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**Unformatted text preview: **6.2 Volume, Page: 372-376 3
v: (3 — :31 of:
i
= -i{3 — ﬁsl: = 9 cubic units Using symmﬂry.
4
V: zlwm—z” 9 gr III
= 2(161' — $9.3)” = “~39 cubic units The :u rvu intcrscct
when: E~23+3=z+1 E—3:+2=n
(x—QHI—lj = [l
.1:=2,l [:4 — 69+ 13:;2 - lEm-I-‘ﬂd'x _1 _a4 13 _ 2
-1152 ,z+-,_-:~" 6£+421|l = ﬁr cubic unila é II
M In PraMms 5-3, we not: thai an cguﬂatem!
triangf: of I“: a in arm a1. Page 254 .- :Ii 5‘; Ema! Appﬁcation: of the In tegrll 5. .. ' = I «- 1] +éfﬁz — 5)]3d‘r ﬂ _' Mic that a semicircle of
' ' - '- 734%)“ : W‘
' . The curves intermi. when
:2 = —‘2£+
#2 + 23—3 = ﬂ
(Ii-3H3: - l]=ﬂ 'f.
— ﬁx“ - 6x2+9zjt
m
1' {coax—Hint:
E in” 3
a. :% l [l-ainizj 6. 2‘t Vaium: Page 255 Hi} 15. Use washers; m h .dn'iiiiiiiiiiii V: I'JIHEE]: —- 3] dz = :r cubic units Us: washers; Um diﬁits; l
V: r[[{:+ 2]“— [1+ 1):] d2:
u : «(:1 + 3:”; = 41' cubic unita 17. Use disks; 1
[viii]? d:- = ‘11”: I; : cubic units
0 Use disks; (.1 + :3)“ a: ‘1
ii
1 II
a. [s‘ + 2.5+ :6] d: I
I ‘i'
+ If + g a: 2,555 cubic units “Pt: I] 3
= 71(zlﬁija dx = gnu"! ‘3 = 221 cubic units
a _ 5;}; Additional Appl'r'uﬁan: of the hate; Us: shelln: 19. Use sheila; '41 I___ I?) a; = a". cubic units Use ahsllﬂ: the curves in at my}? I _
: rcubic um“
11 4
Um.- shells; Tu ﬁml the limits; of iuLugraljou, sul. y = 2:
2i; — ll]J : 131 so J! = ﬂiﬁ. Tllcll.‘ 9:2 :5: J2 -_3[[1'- :u" and the height of the vcrl.iu:n| strip at. I: la. .I.={2+.l2 A gu— 3}?) Thus. using the shell mcthud aw“;
=21] 1— gr“ — 31" a; we : WE 1:2 z 111541530 cubic units Problems .9145, I': ia understood Hm!
given volumes are all in cubic units. .1
a. dish: “[14 — :3.“ ch;
E! Page 25.7 1
b. shells: Erin“ — x] (I:
D 4
11+ wash-2m: arJlly+ l}"t — l2] rlln:
1| -I
= :rj[24 — H]: + :2] a':
ll 11. shells: Er [3+ 2]“ — 1:} d: W4 - 3: r11: 23.. I. sheila: Ear :Iu—a. gI_-.—-‘a. .ﬁ
II. lliﬁlﬂl; I'l-H - y} :Iy
U n. alumna: 2r]{y+ Unfit a y :13; U 4
:l. washers“ «furl-2}? n 2”] #3.
u I
=1‘jl'l - y+ 41M - 5;] fly ll 2
29.. .'. mm: ﬁlm ﬂy
I) I
h. dim: :[u — y“; big
0 I 2
c. awn: EIJU+ 11-534 - y! sly
D u
d. Nashua: rJ-[(r+ 2]2 - 2:] try
.- ' a =tll4 - 93+4v'4 - Ell risr
U - 2
111. I— dialm II“ — :2} d2:
1: page 253 Chapter 5. Additional Applications mm m _.-' :rfz
in. n. shells: 2w] yﬂl — 31:: y] dy
a
:11:
Ex wuﬂ'llut‘u: :rj [1" — silfzy] d'y u
:11: ﬁ ' allclla: MI {3: + lHl - ai|1y}a'y U
.- m ;
1cm: 1:]- [(1 + '2}? -- {r+ 219} 3'1-
L1
1}“: _2
h. sheila: 2r : = «I [5 — Ilillny - aMin u- u
‘ u
h. altullu: ‘3':- _ '11- I-[ln :1! via:
:1 I H '-
u
:. wmlmm: I] Islr: :d: I] H
Ii'_ﬁ_ 4—. n . . I} I
(I. ullulbl: innit I: '
l '1 4r. .
32. n: slmllmﬂlrl.
u :1 b. washers:
m- If :1“ — :3; I
I “ Ill-2?. “HMIILEH V: nlﬂﬁf - [:3]
c. shells: 214:; I '
“f V=wIIEﬂF-[ﬂ’i
:21 a -
:1:_3r¢§ Intersect. where
‘ I = y = i2 : if
cl. washers: “31?: I]
I' 'y = D 'II 9 washers: .V~= {ﬂu/a: _. Scctfon 62. Volume . a. By washers: 1f: it [(a:2 +1}: — 1!] d: b. By washers: V = :r-HIZE)‘z —(—,,:‘;r- 11:21:11; when :I: = I]. —2:
a, By washers:
u
v: If [(—aF—ezji—{éﬂdm
—2
b. By washer-e (salve y = — :2 —‘L1: {or e to ﬁnd1= -2+u'4 — y]: F: Int—2 + J—“H — y: — (Vryﬂag . Use computer eol'tw:u'e or a graphing
calculator to ﬁnd the curves interned. at
[1.133, 0.130} and (3.566, I271]. a. By washers:
3.553 V: r] [ﬂu 1']:i — {lllzﬂfl dz
1.133
b. By washers:
1.271
v = «j [iv/ﬁn? — M‘] e
0.13:: The curves intersect where 8‘ — 1 = 2.3-”.
a” - l = 28"
2' A a: — 2 e 0
a”: 2, —l (reject)
.1: = In 2: y: l a. By washers:
[1.1 2 v: «I [me-=3“ r (a — 1?] d: u
b. By ehel'ie:
In?
V = 211']
n IRE—r A lﬁe= — 1]] d: It The curves interned. where x2 = v— :2 -- 4.1: or Page 255' or by disks:
1 2
V=xj[1n(y+ 1‘3]i dy + xlfln 2 411ng ﬂy e
41. The curves
intersect when
:2 = 3.3, or when
r = I] and a: = 1.
1
a. v: glue”)? — (9)2] a:
n
1
l = 3-bit“ —~ if:
I]
= my _ er] [1): 1w
1
b v _ 2711193” 34"”) iv
a
= him” ya”) dy
u
_ 3 n3 _ 2 ,3: 1
- f” 5f ] o
= 211'
42. The curves intersect when
.I:2 = :3, or when
1: = [I and e : l. Page 260 43. —- ‘1' cf:
1
a
The curves
intersect 5:. when
= 2:, or when
r = I} and .I: = _ If: :3 l _1r.|:3l D+w(r~—-ﬂ-)U2
1 1 1 “($11” 5+2?) .1' ‘E Chapter 6, Additions! Appﬁcm'ons of the Integral The curves
interned: at w
z : 2:, or when
I = l]- and :I: = r
45‘. - it 22102: —- — r
= 5,9
The cross section is a square of sidu 2y- and
ma 4?.
:5 a
v: I 4(9 — than she — z?) a:
— 3 0
= 9(5): — 144 cubic uuiu
Thc cross section ‘15 an equilateral triangle
side ﬂy and area iﬁwy)? = ﬁg? 3
I] 41'. Section 6. 2, Vol‘um e 113. The croaa auction is n11 imsculun righl, triangle
with hymn-.111an 2y illitl SlllL‘ ﬂy with. arm .1 LI.
if: I {11 — 1:2}rI1': '3 [u - :9; ii:
I : “31:9: — = iiii. cubic units 4.9. r111-: cross suction is a ﬁcniiicirclc wiLlI rmlius y
and area. $4.2}? = 24:! cullio ullitii =ﬂli-“J; SI. Sulﬁpom llu: Lrinngulnr brim.- llI'LH one leg If:
on le y-nx'm and :l. vurtux H- on Lllu poniLlVL" 1:- axim MA” = —-I.iu1%= — -"lI1Il HIE equatitm 0|. tlli' Iilli' ii; —~'J=—i:
i: ii 7‘3” 'l'lu: L'l'UHH suction l?! n 51 mm: with iii-in: 3 nml
I alum. I’ly"a : 4H - win-1:]?
ll .Jvﬁ
- r _ _1 2 I “a
—4ﬂl {'2 - if)" - II
P.
L'ml
n.
H
“- 51 The base ol‘ the right triangle has its log: on
LI“: coordinate 11mm with a. vertex at. the
origin. The equation of Llu: line passing
Liuough the hypolnnuiac is y = — I + -1. The
diameter of a typical minioirculnr cross
section of the. solid is y, and the Meal is 'l I 2_il '2
i111ny — any . 53. PuL the, uni-Lea: of f.lh: pyramid at the origin.
[1.3 siclc: will in: along 1.|1u lino y : [li’i’ﬁfdﬂﬂ] z.
Hath aqua",- ﬂicc perpendicular bu the much: will Inn-i: height of y rind width 2y.
’l'wig; Ll“,- mmi uF all I.'|u-. mutanglca l'rmn a: = Ii Lu 1-. = '13:" will givl: Lin: 1-'I;vIlIIIu.'.
$25” 130
_ . '2 __ :575 )2
V_2J.£y dJ: _ 111115": (if
0 LI
'18” :1'5‘ x" _ r- 431}
«(am i = iiﬂlllliiﬂllii n.” 5-1. The crawl xochiuu iii .1 .‘i(|Il-'II'I1. with siilu 3y.
'l‘lli: clumuul ol' mlnnm 15 ii"! = rig: if: = i.liilll(l # i:
V: 23ﬂlqﬂﬂl(l - rim U
u _.i‘_:i_ '1”
- 3'2""(1 " 2mm)
ll. 55. a. us..- disks: V: nlir“’”]‘ rfr
I = 0mm] n3 [I -I
: :I'ltllzll1 = Irlil‘i 4
b. Us: shells; = En‘lrl” if: = 2r(§}zal'l2lil :. Us: washers:
w: n[[y+2}2 — 2%:
= 1[L:"‘”+2)’ - 4w: ﬁt]. a. Useshells: E
.4
V: Eel-2121.13 dr: b?-
D
in. Use shells:
i s: swim: e are} a; : 4—7?“ 0
51'. Use shells and double the portion (by
symmetry] above the reads.
3 las‘ﬁ‘r a: 63.131“? re" 53. V e ﬁlth]ﬁj+2l11.ﬂﬂ]+2{l.ﬂ5]+2(1.03]
+2(0.E|9}+'2(].ﬂl]+2[ﬂ.93]+2(0.99}
+2[e.ssj+2(n.93) +1{u.91}](1.u} dimes} a: 10.05
The volume is approximately 10.05 [1“. 59. V e: h1(1.12)+4{1.09}+2{1.n5)+4(1.03]
+2wssi+4u .u1}+2m.ss)+-1{u.ss)
+2(n.ne)+4[n.93}I-I-1[n.91}](2.0) =§(30.19} re 20,13 rI'he volume is approximately 20.13 In“+ Bil. Use washers with strips perptmdieular to the
y-axis rotated about the reads. The
element of volume is dV: «[0: + «a! — 3F)?
— s — JET-7’91 e
v: swim? + ewe” .— g? + o2 — if) I] Chapter 6. Additianti Appiiestiens of the in: — (b: # Eli-nit]! - 31+} —
ﬂ
= ‘SllrrI-JnE — y! dy = ﬁxbe'jri
[I
= Lingual: (A = 31121 area of circle) til. The cross section is a semicircle with r
arena-:11».1 and volume di" = airy: d1; he] (s _ In anew: _ is” :ems s 2.3; =31; Thusll the volume of the entire sphere is
v: 21:35:19} = gee. Iii. Let B be the top vertex of a reetangular 5-
tetrahedron of side a. The y—axis is v '
and contains 3. The Wis passes tilm
vertex A in the base. Fiﬁ! - .
0 is the projection of H onto the base at
e_e||ter of the square. Draw the herpendi
03' from 0 to a side- mntaining A in the 1
S-
,4
:-
ﬂ!
o base. IEE‘I = ﬁn. The height of the
tetrahedron is H. Also if}?! = '1 .
a l 75
By the Pythagorean theorem.
Hz+%eg= a2 or H: in Let (r, y] be a point on iii. Then, by :9
triangles. ﬁll-g:
7%“— 7 TE Consider a. horizontal element of area with
cross section an equilateral triangle at alti
y. In this triangle. :' corresponds to {‘33 = {uﬁja in the base, so the side oi'
elements] equilateral triangle is «flit, The
element of volume is W = lﬁwﬁzﬁ e = tee—’5; e =%—r§(% ﬁn —- y}dy Section 6.3. Polar Forms and Areas _ ﬁﬁl ﬁa_ ﬂ“ y)‘ o _ _- 3 . .
— 12 eublc niuts 63. Position the equilateral triangular base to reﬂect symmetry as in the ﬁgure below. . Using similar triangles, we see that
y __ of? — a:
;?3uf2 a}? ThusI the cross-sectional triangle at. I! has area
1!
A[:]=%y2=_V:—§(_ﬂz+§ﬂ)2 and the total volume of the solid lignre is of:
Vzﬁjgi—ﬁz-l-ﬁdfdrzﬁﬁfi II} 3
ll
rather than few. the volume of a tetrahedron of side 0. Thus, the ﬁgure cannot
be a tetrahedron. and the conjecture is false. 64. First rotate about the startis. Use disks with vertical strips and double the portion to the
right of the y-axis. The element of volume is ell-’1: tyidz: rro"'ula"il[uE — d1: |1 V :'.1'.‘11_2b:rrjl[u2 — dz if
U
= 21Tt1_2b1[ﬂlz -- aft”; -1 2
_3-rat Now rotate about the fascia. Use disks with
horizontal strips and double the portion
above the e-axis. The element of volume is JV: mead: = Wﬁ—idiibi — ital liil'
t r”: writhing _ 3,?) at U
_ t
=2ﬁt ﬁdgﬂgy — .5933” Page 253 = grail: 65. Use disks with horizontal strips for h 5 y E R. w: n: sly = auri— they r: elm“ r y?) a: truth; — lﬂl’:
t
= gear? — ante + t“) 63. Use disks with horirontal strips for I] 5 5r 5 h. 6.3 Polar Forms and Arms, Pages 331.335 Step 1. Find the simultaneous solution of the
given system of equations. Step 2. Determine
whether the pole lies on the two graphs. Step 3. |Graph the curves to look for other
points dimers-action. Sketch the graph of the polar region whose
area is to be found. Determine rays ti = or
and ﬂ = 13 that bound the region. Compute
the area by the formula .3
A = Ems]? ea
or
I. lemniscate b. circle
1:. nose {3 petals} cl. none [spiral]:
I'— eardioid 1'. line
5. lemniseate h. limacon
I. rose {4 petals) b. lemnis'cate
L circle cl. rose (Hi petals]
e. none {spiral} f. lelnniseate
5. case (3 petals} h. cardioid
a. rose {4 petals) b. circle
1:. Erase-an cl. eardio'uj
f- line I'. line
I. rose {5 petals} h. line ...

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