18.01 Single-Variable Calculus

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18.01Single-VariableCalculus
ProblemSet5
PartIIQuestion3b
Findthevolumeoftheregionx>0,y>0,z>0boundedby
z2
< x + y < z 2
(1)
Examiningthetwo-dimensionalsurfaceatafixedz0,wehavearegionboundedby
€ twolines:
z02
2 y < −x + z0
y > −x +
(2)
Forexample,atz=1,thetwo-dimensionalregionisboundedbythex-andy-axes,the
redlowerboundandblueupperbound:
€
Wenotethatatz=0,andsimilarlyat
z=2,theregioniszero,andthusthe
volumeisboundedbetweenthosetwo
values.Weenvisionthevolumebeing
composedofstacksofthin"plates"from
z=0toz=2.Eachplatehasinfinitesimal
thicknessdz,andsomecalculable
surfaceareaA,andvolumeAdz.We
findthetotalvolumebyintegrating
overz.
Weneedtofindtheareaofasingle
platelocatedataparticularz.
FindingtheAreaofaSinglePlate
Asshowninthefollowingfigure,theregioncanbedividedintotwopieces.Theleft
half,fromx=0tox=z2/2canbesubdividedintomanysmallrectangles,withheight
ytop-ybottomandwidthdx.Wesumtheareasoftheserectanglestofindthetotalarea
ofthelefthalf.Similarly,therightsideoftheplate,fromx=z2/2tozcanbe
subdividedintomanyrectanglesthathave
heightytop-0i.e.,ytopandwidthdx.
Thus,asafunctionofz,thetotalareaofa
singleplateis
z2
2
0
∫ (y
top
)
− y bottom dx +
∫
z
z2
2
y top dx (3)
Using(2)toreplaceyin(3),weget:
€
∫
z2
2
0
⎛
z2 ⎞
⎜ z − ⎟ dx +
2⎠
⎝
∫
z
z2
2
z − x dx (4)
Integrating(4),theareaofasingleplateis
€
z2 ⎛
2⎞
z
z
∫ 0 2 ⎜⎝ z − 2 ⎟⎠ dx + ∫ z 2 z − x dx
2
z2
2
z
⎛
⎛
z ⎞
x2 ⎞
= ⎜ z − ⎟ x + ⎜ zx − ⎟
2⎠ 0 ⎝
2 ⎠ z 2 ⎝
2
2
(5)
(6)
(7)
z3 z4
z2 z3 z4
− + z2 − − +
2 4
2 2 8
2
4
z
z
= −
2 8
=
Finally,weintegratethevolumeofeachplate
€
V=
1
8
∫
2
0
4z 2 − z 4 dz tofindthevolumeofthethree-dimensionalshape:
V=
€
1
8
∫
2
0
4z 2 − z 4 dz
2
1 ⎛ 4z 3 z 5 ⎞
1 ⎛ 32 32 ⎞
V= ⎜
− ⎟ = ⎜ − ⎟
8⎝ 3
5 ⎠0 8⎝ 3
5⎠
⎛ 1 1⎞
= 4⎜ − ⎟
⎝ 3 5⎠
8
=
15
€
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