0— fOH/28cos we: O:"/z4(1+cos29)d0 4[0+—sin29]"/2= 2n (yrhe disk D can be described in polar coordinates as D : {(r, 0) I 0 g r g 3, 0 g 0 g 271’}. Then ffD mydA : f5" f5 (TcosO)(’rsin0)Tdrd6 : (fozw sin90056d0> (f: 1" 3dr) : [2 sin 20]? [£73]: =0. .ffR(w+y) dA: f3"22/2f1 (rc0s0+rsin6)rdrd6:f37r2/2f12r2(cosc9+sin6)drd0 :(f37'2/203080 + sin 9) d0)(1 2 72 dr) : [sin0 — cos 0E7? ﬁrs]: =<—1—o—1+o><§—1>=—aé 9. ffR cos(ar2 +31 2)dA: f07T f: cos( 2)_rdrdt9 —( f0" d6) (fog rcos(r2)dr) :[0];[%sin(r2)]::7r -%(sin9—sin0): —sin9 10. ffo/4 —a:2—y2 dA: fir/2f; 47\/—r2rdrd0:(:/j2 d9)(f02r\/4—r2dr) = [ax/732 H ~ §(4— 7433/2]: = (% + %)(-%(0— 43/5) 2 ‘3'” @D 1-. .1- 1:4: 12 *1 “new xxdemwdr) = [61:23. [—1612]: = MW — e°> = 1(1— e-4> 12. ffR yesz: for/2f05(rsin0)e“°serdrd6:fOSJ/27‘ysir166rwsgd6dr Flrst we 1ntegrate [0/2 7’ 2siHOeTCC’sedﬁ: Letu : rcosO :> du : ~rsin0d0, and f0" /2'r2 sinﬂercosg d0 : ":0 ~re” du : —T[eo — er] : rer A 7'. 'LL:T Then f05 f0"/2 r2 sinﬁer 6°59 d8 dr : f05(rer * r) dr : [ref ~ er — 5’3]: : 465 — 2—23 , where we integrated by parts in the ﬁrst term. 13. R is the region shown in the ﬁgure, and can be described byR:{(r,6)1030£7r/4,1§r£ 2}.Thus ffR arctan(y/ar) ):dA fen/4 ff arctan(tan 9) 7" Cir (119 since y/ar : tan 0. Also, arctan(tan l9) : 6 for 0 S 9 3 77/4, so the integral becomes fJ/4f120rdrd9= "/46d6f1rdr_[;02]3/4 [lr2]2:"_2.%:i7r2' 2 1 3 SECTION 16.4 DOUBLE INTEGRALS IN POLAR COORDINATES ET SECTION 15.4 CI 485 [/me // “1.47 // \$2+y2s4 (as—1)2 +y2 <1 3220.1;20 y>0 7r‘2‘2 2 772 2 6 : 0/ 0T cos6d7'd6 foo/ COS T2cos6drd6 :f0"/21(8cos6)d6 far/21093054 6)d6 7r/2 l w|00 —%[cos3 6sin6+— —(6+sin6cos6)]O =§~%[0+%<%)] : @ne loop is given by the region # {(T, 6) )|_—7r/6 < 6 < 7/6 0 < T < cos36}, sothe areais 7r/6 c0836 7r/61 T: c0536 // dA / /0 Tde6:/ [—19,] d6 D 7r/6 Aw/G 2 T:0 ~7r/6 1r/6 :/ 1cos2 36d622/ ECLM) d6 4/6 2 O 2 2 7T/6 1 1 7r _ g _ ' :— 2[ +651n66]0 12 16. D={(T,6)|OS6§27T,0§T§4+3(3036},50 =ffDdA— f0” “36°59 m1 d6: §"[§r2]:::+3°°59d6:2 0 "2(4+3cos6) d6 ; 0 "2(16+24cos9+9cos 6)d6=%f02"(16+24cos6+9-1+CT°S26)d6 : H166 +24sin6 + ga+§sin2e]:” _ %« 17. By symmetry, ~ _002f7r/4 51116 Td’f‘dg— _ 2fH/4[ T2]:::)in6 d6 :foﬂ/4 sin2 6d6— .a 0”“ :(1 — cos 26) d6 7r/4 top- [6 — —- sin 26]0 [g — sin ll NIH NIH ml: | O + top—- In ._.. 5 O ._1 |.—1 A :1 | [\D V 18. The region lies between the two polar curves in quadrants I and IV, but in quadrants II and III the region is enclosed by the cardiod. In the ﬁrst quadrant, 1 + cos 6 = 3 cos 6 when cos 6 = % => 6 : g, so the area of the region inside the cardiod and outside the circle is A1 :fvr/s/z 1+cos6 TTd d9: fﬂ/3/2 [% 2]T=1+cost9d6 3 cos 6 T23 cos 6 — 2 —1f"/3(/2 1+2c0s6—8cos26)d6= % [6+2sin6—8(%6+ isin26)] —[— g6+sin6—sin26]:/2= (—§f+1-0)-(_% 12.§_;/_§ ...
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