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9.3 Volume
187
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Figure 9.7
Solidwithequilateraltrianglesascross-sections. (JA)
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Figure9.8
Asolidofrotation.
andthevolumeofathin“slab”isthen
(1−x
2
i
)
3(1−x
2
i
)∆x.
Thusthetotalvolumeis
Z
1
−1
3(1−x
2
i
)
2
dx=
16
15
3.
Oneeasywaytoget“nice”cross-sectionsisbyrotatingaplanefigurearoundaline.
For example,infigure 9.8weseeaplaneregionunderacurveandbetweentwovertical
lines;thentheresultofrotatingthisaroundthex-axis;thenatypicalcircularcross-section.
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188
Chapter9 Applications s ofIntegration
Ofcourseareal“slice”ofthisfigurewillnothavestraightsides,butwecanapproxi-
matethevolumeoftheslicebyacylinderordiskwithcirculartopandbottomandstraight
sides;thevolumeofthisdiskwillhavetheformπr2∆x.Aslongaswecanwriterinterms
ofxwecancomputethevolumebyanintegral.
EXAMPLE9.9
Findthevolumeofarightcircularconewithbaseradius10andheight
20. (Arightcircularconeisonewithacicularbaseandwiththetipoftheconedirectly
overthecenterofthebase.) Wecanviewthisconeasproducedbytherotationoftheline
y=x/2rotatedaboutthex-axis,asindicatedinfigure9.9.
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20
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Figure9.9
Aregionthatgeneratesacone;approximatingthevolumebycirculardisks.
(JA)
At a particular r point t on the e x-axis, sayx
i
, the e radius s of f the resulting cone e isthe
y-coordinate of the corresponding point onthe line, , namely y
i
=x
i
/2. Thusthe e total
volumeisapproximately
n−1
i=0
π(x
i
/2)
2
dx
andtheexactvolumeis
Z
20
0
π
x2
4
dx=
π
4
203
3
=
2000π
3
.
Notethatwecaninsteaddothecalculationwithagenericheightandradius:
Z
h
0
π
r2
h2
x
2
dx=
πr2
h2
h3
3
=
πr2h
3
,
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9.3 Volume
189
givingustheusualformulaforthevolumeofacone.
EXAMPLE 9.10
Find the volume of f the e object generated when the area between
y=x
2
andy=xisrotatedaroundthex-axis. Thissolidhasa“hole”inthemiddle;we
cancomputethevolumebysubtractingthevolumeoftheholefromthevolumeenclosed
bythe outer surfaceof thesolid. . In n figure9.10we showthe region thatisrotated,the
resultingsolid with the front half cut away, the cone that formsthe outer surface,the
horn-shapedhole,andacross-sectionperpendiculartothex-axis.
0
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.
Figure9.10
Solidwithahole,showingtheouterconeandtheshapetoberemovedto
formthehole. (JA)
Wehavealreadycomputedthevolumeofacone;inthiscaseitisπ/3.Ataparticular
valueofx,sayx
i
,thecross-sectionofthehornisacirclewithradiusx
2
i
,sothevolumeof
190
Chapter9 Applications s ofIntegration
thehornis
Z
1
0
π(x
2
)
2
dx=
Z
1
0
πx
4
dx=π
1
5
,
sothedesiredvolumeisπ/3−π/5=2π/15.
Aswith theareabetween curves, thereisanalternate approachthat computesthe
desiredvolume“allat once”byapproximating thevolume ofthe actualsolid. . Wecan
approximatethevolumeofasliceofthesolidwithawasher-shapedvolume,asindicated
infigure9.10.
Thevolumeofsuchawasheristheareaofthefacetimesthethickness.Thethickness,
asusual,is∆x,whiletheareaofthefaceistheareaoftheoutercircleminustheareaof
theinnercircle,sayπR
2
−πr
2
.Inthepresentexample,ataparticularx
i
,theradiusRis
x
i
andrisx
2
i
.Hence,thewholevolumeis
Z
1
0
πx
2
−πx
4
dx=π
µ
x
3
3
x
5
5
¶fl
1
0
µ
1
3
1
5
=
15
.
Ofcourse,what wehavedonehere isexactlythesamecalculationasbefore,exceptwe
haveineffectrecomputedthevolumeoftheoutercone.
Supposetheregionbetweenf(x)=x+1andg(x)=(x−1)
2
isrotatedaroundthe
y-axis; see e figure 9.11. It t ispossible, , but t inconvenient, , to o compute the volume of the
resulting solid by y the e method we have used so far. . The e problemisthat there are two
“kinds”oftypicalrectangles: thosethatgofromthelinetotheparabolaandthosethat
touchtheparabolaonbothends. Tocomputethevolumeusingthisapproach,weneedto
breaktheproblemintotwopartsandcomputetwointegrals:
π
Z
1
0
(1+
y)
2
−(1−
y)
2
dy+π
Z
4
1
(1+
y)
2
−(y−1)
2
=
8
3
π+
65
6
π=
27
2
π.
Ifinsteadweconsidera typicalverticalrectangle,but stillrotate aroundthe y-axis,we
getathin“shell”insteadofathin“washer”. Ifweaddupthevolumeofsuchthinshells
we willget anapproximationto the true volume. . What t isthe volumeof sucha shell?
Considertheshellatx
i
.Imaginethatwecuttheshellverticallyinoneplaceand“unroll”
itintoathin,flatsheet. Thissheetwillbealmostarectangularprismthatis∆xthick,
f(x
i
)−g(x
i
) tall, , and 2πx
i
wide (namely, , the circumference e of the shell l before e it was
unrolled).Thevolumewillthenbeapproximatelythevolumeofarectangularprismwith
thesedimensions:2πx
i
(f(x
i
)−g(x
i
))∆x. Ifweaddtheseupandtakethelimitasusual,
wegettheintegral
Z
3
0
2πx(f(x)−g(x))dx=
Z
3
0
2πx(x+1−(x−1)
2
)dx=
27
2
π.
9.3 Volume
191
Not onlydoesthisaccomplishthetaskwith onlyone integral,the integralissomewhat
easier than those e in n the previouscalculation. . Thingsare e not alwaysso neat, , but t it is
oftenthecasethatoneofthetwomethodswillbesimplerthantheother,soitisworth
consideringbothbeforestartingtodocalculations.
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Figure 9.11
Computingvolumeswith“shells”. (JA)
EXAMPLE 9.11
Supposethe areaunder y=−x
2
+1 betweenx =0andx =1is
rotatedaroundthex-axis. Findthevolumebybothmethods.
Diskmethod:
Z
1
0
π(1−x
2
)
2
dx=
8
15
π.
Shellmethod:
Z
1
0
2πy
p
1−ydy=
8
15
π.
Exercises
1. Verifythatπ
Z
1
0
(1+
y)
2
−(1−
y)
2
dy+π
Z
4
1
(1+
y)
2
−(y−1)
2
=
8
3
π+
65
6
π=
27
2
π.
2. Verifythat
Z
3
0
2πx(x+1−(x−1)
2
)dx=
27
2
π.
3. Verifythat
Z
1
0
π(1−x
2
)
2
dx=
8
15
π.
4. Verifythat
Z
1
0
2πy
p
1−ydy=
8
15
π.
5. Useintegrationtofindthevolumeofthesolidobtainedbyrevolvingtheregionboundedby
x+y=2andthexandyaxesaroundthex-axis. 
6. Findthevolumeofthesolidobtainedbyrevolvingtheregionboundedby y y=x−x
2
and
thex-axisaroundthex-axis. 
192
Chapter9 Applications s ofIntegration
7. Findthevolumeofthesolidobtainedbyrevolvingthe e regionboundedby y=
sinx,the
y-axis,andthelinesy=1andx=π/2aroundthex-axis. 
8. LetSbetheregionofthexy-planeboundedabovebythecurvex
3
y=64,belowbytheline
y=1,ontheleftbythelinex=2,andontherightbythelinex=4. Findthevolumeof
thesolidobtainedbyrotatingSaround(a)thex-axis,(b)theliney=1,(c)they-axis,(d)
thelinex=2. 
9. Theequationx
2
/9+y
2
/4=1describesanellipse. Findthevolume e ofthesolidobtained
byrotatingtheellipsearoundthex-axisandalsoaroundthey-axis.Thesesolidsarecalled
ellipsoids;oneisvaguelyrugby-ballshaped,oneissortofflying-saucershaped,orperhaps
squished-beach-ball-shaped. 
Figure 9.12
Ellipsoids.(JA)
10. Useintegrationtocomputethevolumeofasphereofradiusr. . Youshouldofcoursegetthe
well-knownformula4πr
3
/3.
11. Ahemisphericbowlofradiusrcontainswatertoadepthh. . Findthevolumeofwaterinthe
bowl. 
12. Thebaseofatetrahedron(atriangularpyramid)ofheighthisanequilateraltriangleofside
s.Itscross-sectionsperpendiculartoanaltitudeareequilateraltriangles. Expressitsvolume
V as s anintegral,andfindaformulaforV V intermsofhands. Verify y thatyouranswer is
(1/3)(areaofbase)(height).
13. Thebaseofasolidistheregionbetweenf(x)=cosxandg(x)=−cosx,−π/2≤x≤π/2,
anditscross-sectionsperpendiculartothex-axisaresquares. Findthevolumeofthesolid.
The average ofsomefiniteset ofvaluesisafamiliarconcept. . If,forexample,theclass
scoresonaquizare10,9,10,8,7,5,7,6,3,2,7,8,thentheaveragescoreisthesumof
thesenumbersdividedbythesizeoftheclass:
averagescore=
10+9+10+8+7+5+7+6+3+2+7+8
12
=
82
12
≈6.83.
Suppose that between t =0 andt =1 the speed ofan object issin(πt). . What t is s the
averagespeedoftheobjectoverthattime?Thequestionsoundsasifitmustmakesense,
9.4 Averagevalue e ofa function
193
yetwecan’tmerelyaddupsomenumberofspeedsanddivide,sincethespeedischanging
continuouslyoverthetimeinterval.
Tomakesenseof“average”inthiscontext,wefallbackontheideaofapproximation.
Considerthespeedoftheobjectattenthofasecondintervals: sin0,sin(0.1π),sin(0.2π),
sin(0.3π),...,sin(0.9π).Theaveragespeed“should”befairlyclosetotheaverageofthese
tenspeeds:
1
10
X9
i=0
sin(πi/10)≈
1
10
6.3=0.63.
Ofcourse,ifwecomputemorespeedsatmoretimes,theaverageofthesespeedsshould
beclosertothe“real”average.Ifwetaketheaverageofnspeedsatevenlyspacedtimes,
weget:
1
n
nX−1
i=0
sin(πi/n).
Heretheindividualtimesaret
i
=i/n,sorewritingslightlywehave
1
n
nX−1
i=0
sin(πt
i
).
Thisisalmost the sort t ofsumthat t we knowturns s intoan n integral; what’sapparently
missingis∆t—butinfact,∆t=1/n,thelengthofeachsubinterval.Sorewritingagain:
n−1
i=0
sin(πt
i
)
1
n
=
n−1
i=0
sin(πt
i
)∆t.
Nowthishasexactlytherightform,sothatinthelimitweget
averagespeed=
Z
1
0
sin(πt)dt=−
cos(πt)
π
1
0
=−
cos(π)
π
+
cos(0)
π
=
2
π
≈0.6366≈0.64.
It’snotentirelyobviousfromthisonesimpleexamplehowtocomputesuchanaverage
ingeneral. Let’slookat t asomewhat more complicatedcase. . Supposethat t thevelocity
ofanobjectis16t
2
+5feetpersecond. Whatistheaveragevelocitybetweent=1and
t=3? Againwesetupanapproximationtotheaverage:
1
n
nX−1
i=0
16t
2
i
+5,
194
Chapter9 Applications s ofIntegration
wherethevaluest
i
areevenlyspacedtimesbetween1and3.Onceagainweare“missing”
∆t,andthistime1/nisnotthecorrectvalue. Whatis∆tingeneral? Itisthelengthof
asubinterval;inthiscase wetaketheinterval[1,3]anddivideitinto nsubintervals,so
eachhaslength(3−1)/n=2/n=∆t. Nowwiththeusual“multiplyanddividebythe
samething”trickwecanrewritethesum:
1
n
n−1
i=0
16t
2
i
+5=
1
3−1
n−1
i=0
(16t
2
i
+5)
3−1
n
=
1
2
n−1
i=0
(16t
2
i
+5)
2
n
=
1
2
n−1
i=0
(16t
2
i
+5)∆t.
Inthelimitthisbecomes
1
2
Z
3
1
16t
2
+5dt=
1
2
446
3
=
223
3
.
Doesthisseemreasonable? Let’spictureit:infigure9.13isthevelocityfunctiontogether
withthehorizontalliney=223/3≈74.3.Certainlytheheightofthehorizontallinelooks
atleastplausiblefortheaverageheightofthecurve.
0
1
2
3
0
25
50
75
100
125
150
.........................
...................
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...........
.........
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......
......
......
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...
Figure9.13
Averagevelocity.
Here’sanother wayto o interpret “average” that maymake our computation appear
evenmorereasonable. Theobjectofourexamplegoesacertaindistancebetweent=1
andt=3.Ifinsteadtheobjectweretotravelattheaveragespeedoverthesametime,it
wouldgothesamedistance.Atanaveragespeedof223/3feetpersecondfortwoseconds
theobjectwouldgo446/3feet. Howfar r doesitactuallygo? ? Weknowhowtocompute
9.5 Work
195
this:
Z
3
1
v(t)dt=
Z
3
1
16t
2
+5dt=
446
3
.
Sonowweseethatanotherinterpretationofthecalculation
1
2
Z
3
1
16t
2
+5dt=
1
2
446
3
=
223
3
is:totaldistancetraveleddividedbythetimeintransit,namely,averagespeed.
It may y also be e helpful l to o interpret the e average geometrically. . The e area under r the
velocitycurveis
Z
3
1
16t
2
+5dt=
446
3
.
Theareaunder y=223/3overthe same interval[1,3]issimplytheareaofarectangle
thatis2by223/3witharea446/3.Sotheaverageheightofafunctionistheheightofthe
horizontallinethatproducesthesameareaoverthegiveninterval.
Exercises
1. Findtheaverageheightofcosxovertheintervals[0,π/2],[−π/2,π/2],and[0,2π]. 
2. Findtheaverageheightofx
2
overtheinterval[−2,2]. 
3. Findtheaverageheightof1/x
2
overtheinterval[1,A]. 
4. Findtheaverageheightof
1−xovertheinterval[−1,1].
5. Anobjectmoveswithvelocityv(t)=−t
2
+1feetpersecondbetweent=0andt=2.Find
theaveragevelocityandtheaveragespeedoftheobjectbetweent=0andt=2. 
6. Theobservationdeckonthe102ndfloor r of theEmpire StateBuildingis 1,224feet above
the ground. . If f a steel ball is dropped from the observation deck its velocity at t time e t is
approximately v(t)=−32tfeetper second. . Findtheaveragespeedbetweenthetimeitis
droppedandthetimeithitstheground,andfinditsspeedwhenithitstheground. 
Afundamentalconcept in classicalphysicsiswork: : Ifanobjectismovedinastraight
lineagainstaforceF foradistancestheworkdoneisW W =Fs.
EXAMPLE 9.12
How much workisdone e in lifting a 10 pound weight t vertically a
distanceof 5feet? ? The e force due to gravityona 10pound weightis10poundsat the
surface of theearth,andit doesnot change appreciablyover 5feet. . Theworkdone e is
W =10·5=50foot-pounds.
196
Chapter9 Applications s ofIntegration
Inrealityfewsituationsaresosimple.Theforcemightnotbeconstantovertherange
ofmotion,asinthenextexample.
EXAMPLE9.13
Howmuchworkisdoneinliftinga10poundweightfromthesurface
of the earth to an orbit 100 miles s above e the surface? ? Over r 100 milesthe force due to
gravitydoeschangesignificantly,soweneedtotakethisintoaccount. Theforceexerted
ona 10poundweight at adistance r fromthecenter ofthe earth isF F =k/r
2
andby
definitionitis10whenristheradiusoftheearth(weassumetheearthisasphere).How
canweapproximatetheworkdone? Wedividethepathfromthesurfacetoorbitinton
smallsubpaths.Oneachsubpaththeforceduetogravityisroughlyconstant,withvalue
k/r
2
i
atdistancer
i
. The e worktoraisethe objectfromr
i
tor
i+1
isthusapproximately
k/r
2
i
∆randthetotalworkisapproximately
n−1
i=0
k
r
2
i
∆r,
orinthelimit
W =
Z
r
1
r
0
k
r2
dr,
wherer
0
istheradiusoftheearthandr
1
isr
0
plus100miles.Theworkis
W=
Z
r
1
r
0
k
r2
dr=−
k
r
r
1
r
0
=−
k
r
1
+
k
r
0
.
Usingr
0
=20925525feet wehave r
1
=21453525. Theforceonthe10poundweightat
thesurfaceoftheearthis10pounds,so10=k/20925525
2
,givingk=4378775965256250.
Then
k
r
1
+
k
r
0
=
491052320000
95349
≈5150052 foot-pounds.
Notethatifweassumetheforceduetogravityis10poundsoverthewholedistancewe
wouldcalculatetheworkas10(r
1
−r
0
)=10·100·5280=5280000,somewhathighersince
wedon’taccountfortheweakeningofthegravitationalforce.
EXAMPLE 9.14
How much work is done e in lifting a a 10 0 kilogramobject t fromthe
surface of f the e earth h to o a distance D D from m the center of the earth? ? This s is the same
problemasbeforeindifferentunits,andwearenotspecifyingavalueforD. Asbefore
W=
Z
D
r
0
k
r2
dr=−
k
r
D
r
0
=−
k
D
+
k
r
0
.
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