9.5 Work
197
While“weightinpounds”isameasureofforce,“weightinkilograms”isameasureofmass.
ToconverttoforceweneedtouseNewton’slawF=ma.Atthesurfaceoftheearththe
accelerationduetogravityisapproximately9.8meterspersecondsquared,sotheforceis
F=10·9.8=98.Theunitshereare“kilogram-meterspersecondsquared”or“kgm/s
2
”,
alsoknownasaNewton(N),soF=98N.Theradiusoftheearthisapproximately6378.1
kilometersor6378100meters.Sincetheforceduetogravityobeysaninversesquarelaw,
F=k/r
2
and98=k/6378100
2
,k=3.986655642·10
15
and
W =−
k
D
+6.250538000·10
8
Newton-meters.
AsD increasesW W ofcoursegetslarger,sincethequantitybeingsubtracted,−k/D,gets
smaller. Butnotethat t theworkW W willnever r exceed6.250538000·10
8
,andinfactwill
approach this s value e as s D getslarger. . In n short, , with h a finite e amount of f work, namely
6.250538000·10
8
N-m,wecanliftthe10kilogramobjectasfaraswewishfromearth.
Nextisanexampleinwhichtheforceisconstant,buttherearemanyobjectsmoving
differentdistances.
EXAMPLE 9.15
Supposethatawater tankisshapedlikearightcircularconewith
the tipat thebottom,andhasheight 10metersand radius2 metersat thetop. . If f the
tankisfull,howmuchworkisrequiredtopumpallthewateroutoverthetop? Herewe
havealargenumberofatomsofwaterthatmustbelifteddifferentdistancestogettothe
topofthetank. Fortunately,wedon’treallyhavetodealwithindividualatoms—wecan
consideralltheatomsatagivendepthtogether.
Toapproximatethework,wecandividethewaterinthetankintohorizontalsections,
approximatethevolumeofwaterinasectionbyathindisk,andcomputetheamountof
workrequiredtolifteachdisktothetopofthetank. Asusual,wetakethelimitasthe
sectionsgetthinnerandthinnertogetthetotalwork.
Atdepthhthecircularcross-sectionthroughthetankhasradiusr=(10−h)/5,by
similartriangles,andareaπ(10−h)
2
/25.Asectionofthetankatdepthhthushasvolume
approximatelyπ(10−h)2/25∆handsocontainsσπ(10−h)2/25∆h kilogramsofwater,
whereσ isthedensityofwaterinkilogramspercubicmeter;σ≈1000. . Theforcedueto
gravityonthismuchwateris9.8σπ(10−h)
2
/25∆h,andfinally,thissectionofwatermust
beliftedadistanceh,whichrequiresh9.8σπ(10−h)
2
/25∆hNewton-metersofwork. The
totalworkistherefore
W=
9.8σπ
25
Z
10
0
h(10−h)
2
dh=
980000
3
π≈1026254 Newton-meters.
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198
Chapter9 Applications s ofIntegration
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.........................................................................................................................
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10
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2
..............................
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...
Figure9.14
Cross-sectionofaconicalwatertank.
A spring has a a “natural l length,” itslength if nothing is stretching or r compressing
it. Ifthespringiseitherstretchedor r compressedthespringprovidesanopposingforce;
accordingtoHooke’sLawthemagnitudeofthisforceisproportionaltothedistancethe
springhasbeenstretchedorcompressed: F=kx. . Theconstantofproportionality,k,of
coursedependsonthespring. Notethatxhererepresentsthechangeinlengthfromthe
naturallength.
EXAMPLE 9.16
Suppose k=5for a given spring thathasa naturallength of0.1
meters. Supposea a force isapplied thatcompressesthespring tolength0.08. . What t is
the magnitude ofthe force? ? Assuming g that the constant t khasappropriatedimensions
(namely,kg/s2),theforceis5(0.1−0.08)=5(0.02)=0.1Newtons.
EXAMPLE9.17
Howmuchworkisdoneincompressingthespringinthepreviousex-
amplefromitsnaturallengthto0.08meters? From0.08metersto0.05meters?Howmuch
workisdonetostretchthespringfrom0.1metersto0.15meters? Wecanapproximate
theworkbydividing thedistancethat thespringiscompressedinto smallsubintervals.
Then theforce exertedbythe springisapproximatelyconstant over the subinterval,so
theworkrequiredtocompressthespringfromx
i
tox
i+1
isapproximately5(x
i
−0.1)∆x.
Thetotalworkisapproximately
n−1
i=0
5(x
i
−0.1)∆x
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9.5 Work
199
andinthelimit
W =
Z
0.08
0.1
5(x−0.1)dx=
5(x−0.1)
2
2
0.08
0.1
=
5(0.08−0.1)
2
2
5(0.1−0.1)
2
2
=
1
1000
N-m.
Theother valuesweseeksimplyusedifferentlimits. . Tocompressthespringfrom0.08
metersto0.05meterstakes
W =
Z
0.05
0.08
5(x−0.1)dx=
5x
2
2
0.05
0.08
=
5(0.05−0.1)
2
2
5(0.08−0.1)
2
2
=
21
4000
N-m
andtostretchthespringfrom0.1metersto0.15metersrequires
W =
Z
0.15
0.1
5(x−0.1)dx=
5x
2
2
0.15
0.1
=
5(0.15−0.1)
2
2
5(0.1−0.1)
2
2
=
1
160
N-m.
Exercises
1. Howmuchworkisdoneinliftinga100kilogramweightfromthesurfaceoftheearthtoan
orbit35,786kilometersabovethesurfaceoftheearth? 
2. Howmuchworkisdoneinliftinga100kilogramweightfromanorbit1000kilometersabove
thesurfaceoftheearthtoanorbit35,786kilometersabovethesurfaceoftheearth? 
3. Awatertankhastheshapeofacylinderwithradiusr=1meterandheight10meters. If
thedepthofthewateris5meters,howmuchworkisrequiredtopumpallthewateroutthe
topofthetank? 
4. Supposethetank k ofthepreviousproblemislyingonitsside,sothatthecircularendsare
vertical,andthat ithas the same amountof wateras before. . Howmuchwork k is required
topumpthewateroutthetopofthetank(whichisnow2metersabovethebottomofthe
tank)?
5. Awatertankhas s theshapeofthebottomhalfofaspherewithradius r=1meter. . Ifthe
tankisfull,howmuchworkisrequiredtopumpallthewateroutthetopofthetank? 
6. Aspringhasconstantk=10kg/s
2
. Howmuchwork k isdoneincompressingit1/10meter
fromitsnaturallength? 
7. Aforceof2Newtonswillcompressaspringfrom1meter(itsnaturallength)to0.8meters.
Howmuchworkisrequiredtostretchthespringfrom1.1metersto1.5meters? 
8. A20meterlongsteelcablehasdensity2kilogramspermeter,andishangingstraightdown.
Howmuchworkisrequiredtolifttheentirecabletotheheightofitstopend? 
9. Thecableinthepreviousproblemhasa100kilogrambucketofconcreteattachedtoitslower
end. Howmuchworkisrequiredtolifttheentirecableandbuckettotheheightofitstop
end? 
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200
Chapter9 Applications s ofIntegration
Supposea beamis10meterslong,andthatthereare threeweightsonthe beam: : a10
kilogramweight3metersfromtheleftend,a5kilogramweight6metersfromtheleftend,
anda4kilogramweight8metersfromthe leftend. . Whereshould d afulcrumbeplaced
sothatthebeambalances? Let’sassignascaletothebeam,from0attheleftendto10
attheright, sothat t we candenote locationsonthe beamsimplyasx coordinates; ; the
weightsareatx=3,x=6,andx=8.
Supposetobeginwiththatthefulcrumisplacedatx=5. Whatwillhappen? Each
weightappliesaforcetothebeamthattendstorotateitaroundthefulcrum;thiseffect
is measured d by a quantity called torque, proportional to the e mass timesthe distance
fromthefulcrum. Ofcourse,weightson n different sidesof thefulcrumrotate the beam
inoppositedirections. Wecandistinguishthisbyusingasigneddistanceintheformula
for torque. . Sowith h the fulcrumat 5,the torquesinducedby thethreeweightswillbe
proportionalto (3−5)10 0 = = −20, (6−5)5 5 = = 5, and (8−5)4 4 = = 12. . For r the e beam m to
balance,thesumofthetorquesmust bezero;sincethe sumis−20+5+12=−3,the
beam rotates counter-clockwise, , and to o get t the e beam m to o balance we need d to o move the
fulcrumtotheleft. Tocalculateexactlywherethefulcrumshouldbe,welet¯xdenotethe
locationofthefulcrumwhenthebeamisinbalance.Thetotaltorqueonthebeamisthen
(3−¯x)10+(6−¯x)5+(8−¯x)4=92−19¯x.Sincethebeambalancesat¯xitmustbethat
92−19¯x=0or ¯x=92/19≈4.84,thatis,thefulcrumshouldbeplacedatx=92/19to
balancethebeam.
Nowsupposethatwehaveabeamwithvaryingdensity—someportionsofthebeam
containmoremassthanotherportionsofthesamesize. Wewanttofigureoutwhereto
putthefulcrumsothatthebeambalances.
EXAMPLE 9.18
Supposethe beamis10 meterslong andthat thedensityis1+x
kilogramspermeteratlocationxonthebeam.Toapproximatethesolution,wecanthink
of the beamasa sequence ofweights“on” a beam. . For r example, , we e can think k of the
portionofthebeambetweenx=0andx=1asaweightsittingat x=0,theportion
betweenx=1andx=2asaweightsittingatx=1,andsoon. Wethenapproximate
themassoftheweightsbyassumingthateachportionofthebeamhasconstantdensity.
Sothemassofthefirstweightisapproximatedasm
0
=(1+0)1=1kilograms,namely,
(1+0) kilogramsper r meter times1 1 meter. . The e second weight is s m
1
=(1+1)1 = = 2
kilograms,andsoontothe tenthweightwithm
9
=(1+9)1=9kilograms. Sointhis
casethetotaltorqueis
(0−¯x)m
0
+(1−¯x)m
1
+···+(9−¯x)m
9
=(0−¯x)1+(1−¯x)2+···+(9−¯x)9.
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9.6 Center r ofMass
201
Ifwesetthistozeroandsolvefor ¯xweget ¯x=6. . Ingeneral,ifwedividethebeaminto
nportions,themassofweightnumberiwillbem
i
=(1+x
i
)(x
i+1
−x
i
)=(1+x
i
)∆x
andthetorqueinducedbyweightnumberiwillbe(x
i
−¯x)m
i
=(x
i
−¯x)(1+x
i
)∆x. The
totaltorqueisthen
(x
0
−¯x)(1+x
0
)∆x+(x
1
−¯x)(1+x
1
)∆x+···+(x
n−1
−¯x)(1+x
n−1
)∆x
=
nX−1
i=0
x
i
(1+x
i
)∆x−
n−1
i=0
¯x(1+x
i
)∆x
=
nX−1
i=0
x
i
(1+x
i
)∆x−¯x
nX−1
i=0
(1+x
i
)∆x.
Ifwesetthisequaltozeroandsolvefor ¯xwegetanapproximationtothebalancepoint
ofthebeam:
0=
nX−1
i=0
x
i
(1+x
i
)∆x−¯x
nX−1
i=0
(1+x
i
)∆x
¯x
nX−1
i=0
(1+x
i
)∆x=
nX−1
i=0
x
i
(1+x
i
)∆x
¯x=
nX−1
i=0
x
i
(1+x
i
)∆x
n−1
i=0
(1+x
i
)∆x
.
The numerator r of f this s fraction n iscalled the moment t of f the e systemaround zero, and
thedenominatorhasaveryfamiliarinterpretation. Consideronetermofthesuminthe
denominator: (1+x
i
)∆x. Thisisthedensitynearx
i
timesashortlength,∆x,whichin
otherwordsisapproximatelythemassofthebeambetweenx
i
andx
i+1
. Whenweadd
theseupwegetapproximatelythemassofthebeam.
Noweachofthesumsinthefractionhastherightformtoturnintoanintegral,which
inturngivesustheexactvalueof¯x:
¯x=
Z
10
0
x(1+x)dx
Z
10
0
(1+x)dx
.
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202
Chapter9 Applications s ofIntegration
Thenumeratoristhemomentofthesystem:
Z
10
0
x(1+x)dx=
Z
10
0
x+x
2
dx=
1150
3
,
andthedenominatoristhemassofthebeam:
Z
10
0
(1+x)dx=60,
andthebalancepoint,officiallycalledthecenterofmass,is
¯x=
1150
3
1
60
=
115
18
≈6.39.
Itshouldbeapparentthattherewasnothingspecialaboutthedensityfunctionσ(x)=
1+xor thelengthofthe beam,or eventhat theleftendofthebeamisat theorigin.
Ingeneral,ifthedensityofthebeamisσ(x)andthebeamcoverstheinterval[a,b],the
momentofthebeamaroundzerois
M
0
=
Z
b
a
xσ(x)dx
andthetotalmassofthebeamis
M=
Z
b
a
σ(x)dx
andthecenterofmassisat
¯x=
M
0
M
.
EXAMPLE9.19
Supposeabeamliesonthex-axisbetween20and30,andhasdensity
functionσ(x)=x−19.Findthecenterofmass. Thisisthesameasthepreviousexample
exceptthatthebeamhasbeenmoved.Notethatthedensityattheleftendis20−19=1
andat the right end is30−19 =11, , asbefore. . Hence e the center of massmust be at
approximately20+6.39=26.39. Let’sseehowthecalculationworksout.
M
0
=
Z
30
20
x(x−19)dx=
Z
30
20
x
2
−19xdx=
x3
3
19x2
2
30
20
=
4750
3
M=
Z
30
20
x−19dx=
x
2
2
−19x
30
20
=60
M
0
M
=
4750
3
1
60
=
475
18
≈26.39.
9.6 Center r ofMass
203
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0
1
0
1
x
i
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(¯x,¯y)
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0
1
x
i
m
i
Figure 9.15
Centerofmassforatwodimensionalplate.
EXAMPLE 9.20
Supposeaflatplateofuniformdensityhastheshapecontainedby
y=x2,y=1,andx=0,inthefirstquadrant.Findthecenterofmass.(Sincethedensity
isconstant,thecenterofmassdependsonlyontheshapeoftheplate,notthedensity,or
inotherwords,thisisapurelygeometricquantity. Insuchacasethe e center ofmassis
calledthecentroid.)
Thisisatwodimensionalproblem,butitcanbesolvedasifitweretwoonedimensional
problems: we e needtofindthe x andy coordinatesof thecenter ofmass, , ¯x x and ¯y, , and
fortunatelywe candotheseindependently. . Imagine e lookingat the plate edge on,from
belowthex-axis. The e plate willappear to be abeam,andthe massofa short section
of the“beam”,saybetween x
i
and x
i+1
,isthemassofastrip ofthe plate between x
i
andx
i+1
. Seefigure9.15showingtheplatefromaboveandasitappearsedgeon. Since
theplatehasuniformdensitywemayaswellassumethat σ=1. . Thenthemassofthe
platebetweenx
i
andx
i+1
isapproximatelym
i
=σ(1−x
2
i
)∆x=(1−x
2
i
)∆x.Nowwecan
computethemomentaroundthey-axis:
M
y
=
Z
1
0
x(1−x
2
)dx=
1
4
andthetotalmass
M=
Z
1
0
(1−x
2
)dx=
2
3
andfinally
¯x=
1
4
3
2
=
3
8
.
204
Chapter9 Applications s ofIntegration
Next we do o the e same thing to find d ¯y. . The e mass of the e plate between y
i
and y
i+1
is
approximatelyn
i
=
y∆y,so
M
x
=
Z
1
0
y
ydy=
2
5
and
¯y=
2
5
3
2
=
3
5
,
sincethetotalmassMisthesame. Thecenterofmassisshowninfigure9.15.
EXAMPLE 9.21
Findthecenterofmassofathin,uniformplatewhoseshapeisthe
region between y = cosx and the x-axis s between n x = −π/2 and x = π/2. . It t isclear
that ¯x=0,butforpracticelet’scomputeitanyway. . Wewillneedthetotalmass,sowe
computeitfirst:
M=
Z
π/2
−π/2
cosxdx=sinx
π/2
−π/2
=2.
Themomentaroundthey-axisis
M
y
=
Z
π/2
−π/2
xcosxdx=cosx+xsinx
π/2
−π/2
=0
andthemomentaroundthex-axisis
M
x
=
Z
1
0
y·2arccosydy=y
2
arccosy−
y
p
1−y2
2
+
arcsiny
2
1
0
=
π
4
.
Thus
¯x=
0
2
, ¯y=
π
8
≈0.393.
9.7 Kineticenergy; ; improper integrals
205
Exercises
1. Abeam10meterslonghasdensityσ(x)=x
2
atdistancexfromtheleftendofthebeam.
Findthecenterofmass ¯x.
2. Abeam10meterslonghasdensityσ(x)=sin(πx/10)atdistancexfromtheleftendofthe
beam.Findthecenterofmass¯x. 
3. Abeam4meters s longhasdensity σ(x)=x
3
atdistance xfromtheleft endofthebeam.
Findthecenterofmass ¯x.
4. Verifythat
Z
2xarccosxdx=x
2
arccosx−
x
1−x2
2
+
arcsinx
2
+C.
5. Athinplateliesintheregionbetweeny=x
2
andthex-axisbetweenx=1andx=2.Find
thecentroid. 
6. Athinplatefillstheupperhalfoftheunitcirclex
2
+y
2
=1. Findthecentroid. 
7. Athinplateliesintheregioncontainedbyy=xandy=x
2
. Findthecentroid. 
8. Athinplateliesintheregioncontainedbyy=4−x
2
andthex-axis.Findthecentroid. 
9. Athinplateliesintheregioncontainedbyy=x
1/3
andthex-axisbetweenx=0andx=1.
Findthecentroid. 
10. Athinplateliesintheregioncontainedby
x+
y=1andtheaxesinthefirstquadrant.
Findthecentroid. 
11. Athinplatelies s intheregionbetweenthe circle x
2
+y
2
= 4andthecircle x
2
+y
2
= 1,
abovethex-axis.Findthecentroid.
12. Athinplateliesintheregionbetweenthecirclex
2
+y
2
=4andthecirclex
2
+y
2
=1in
thefirstquadrant. Findthecentroid. 
Recallexample 9.14inwhichwe computedthe workrequiredto lift anobjectfromthe
surfaceoftheearthtosomelargedistanceDaway.SinceF=k/xwecomputed
Z
D
r
0
k
x2
dx=−
k
D
+
k
r
0
.
WenoticedthatasDincreases,k/Ddecreasestozerosothattheamountofworkincreases
tok/r
0
.Moreprecisely,
lim
D→∞
Z
D
r
0
k
x2
dx= lim
D→∞
k
D
+
k
r
0
=
k
r
0
.
Wemightreasonablydescribethiscalculuationascomputingtheamountofworkrequired
tolifttheobject“toinfinity,”andabbreviatethelimitas
lim
D→∞
Z
D
r
0
k
x2
dx=
Z
r
0
k
x2
dx.
206
Chapter9 Applications s ofIntegration
Suchanintegral,withalimit ofinfinity,iscalledanimproper r integral. . Thisisabit
unfortunate,sinceit’snotreally“improper”todothis,norisitreally“anintegral”—itis
anabbreviationforthelimitofaparticularsortofintegral.Nevertheless,we’restuckwith
theterm,andtheoperationitselfisperfectlylegitimate. Itmayatfirstseemoddthata
finiteamountofworkissufficienttoliftanobjectto“infinity”,butsometimessurprising
thingsareneverthelesstrue,andthisissuchacase. Ifthevalueofanimproper r integral
isafinitenumber,asinthisexample,wesaythattheintegralconverges,andifnotwe
saythattheintegraldiverges.
Here’sanotherway,perhapsevenmore surprising,to interpretthiscalculation. . We
knowthatoneinterpretationof
Z
D
1
1
x2
dx
isthe area under y =1/xfromx =1 tox =D. . Ofcourse,asD D increasesthisarea
increases. Butsince
Z
D
1
1
x2
dx=−
1
D
+
1
1
,
whiletheareaincreases,itneverexceeds1,thatis
Z
1
1
x2
dx=1.
Theareaoftheinfiniteregionundery=1/xfromx=1toinfinityisfinite.
Consider aslightlydifferentsortofimproperintegral:
Z
−∞
xe
−x
2
dx. Therearetwo
wayswemight tryto computethis. . First,wecouldbreakit t upintotwo more familiar
integrals:
Z
−∞
xe
−x
2
dx=
Z
0
−∞
xe
−x
2
dx+
Z
0
xe
−x
2
dx.
Nowwedotheseasbefore:
Z
0
−∞
xe
−x
2
dx= lim
D→∞
e
−x
2
2
0
D
=−
1
2
,
and
Z
0
xe
−x
2
dx= lim
D→∞
e
−x
2
2
D
0
=
1
2
,
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