22HemisphereTakeStobetheunitupperhemispherede nedbyx2y2z21z0AccordingtotheorientationconventionthenormalntoSshouldbeorientedupwardpointingawayfromtheoriginThatmeansnxyz p x2y2z ID: 437325
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Stokes'theoremclaimsthatifwe\capo "thecurveCbyanysurfaceS(withappropriateorientation)thenthelineintegralcanbecomputedasZC~Fd~r=ZZScurl~F~ndS:Nowlet'shavefun!Moreprecisely,letusverifytheclaimforvariouschoicesofsurfaceS.2.1DiskTakeStobetheunitdiskinthexy-plane,denedbyx2+y21,z=0.Accordingtotheorientationconvention,thenormal~ntoSshouldbeorientedupward,sothatinfact~n=(0;0;1).ZZScurl~F~ndS=ZZS(�x;0;z�1)(0;0;1)dS=ZZS(z�1)dS=ZZS(�1)dS=�Area(S)= � :2.2HemisphereTakeStobetheunitupperhemisphere,denedbyx2+y2+z2=1,z0.Accordingtotheorientationconvention,thenormal~ntoSshouldbeorientedupward,pointingawayfromtheorigin.Thatmeans~n=(x;y;z) p x2+y2+z2=(x;y;z).LetusparametrizeSinsphericalcoordinates,2 2.3ParaboloidTakeStobethepartoftheparaboloiddenedbyz=1�x2�y2,x2+y21.Accordingtotheorientationconvention,thenormal~ntoSshouldbeorientedupward,pointingawayfromthez-axis.LetusparametrizeSusingxandyasparameters,withdomainofparametrizationDtheunitdiskx2+y21:~r(x;y)=(x;y;1�x2�y2):Thetangentvectorsare~rx=(1;0;�2x)~ry=(0;1;�2y)sothatanormalvectorisgivenbythecrossproduct~rx~ry=(2x;2y;1):Checktheorientation:Thisnormalvectorpointsupandawayfromthez-axis.Ok!Noneedto ipit.ZZScurl~F~ndS=ZZDcurl~F(~rx~ry)dxdy=ZZD(�x;0;z�1)(2x;2y;1)dxdy=ZZD�2x2+z�1dxdy=ZZD�2x2+(1�x2�y2)�1dxdy=ZZD�3x2�y2dxdy=�Z20Z10�3r2cos2+r2sin2rdrd=�Z20Z10�3r3cos2+r3sin2drd=�Z203 4cos2+1 4sin2d=�3 4()+1 4()= � :2.4ConeTakeStobethepartoftheconedenedbyz=1�p x2+y2,x2+y21.Accordingtotheorientationconvention,thenormal~ntoSshouldbeorientedupward,pointingawayfromthe4 Accordingtotheorientationconvention,thenormal~ntoS1shouldpointawayfromthez-axis.Thatmeans~n=(x;y;0).LetusparametrizeS1incylindricalcoordinateswith0z5and02:~r(z;)=(cos;sin;z):ZZS1curl~F~ndS=ZZS1(�x;0;z�1)(x;y;0)dS=ZZS1�x2dS=�Z20Z50(cos2)(1)dzd=�5Z20(cos2)d=�5:Thenormal~ntoS2shouldpointup.Thatmeans~n=(0;0;1).ZZS2curl~F~ndS=ZZS2(�x;0;z�1)(0;0;1)dS=ZZS2(z�1)dS=ZZS2(4)dS=4Area(S2)=4:Combiningthetwoparts,weobtainZZScurl~F~ndS=ZZS1curl~F~ndS+ZZS2curl~F~ndS=�5+4= � :3Closedsurfaces3.1ClosedcurvesRecallthefollowingfromchapter13.Denition3.1.Aclosedcurveisacurvethatendswhereitstarted.Inotherwords,aclosedcurveChasnoendpoints oatingaround;itformsaloop.Anotherwaytosaythisisthatitsboundaryisempty:@C=;.Ingeneral,theboundaryofacurve6 toeachpartyieldsZZScurl~F~ndS=ZZS1curl~F~ndS+ZZS2curl~F~ndS=ZC~Fd~r�ZC~Fd~r=0wheretheoppositesignscomefromtheorientationconvention.Infact,property(2)characterizescurls:Avectoreldisthecurlofsomevectoreldifandonlyifitsintegralalonganyclosedsurfaceiszero.Property(2)isnotasmysteriousasitseems.Thekeyisthatcurlsareveryspecial.Mostvectoreldsarenotthecurlofavectoreld.4Whichvectoreldsarecurls?Wehaveseenthatvectoreldsoftheformcurl~Fare(relatively)easytointegratealongsurfaces.Buthowdoweknowifagivenvectoreldisthecurlofsomevectoreld?Hereisanecessarycondition.Proposition4.1.Let~Fbeaniceenoughvectoreld(twicecontinuouslydi erentiable).Thenwehavediv(curl~F)0.Inwords:acurlisalwaysincompressible.Proof.Write~F=(F1;F2;F3)andabbreviatethepartialdi erentiationoperatorsas@1=@ @x;@2=@ @y;@3=@ @z:Thenthecurliscurl~F=~i~j~k@1@2@3F1F2F3=~i(@2F3�@3F2)�~j(@1F3�@3F1)+~k(@1F2�@2F1)=(@2F3�@3F2;@3F1�@1F3;@1F2�@2F1):Itsdivergenceisdivcurl~F=@1(@2F3�@3F2)+@2(@3F1�@1F3)+@3(@1F2�@2F1)=@1@2F3�@1@3F2+@2@3F1�@2@1F3+@3@1F2�@3@2F1=(@1@2F3�@2@1F3)+(@2@3F1�@3@2F1)+(@3@1F2�@1@3F2)0+0+0=0: 8