hw-4-sec-6.2-solns

hw-4-sec-6.2-solns - G t 2. 438 El CHAPTERS APPLICATIONS...

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Unformatted text preview: G t 2. 438 El CHAPTERS APPLICATIONS OFlNTEGRATlON 56. The curve the line will determine a region when they intersect at two or more points. we solve the equation 1/ (r2 —i— 1) 2 mac => 2 i'=a:(m:v2+ , :> r(m.r —:m)—.r=0 é x(mx2+m—1)= 2 $=Oormrg+m—1=O => * 1 1 ~ . . . . . m I — ~ 1. Note that if m = 1, this has only the solution as = 0, and no region 1=Oorr2:———- => 771 is determined. But if 1/‘m — 1 > 0 ¢> 1/ m > 1 <=> 0 < m < 1, then there are two solutions. [Another way of seeing this is to observe that the slope of the tang t to y = 1‘ /' (x2 + 1) at the origin is y’(0) = 1 and therefore we must have 0 < m < 1.] Note that we cannot just integrate b ’een the positive and negative roots, since the curve and the line cross at the origin. Since mm and w,"(x2 + 1) are both odd functr ._ the total area is twice the area between the curves on the interva [0, t/1/m ~— 1]. So the total area enclosed is /‘/1/m‘1 0 2 [—1— —ma:] dr:2§%ln(x2+1)——%mcc2} V Urn—1 = {ln(1 —1+1)—-m(1/m—1)] —(ln1—0: 12-1-1 0 \ ll ln(1/m) — 1 + m = m — lnm —— 1 6.2 Volumes 1. A cross-section IS a disk With radius 2 — so its area 15 Ah) 2 'F(2 — . £2A(m)dm=/127r(2— $192 da: 2 17/ (4-2z+i—xz)dx 1 V ll I! ll :4 ’E‘ H l a 2 Fl(8‘4+383)—(4—1+315)i H H 7r(1+-17§) = §7r @A cross—section is a disk with radius 1 —- 3:2, so its area is A(m) = 1r(1 — x2)2. V=/_:A(ac)dac=/1 7r(1—;c2)2da: ——1 51V SECTION 6.2 VOLUMES C1 439 @A cross-section is a disk with radius 1/35, so its area is : «(l/x)? V2/12A(x)dx=/12n(i—)2 dzr H :1 bx H Rd H E: II a I is IH L...__u >4 N =wkswau=§ a 4. A cross-section is a disk with radius \/ 25 — x2, so its area is 31(1) 2 7r(\/ 25 — 1‘2 )2. a; V=/24A(a:)d:v=/:7r(\/2—5——x2)2dx 4 = /2(25-m2)dx=1r[25rc—%13}: =flU%—%%%w-%H=%r ' @A cross—section is a disk with radius 2 \/y, so its area is A(y) = «(2 fly. V=/09A(y)dy=/09fi(2\/g)2dy=47A9ydy = 47r[§y2]§ : 21(81) 2 162a 6. A cross-section is a disk with radius y — 312, so its area is Aify') = x( y — 3/2)"). V = 101A<y>dy = I; my — y2)'-’dy l A ’E l i.—- + J: H I4 7. A cross-section is a washer (annulus) with inner radius :33 and outer radius 1‘ so its area is ACE) = “(55V — W($3)2 = 7(1‘2 — x6) SECTiON 6.2 VOLUMES D 441 A cross-section is circular with radius 4 —— 1.2, so its area is A(r) = 7r(4 — $2)2 2: 7r(16 -— 8x2 + .724). 13. A cross—section is a washer with inner radius (1 + sec at) —- 1 2 sec :5 and outer radius 3 — = 2, so its area is 14(1) 2 7r[22 — (sec art-)2} = 77(4 — sec2 :6). 7/3 1r/3 -—/ A(x) dz 2/ 7r(4—sec2 1‘) d1 ..,./3 —~7r/3 7r I 3 = 27r / (4 — sec2 x) due [by symmetry] 0 1/3 = 27 [4x —- tansr] =27r(‘4§“'-\/§) 0 442 C1 CHAPTERS APPLICATIONS OF INTEGRATION y = J55 2; 1 = y2. so the outer radius is 2 — yz. V : fol «[(2 — y2)2 - (2 —— y)? dy :7rf01[(4—4y2+y4)—(4—4y+y2)3dy =wfo‘ (y4~5y2+4y)dy =riéys — 31/3 +292]: +2)=%7r DJIU‘ we- ® y = 3:2 => q: = fl for 1' Z 0. The outer radius is the distance from I z: —1 to .r. = fl and the inner radius is the distance from :c = «1 to I = y2. , 1 i' “:2 '7 , 12 1 2 2 2 V =/ W{L\/§~(-1)j * y‘—\—1)3}dy=fi/ [(x/EH) -(y +1)]dy O 0 :w/Ol(31+2x/3+1~y“‘—2y2—1)dy=7‘£,1 (y+2fi-y4'2y2)dy V-“ 18. For 0 S y < 2, a cross-section is an annuius with inner radius 2 — 1 and outer radius 4 —— 1, the area of which is A1 (y) = 7r(4 — 1)? — 7r(‘2 — 1)2. For 2 g y g 4, a cross-section is an annulus with inner radius y — 1 and outer radius 4 — 1, the area ofwhich is Az(y) = 7r(4 — 1)2 — fly —— 1)2. V: f:A(y)dy=7rf02 [(4~ 1)2 —— (2 — 1)2}dy+7rf: [(4— 1)2 # (y —1)2]dy = 7r[8y]: + «f;(8 + 2y -— yz) dy = 167r—t-7r[8y+y2 - éyg’]: ii i m; :167r+7r[(32+16~§3é _(16+4.§)1j "t it :—‘7l' 3 1 19. 9h; about 0A (the line y = 0): V = / A(I) dzc = o 20 9R1 about 00 (the line x = 0): SECTION 6.2 VOLUMES D 443 21. 9h about AB (the line :5 2 1): 1 2 1 1 V = A(y) dy :/ «(I —— dy = 7r] (1 _ 2y1/3 + yz/s) dy : 7+; _~ gyzi/B + gys/s] o o o 22. @1 about BC (the line y = 1): V = fol [1(2) dx : f01[7r(1)2 — 7r(1 ~— 1‘3)? d3: = 7rfo1 {1 — (1 -— 2x3 + (1:6)}d1,‘ = wf01<2w3~ mm = was" — w; = we — a) = 23. 9%; about 0A (the line y = O): 1 Vz/01A($)dx=/ol[7r(1)2-7r(V/;)2]dz=n-A(1~I)d93:7r[$_%$2];:fi(1_%)z Ma 1 gig about OC’ (the line at 2 O): V = fol A(y) dy = f0} 77(y2)2 dy = w fol y“1 dy = agys] : 25. gig about AB (the line a: = 1): 1 1 V: A(y)dy= [wtlf-wtl—yzf L.—_J 1 1 dy=rx/ il—(1~2y2+y4)]dy=fi/(2y2~y4)dy 0 O 1 —- 23-175 — 3-1 ~11 ‘7r[3y 5y]0_7(3 5)‘ 57" 26. 9712 about BC (the line 3/ = 1): 2 V V:IOIA($)da:=f017r(1—\G) dx=7rf01(l—2a°1"2+r)d. zaér— gx3/2+%m2] =7r(1~—§+%) = g 27. gig about 0A (the line y = 0): 1 1 2 39 1 6x r 2 711 V=/0A(a:)d:v=/0 --’:T(.1‘)‘JdIZTA(I-I)dl'=fiiér —%Ié0=fi%—%):%7fl Note: Let 9R : Ugh + mg + gig. If we rotate 91 about any of the segments 04. OC. AB, or BCX we obtain a right circular cylinder of height 1 and radius 1. Its volume is 7rr2h = 7r(1)2 - 1 = 774 As a check for Exercises 19, 23, and 27, we can add the :t5: answers, and that sum must equal 7r. Thus. + '_'2' ‘r E = (‘ ‘14‘ )7 = :. 28. 9113 about 00 (the line 1' = 0): t ‘— 2 . v 5‘; v =10‘Atwdy = I; [4%) ~ ruff] dy = who“ — ya dy = Fla 3— as] = we — %) = er Note: See the note in Exercise 27. For Exercises 20, 24, and 28., we have + + f = w. 29. gig about AB (the line :I: = 1): 1 1 22 :3" 2 1" a 4 13 23 V=/A(y)dy=/ {W(1~y)—W(1~ y)]dy=fi/ ill—2y‘+y)-(1—2y/+y/)]dy 0 0 0 1 1- ' / 5;.» =W/(-2y2+y4+2y1"3 ~y2/3ldy=W§—§y3+ éy°+§y4‘3‘ 953/ ’3} 0 _ Note: See the note in Exercise 27. For Exercises 21. 25. and 29, we have 130- + + 133—; = (ii—ls‘lfifléfir = 7r. 444 D CHAPTERS APPLICATlONS OFlNTEGRATiON 30- gia about BC (the line y = 1): V = f3 Amdw =11? [w —x3>2 -«(1 — way] d1 :7 4 + 1x7 + 3333/2 — l.122 7 3 2 do I; {(1 e 213 + m6) — (1 — 2x“? m] dz = fif01(-2:133 +x5 + 2151/2 — x) dx = «Pear Note: See the note in Exercise 27. For Exercises 22, 26, and 30. we have + 7r/4 31. V : 7r] (1 ~ tam3 :02 dx 0 32. y = (as — 2)”1 and 8:1: —- y = 16 intersect when (m-2)4=8x—16=8(a:—2) <=> (x—2)4—8(x—2)=0 <=> (a:——2)[(x~2)3~8}:0 <2 zc~2=0 or I—222 <=> 32:20r4. y=(ac~2)4 => :c—2=:&:{/§ => SECTION 6.2 VOLUMES D 445 (3, 2J2) 27r V: / {(4— cosx)2 — [4—- (2 —cosa:)]2}da: o 27r : 7r/ [(4 — cosac)2 —— (2 + €083.02] dx 0 37. 4 y = 2 + r2 c051 and y = I4 + 1' + 1 intersect at y=xa+x+l x:a::—1.288and.r=bz0.884. b V = w/ {(2 + rzcosxf — (x4 + r + ifldx x 23.780 0 y= 2+x2cosx ~2 2 0 38. We see from the graph in Exercise 6.1.36 that the r—coordinates of the points of intersection are x = 0 and at = a z 1.17, with 3:1: —— 2:3 > 3:4 on the interval (0, a), so the volume of revolution is a ‘1 . 77/ [(32: — we)? ~(x4)2]d1‘ = 7r/ (9:02 — 6:10“1 +:c6 ~— 1‘5) d1 = «[313 — 355—15 + $37 - $33918l ~ 6.74 o 0 ...
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hw-4-sec-6.2-solns - G t 2. 438 El CHAPTERS APPLICATIONS...

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