四维球的表面积、体积求解

TheVolumeandSurfaceAreaofa4DSphereFeixiang,XUAug30,2016AbstractInthispaper,I'mtryingtocalculatethevolumeofspheresfromonedimensiontofourdimensionusingbasicintegralcalculationstepbystep.ThenIderivethesurfaceareaof4Dspherefromthevolume.1TheVolumeCalculation1.11DcaseHere,weconsideralinesegmentwithlength2R.Takeitsmidpointascenter.1.22DcaseIn2-dimensionalcase,asphereisacircle.Now,Itrytoexpandthesegmentintoacircle.First,weconstructtheper-pendicularlineofthelinesegmentthroughitsmidpoint.Thenwemovethesegmentalongtheperpendicularlineupward.Atthesametime,weshrinkthesegmenttokeepthedistancebetweenthesegment'sendpointandthecenterun-changed,whichisR.Obviously,whenmovingadistanceR,thesegmentshrinksintoapoint.Theareawhichtheshrinkingsegmenthasbeenscanningbecomesahalfofa2-dimensionalsphere,namelyasemi-circle.Connectthemovingsegment'sendpointandthecenterintoasegment,assumetheanglebetweenthenewsegmentandtheinitialsegmentis.variesfrom0to/2.Thenwecancalculatethecircle'sarea(whichcalledthevolumeofthe2Dsphere).1 Figure1Intheprocessofmoving,thelengthofsegmentis2Rcos,considerthatitmovesalittledistanced(Rsin),sotheareaitscansis2Rcos
d(Rsin).Thusthesemi-circle'sareaisZ22Rcosd(Rsin)=R2=2:0Sothecircle'sareaisR2,whichisthevolumeofthe2Dsphere.1.33DcaseInasimilarway,wecanexpandthecircleintoasphere.First,constructtheperpendicularlineofthecirclethroughitscenter.Thenmovethecirclealongtheperpendicularlineupwardandshrinkthecircletokeepthedistancebetweenthecircle'sedgeandthecenterunchanged,whichisRaswell.Thespacewhichtheshrinkingcirclehasbeenscanningbecomesahalfofa3-dimensionalsphere.Similarly,theradiusofthemovingcircleis2Rcos,considerthatitmovesalittledistanced(Rsin),sothespaceitscansis(Rcos)2
d(Rsin).Thusthehemisphere'svolumeisZ22(Rcos)2d(Rsin)=R3:03Sothesphere'svolumeis4R3.32 1.44DcaseFromthederivationabove,wecaneasilyderivethatthevolumeofthe4DhemisphereisZ241(Rcos)3d(Rsin)=2R4:034Sothe4Dsphere'svolumeis12R4.22TheSurfaceAreaCalculation2.1LowdimensionscaseInthetwodimension,the2Dsphere'svolume(thecircle'sarea)isR2,thesurfacearea(thecircle'sperimeter)is2R.WecaneasilyndoutthatthesurfaceareaisthederivativeofthevolumerelativetoR:d(R2)=dR=2R:Inthethreedimension,thesphere'svolumeis4R3,thesurfaceareais34R2.WecaneasilyndoutthatthesurfaceareaisthederivativeofthevolumerelativetoRaswell:432d(R)=dR=4R:3Bysuchanalogy,inthefourdimension,thesphere'svolumeis12R4,then2thesurfaceareashouldbe:12423d(R)=dR=2R:2Hereistheproofbelow.2.2Proof2.2.12DCaseNowwehaveacirclewithradiusR,increasetheradiusby
R,thenwegetanannulus.Theareaofannulusis(R+
R)2R2,when
Risverysmall,wecanconsidertheannulusasarectanglewithwidth
R.Thenthelengthoftherectangleis[(R+
R)2R2]=
R.When
R!0,thelengthistheperimeterofthecirclewithradiusR,whichis:(R+
R)2R2lim=2R:
R!0
RSotheperimeterofthecirclewithradiusRis2R.3 2.2.23DCaseNowwehaveaspherewithradiusR,increasetheradiusby
R,thenwegetasphericalshell.Thevolumeofsphericalshellis4(R+
R)34R3,33when
Risverysmall,wecanconsidertheannulusasacuboidwithheight
R.Thenthebaseareaofthecuboidis[4(R+
R)34R3]=
R.When33
R!0,thebaseareaisthesurfaceareaofthespherewithradiusR,whichis:4(R+
R)34R3lim33=4R2:
R!0
RSothesurfaceareaofthespherewithradiusRis4R2.2.2.34DCaseSimilarly,wecanprovethatthesurfaceareaofa4DspherewithradiusRis22R3.3ConclusionSpherewithradiusRVolumeSurfaceArea1D/2DR22R3D4R34R234D12R422R32Table1:Conclusion4InferenceInsection1,Isolvedthevolumeproblemfromlowerdimensiontohigherdimension,herecomesaninferenceofsolvingndimension:Z2V=V(cos)n1d(Rsin):nn104