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**Unformatted text preview: **F-1{1 + :2} a: (”51311: = 12 _ 1 + man-.311 = W
= «or—w . = (1%=IF+%+[§€T4]R _, Page 269 l1 .— r=sinﬂ+cuaﬂ 'r2=rainﬂ'+rcnaﬂ L‘ - The curve 1' = lin’gis traced out-fq'tHI-hé.
I] 55 <2r. We ﬁn. the-w: length banning sing sing? + ms: 55
= M: =4
1113- IL
5 = I Eva-(5)3111 E1“! (5]: + [(5).]: d9
9
I}:
= I aruzsfsin a .19
ﬂ .
= 503(th all 1r": = 25' - 5.4, Arc [myth ind Surface Am “92]? + (2“? d3.
9 a + 4 as = gm +413f3|: W: —3 ai—wL gun—“.- I] Page 2?? It appears the arc length is appmjmmly 1_33_
29'- {(3}: 5(312:l+;{— 11:- ]|=.:2~m
1+Lf’{z)]! =1+ 391— %-I—IE?-= {:2 + ﬁr:
1* S= 21 £145 + Inﬂux“ + 4—511: = arr Bar“ + I31+ ﬁr+ far—3F: : 2W 112—35?" ]I:=1151151r T
3
:‘h + if [11]} }'~’ d: :1? + £5111: = 21r[;:"+;1ur]11 = 23M! +%1n 3]
Numerical approximation is 12M
amfhhﬁﬂ-h4ﬁ :11: in + (1’33 .. 139—2”): dy
= (1+:yﬂ3}2 _ %+ (iy-aﬂl)! d‘y
= 1111*” + if“)? 11 yarn — 2.13 ”gnaw-ht” b. .92 2x —211' Hr—|p lull—q + 39' d! 3. About Lin: y—Im'n. distance is —x: 3=2f—J1‘[_U(%F5I3_ MHXW" 19—21(3)” I]
= —211' (:9 ”3 -%-#”W® alt—h... __ —21r(3% 1013 3 '1 _ Hawajaﬂl II
I
h.
.1
I
Ii 3
R
I:
:11
M
i: Chipm- 6. Additrbm.‘ Applications of the Intggral' I in. About the n-axis, diatanm in 3.1: I
Frying” + iii—2”") in = 211' (Na-1» if“) 119. =2ﬂgqu + 3 m4}3)|1=__ﬂ_9r R: 3.5343
c. About. the lint: y- _ -1, distance is y +'
1 S= 21f[11+ Him”l +1-11-2’31d11
D 2 Wings 1,213+ = Eﬂg’yﬂﬁl +3 3 ”5:3 ﬂh‘ﬁh all. 1 £1.13 +iy—2fal d” §M¢fﬂ 1 _.1 ﬁns”! =% :1 12.0156 31.1111: -111; a;=‘{1+(—§=dx=§m d3 = 21m 115‘ S: gm'l‘ir dz
=W£13113 =31r‘f1l} RS 29.3E38 1fl+tdr is = 211‘: d5 = ‘2‘11'3‘1'1 + a: [it
3 S=21rJl1lfl+md= Lelu=l+x -4w(=c+ 11mm: — 21 3 232,
_ —“11—"- 32. .1; = :1“ «is; 111 = ﬂ = 1-5- 25 43.59
33. «£11 =[%r"1'ﬂ — ﬁrst“) (is: its: {1+(1I—1f2 __ %:l!=:l=dz = 12-112 + 1,112; .11
_1T=(1:—1,i2+%11;2] at: = 5:11:11” + cm} .1:
1 Station 6.4, An: Length Ind Surface Area .—. 1'5: 3: 1
gig — 1%] as 27.12
—(4— 31‘1” dz . 35. Set: the solution to Problem 35. I.
p- M]. 11
= 2.1m 1
_ g; .1333} 3”: “113d; {1 _ £2153 }34f'2(2= —1f3 d3) = -a<2)t§}nu ~ :“r'anmlg =21 57. y = tans, y'zmaz.
= 111 + 506:]: (11:.
1
5 = 2wItanxV1+ seclr d:
u
a: 3.6322 by calculatar
3
‘33. L=I1f1+(y']1 .1.
u f
*1 1' (£11 01+ {1'11}!
0.0 3.7 2.8328
[1.3 3.9 4.0202
0.5 4.1 4.2202
[1.9 4.1 4.2202
1.2 4.2 4.31.74
1.5 4.4 4.5122
1.3 4.5 4.2074
2.1 4.1] 5.0011]
2.4 5.2 5.2953
2.7 5.5 5.5902
3.0 0.0 5.0825 Page 2.?3 L as 113.3323 + 29.0252] + + 2115.511112]
+ B.ﬂ528]{0.3} s. 14.11543 39. s: T212111 + {1'55}: .1.
"r. r'rz.) =.a.fl+lr'(=n= 0.0 5.7 0.0000
0.3 3.9 1.2079
0.5 4.1 2.5321
0.9 4.1 2.7052
1.2 J1.2 5.1309
1.5 4.4 6.7553
1.8 4.5 3.4?33
2.1 4.9 10.5021
2.4 5.2 12.70 87
2.? 5.5 15.05 35
3.0 5.0 13.2434 5' s: 3qu + 40.2019) + 2(2.5:121} +
+4(15.0935) Ham-110.3) :2: 141.3974
— ' _ 1" _
40. y-fny _-E'.d.aﬁ 1+ﬁd‘z
d5 = 2n:- da = 211% r]———h—M d1:
{1
S: 21(ﬁjfﬁi + PI: 11:
n
_ um 41. Let the an: of the graph be aulndivided into
small elements, each approximated by .213.
Each of 111m in the hypotenuse of a right
triangle With legs 22.: and ﬂy. 11111111.I 01”}2 +151 2 = .111 11%): .1: .13 = 2n.’1+(:—::)2 .1... Thus,
5: £12“: 1 + dim, A:
ﬂ-iy-l-Oﬂ- 1:! E; 1' . .21
5: 2a- ally-loo l5:31:11'1 «reg—3:): Ark = 2x I: 1+6?!) dz Page 2.74 42. = 2n- fn Ill l+[f'[:]]2rf: By the MVT there exists an a": such that
. {(It— i—ﬂ‘ﬂ ﬁn
1' in") = W = 1?. its -._- ”qﬁrijlg-I-(Ayﬂa: 1%): ﬁlth
= all + [rain]? at a. Let Pr— 1 and Pt be the points with
coordinates [it]: _ 1, n_ 1] and {ab Ft)?
respectively. When the line segment
P*_1P1_ in revolved about an: Nixie, it
generates the i'ruatnm of a cane with radii
r1 = n_ 1 and r2 = 3* and slant height
I? = £1- In the text, We showed that such
a fruatum has surface area 'IrIIr'1 + r2}! = r{y*_1 + ydli
b. Assume rlgri. Then r1=§r1+%r:¢_:§r1+l£n2
and r=1r+1r >1r+ r 2 52 E? -I‘t '52
aothat ‘ 5 tin-H‘s} S r A similar argument applies if r: 2 r1.
Since 1-1 = y*_1 and “I- — y” weaee that
{ﬂew 1 + “1' Is a number between 3"]:—
and pi. and by applying the intermediate
value theorem, i‘ii’r—I + 5'9 = 11¢le for some nnmber ck between :k- 1 and 1.}.
Thus, lit—1 + ilk: 2flzck}
e. The slant height t", of the line segment
P*_ 113* is {re — rh_1]i + {F1- — lift—ll2 = {an}? + [Hal - fiat—I}: l‘IiVT, we have fife: : Ifitn1l_ I,“ if) for some number a: between :*_1 and
It. Time, fin) — “It—Ill —.f (”t lire — 171—1) Chapter 6, Additional Appﬂraﬁnns efthe integral = f 1:;th and Er: fiﬁrtla + [Ital — .l'itr—lla
= f1 + ”1::le ‘33:: It [allows that the surface area generated
by revolving the line segment Pi_1P,‘
about the .II-axia ia ﬁst = mitt—1 + Filer = renew: + Lr'ImI‘H” be where eﬁ is the number formed in part b.
By adding all such areas associated with
the paetitloin P'1 we obtain an approxi-
mation for the total surfaee area. of retrain
tinn Send by taking the limit as II I’ll—III!I
we are let the deﬁne S by =“gilerHE i': Earn-ah! 1+[r [=3] 1]“ be 11. :E_1(cl.£:*—oz&=:rauci—Ixmd n-woonallPIImﬂ. = linlw ): azmkhh + [Haj]1t a: = 2:! m) 1 + If uni a 1H. (r-RJ=+‘y'=r3er1r=1lr=-(e~ R3
r: = a? _ (z — eﬂ'mmrz — R] '
=02 _ (a — lam-ma — m d3: 21:1l1+{y']2 tit : 2r: 1 +[Flea—iffm] d: _ P-(s-maux— R)“
bill: will? I" :21: d;
;;f2 — (I: — R! ﬁr
19:2er 7' '26: R— :2 — (a- R}
Letu=.r- R;du=d1;z=u+ﬁ;if
e=R+rthenu=nandifz=Rmn
thena= —r: = 211'] 732.13%“
r r
._ d
-21r1:[ rgfuz-l-RI [Eu—n]
-r -r
= 21r[-—{:-3 it)”: + ruin—‘1“. j" This corresponds to rotating the upper semi-
circle about the y—axis. For the full circle, 3: 432m.
45. y = 0:1“ + He'll-“1
g“ = 21.09““ + 2(1 — unit; ‘2'
1+ (9')? =1+[2nc?“‘1+ 2(1 r renal—1"]? = 1 Hawaii?" ' ‘3 — anon — non 6.5
+4“ — #027394"!
= mum-n +t 1-
+ an — n]3D1:2u_h]
Since Snﬁn — HOD =%
= [snafu-1 + 201 — nine ““11
Thus. the arc length is
t
L: Inna?“ + 2n» — noel-2'] a 2'
= [219'- — new-“i1“
= CHEW 02"] _ DH!“ -n] “in-Ht]!
4t. 9' = ”2 — :2;
- :3
1+(y)“=1+ '3'— 1“
= :4 +1+___.(r2+ 2)?
a 33 g :t ‘3' :5
_ z” 2 _ :3 3 _ 9
2
For the general fomulo, if y = Ct“ + 3:1 _ " 3 1 + (9')” =1+[mr:='*—1 + (2 — one! H]!
[now—1 — {n - ewe-“F 4?. We are given Section 6.5. Physical Applications: Work, Liquid Force. and Cell-timid: Hat} =_ full + [IT-2)]: dt = ]n[aec .r + tan 2) Page 2375 0
[or all a: in [1}. 1]. Differentiathtg both sides of
this equation! we obtain 2
r/——- + we) a = m—Wtr: re == m 1+ U‘tr)!” = weir
[)"(zﬂlI .—- seat:2 ’5 — 1 = tang:
.Iffliz} = :l:t.u.n :
ﬁt] = :Flnlcos :H— G Since ﬁll] = l] [the curve 9 7: ﬁx} pnaees
through the origin]. 0 = ﬂ and
ﬂat] = :hlnlcos :rL Physical Applications: Work, Liquid Force,
and Gmtmsids, Page: 495401 The work W done on 3 object moving a
distance d in a straight line against a constant
farce Fis W = Pd. If the force varies with .r
for e 5 :5 t, the work is given by the Integral ‘
W= {life} d: I
Fluid pressure in force per unit area exerted
by a ﬂuid. The ﬂuid force on a unit
haritontni plate of area. A at depth h is phtl.
where p in the densit}r of the ﬂuid. For a.
plate submerged vertically from depth : = a
to a: = t, the ﬂuid force is given by the
integrai F = ighlﬂﬂzj d1: I
where Mr] is the depth at sand 12(3) is the
corresponding length of a typical horieontal
approximating strip. Dentinr p is a ratio of mass per unit length
or per unit area1 or per unit volume. if the
density varies in terms or a variable, ﬁnd the
element ol’ mass and then sum up (integrate).
Let f and g be continuous and satisfy 1(a)? ﬁx} on the interval [n. t], and consider
a thin plate [lamina] of uniform denuit;r p
that covers the region R between the graphs
of y : {[3] and y = 9(a) on the interval [n, t]. Then the centroid of R is the point
(1,?) such that...

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