《高数双语》课件section 11.4.pptx

1、Section 11.41Greens Theorem2Theorem Suppose there is a closed bounded domain(1)(,),(,)().P x y Q x yC bounded by a piecewise smooth simple curve(C),and functions 2R Then the following relation holds:,CQPdP x y dxQ x y dyxy indicates that the integration is C where in the positive direction of(C).3Fl

2、ux Density at a Point:DivergenceThis idea is that theWe need a new ideas for Greens Theorem.flux density of a vector field at a point,which in mathematics is calledWe obtain it in the following way.the divergence of the vector field.is the velocity field of a fluidF(,)(,)i(,)jx yP x yQ x ySuppose th

3、at flow in the plane and that the first partial derivatives of P and Q arebe a point in(W)andcontinuous at each point of a region(W).(,)x yLetlet A be a small rectangle as in the figure.The sides of the rectangle,parallel to thecoordinate axes.4Flux Density at a Point:DivergenceThe rate at which flu

4、id leaves the rectangle across the bottom edge is approximate F(,)(j)(,).x yxQ x yx Similarly,we haveExit Rates:Top:Left:F(,)(j)(,)x yyP x yy Bottom:F(,)(j)(,)x yxQ x yx Right:F(,)i(,)xx yyP xx yy F(,)j(,)x yyxQ x yyx 5Flux Density at a Point:Divergence (,)(,)PP x yyP x yxxyy Top and Bottom:Right an

5、d Left:(,)(,)QQ x yyQ x yxyxy Then,we have Flux across rectangle boundaryrectangle areaPQxythereforeFlux across rectangle boundary,PQx yxy flux density of F at the point(,)x ydiv F.6The Flux Divergence or Normal Form of Greens TheoremTheorem The outward flux of a field FijPQacross a simple over the

6、region()div Fclosed curve C equals the double integral of enclosed by C.()()()F n.CCPQdsPdyQdxdxdyxy Divergence integral Outward fluxThe divergence form of Greens Theorem in the plane states that thenet outward flux of a vector field across a simple closed curve can becalculated by integrating the d

7、ivergence of the field over the regionenclosed by the curve.7Divergence in Three DimensionsThe divergence of a vector field F(,)2ijk.x y zxzxyzis the scalar functionF(,)(,)i(,)j(,)kx y zP x y zQ x y zR x y zExample Find the divergence of Solution The divergence of F isF(2)()()xzxyzxyz21.zxdiv FF.PQR

8、xyz Finish.8Operational Rules for Divergences There are some useful rules for the divergences:8.(A)ACCwhere C is a constant.(AB)BAAB.or1.div(A)div ACC 2.div(AB)div Adiv Bwhere u is a derivable scalar function.3.div(A)div Agrad,uuu Aorrot(rotA)grad(divA)A 4.div(AB)B rot AA rot B5.div(rotA)0 6.rot(gra

9、d)0u is called Laplace expression.222222xyz where 7.div(grad),uu Gauss Formula9Theorem Suppose that a region(1)(,),(,)().Q x y z R x y zCV a piecewise smooth closed simple surface(S),and functions 3RV Then the following relation holds:()(),VSPQRdVPdydzQdzdxRdxdyxyz points to the outside of (V).Swher

10、e the normal vector ofis bounded by a piecewise(,),P x y zGauss Formula10Proof We only prove()()VSRdVRdxdyz 21(,)(,)()()12(),(,),(,)xyxyzx yzx yVDDRRdVdz dzzR x y zx yR x y zx yd 123()()()()()SRdxdyRdxdy where the normal vector is downwards on 1,upwards on 2,and outwards on 3.Gauss Formula11Proof(co

11、ntinued)We only prove()()VSRdVRdxdyz 1231()()()2()(),(,)0,(,)xyxyDDRdxdyR x y zx y dRdxdyRdxdyR x y zx y d ()12(),(,),(,)xySDRdxdyR x y zx yR x y zx yd ()()VSRdVRdxdyz Then,Gauss Formula12()()VSPQRdVPdydzQdzdxRdxdyxyz FijkPQRThe flux of a vector field across a closed oriented surface S in the direct

12、ion of the surfaces out ward unit normal field n is()()FSF nSSdd SPQRdivFFxyz ()()()FSF nF.SSVdddV SDivergenceDivergence TheoremTheorem Divergence TheoremThe flux of a vector field F across a closed oriented surface S in the direction of the surfaces out ward unit normal field n equals the integral

13、over the region(V)enclosed by the surface:Fof()()()FSF nF.SSVdddV SThe Divergence Theorem says that undersuitable conditions,the outward flux of a vector field across a closed surface(orientedoutward)equals the triple integral of thedivergence of the field over the region enclosedby the surface.Page

14、 361 Gauss formulaGauss formula()()()FSF nF.SSVdddVS Flux of a vector field F across a closed oriented surface S(,)FP Q R()()2)()SVPQRPdydzQdzdxRdxdydVxyzwhere the normal vector of(S),n,points the outside of(V).()()1)(,)()SVPQRP Q R ndSdVxyzwhere the normal vector of(S),n,points the outside of(V).()

15、()3)(coscoscos)()SVPQRPQRdSdVxyz,(,)x y zare the angles between the outward normal vector aton the surface and the three axes.15Gauss FormulaExample Evaluate33322()(),SIx dydzy dzdxzxydxdy where(S)is1)the outside of the sphere 2222;xyzRSolution()()SVPQRPdydzQdzdxRdxdydVxyz 2)the upper side of the up

16、per hemisphere 222.zRxy 33322()222()()333SVIx dydzy dzdxzxydxdyxyzdV 1)2220003sinRddrrdr512.5R 16Gauss FormulaSolution()()SVPQRPdydzQdzdxRdxdydVxyz 2)(S)is the upper side of the upper hemisphere 222.zRxy1133322()33322()33322()()()()upupSSSSIx dydzy dzdxzxydxdyx dydzy dzdxzxydxdyx dydzy dzdxzxydxdy 2

17、)425350066.552RRRddR SupS1 1122222()()333(0)VSxyzdVxydxdy 17Gauss Formulaand(S)is the 23(4)4.aaaExample EvaluateFijkxyzwheresphere2222.xyzaSolution(I)The outer unit normal to(S),calculated from the gradient of 2222(,),f x y zxyzaisijk.xyza 2222ijkn4()xyzxyz Therefore,.ad SF nd S222xyzda S2ada SHence

18、,()F n,Sd S()F nSd S()Sad S()Sad S18Gauss FormulaSolution(II)The divergence of F is3.F()()()xyzxyzSo,by Divergence Theorem,we have34.a ()()F nFSVddV S()3VdV()3VdV 3433a and(S)is the Example EvaluateFijkxyzwheresphere2222.xyza()F n,Sd S19Gauss FormulaExample Evaluate()()(),SIxy dxdyyz xdydz where(S)i

19、s the boundary surface of the region(V)=221,03.xyzSolution()()FSSSdPdydzQdzdxRdxdywe have(),0,Pyz x QRxySince,0,0.PQRyzxyzTherefore,by the Divergence Theorem,we obtain()()VIyz dxdydz()(sin)Vrz rdrd dz213000(sin)ddrrz rdz 9.2 20Stokes TheoremStokes Theorem says that,under conditions normally met in p

20、ractice,the circulation of a vector field around the boundary of an orientedsurfaces unit normal vector field n equals the integral of the normalsurface in space in the direction counterclockwise with respect to thecomponent of the curl of the field over the surface.The orientation of the bounding c

21、urve C gives it a right handed relation to the normal field n.Circulation DensityThe positive orientation of the circulationdensity for the plane is the counterclockwiserotation around the vertical axis,looking downward on the xy plane from the tip of the vertical unit vector k.The circulation value

22、 is actually the k component of a general circulation vector,called the curl of the vector field F.curl Fk.QPxyCirculation Density at a Point:The k Component of CurlCirculation DensityCirculation Density:Curl As we had seen,the k component of the circulation density or curl of is the scalar FijPQa t

23、wo dimensional field(,)x yat a point In three dimensions,the circulation around a point P.QPxy quantity This vector is normal to the planein a plane is described with a vector.of the circulation and points in the directionthat gives it a right hand relation to thecirculation line.Circulation Density

24、It turns out that the vector of greatest circulation in a flow with velocity isFijkPQRfield curl Fijk.RQPRQPyzzxxyIf we let ijk,xyz iscurl Fthe F:FijkxyzPQR ijkRQPRQPyzzxxycurl F.Circulation Density:Curl 24Circulation DensityExample Find the curl of 22F()i4 jk.xyzxSolution curl FF ijkxyzPQR 2224ijxz

25、xyxyzzx 24kzxyxy4i2 jk.x Stokes FormulaSchool of Science,BUPT25Theorem Suppose a region()Cand functions 3RG is a piecewise smooth and directed simple closed curve in The above formula is called Stokes formula.(1),().P Q RCG(),G()Sis a piecewise smooth oriented surface in()Gand the direction of(C)and

26、 the normal vector of the surface(S)whose boundaryis(),Caccord with the right hand rule.Then()()CSRQPRQPPdxQdyRdzdydzdzdxdxdyyzzxxy Stokes FormulaTheorem Stokes TheoremThe circulation of a vector field FijkPQRthe direction counterclockwise with respect to thearound the boundary C of a oriented surfa

27、ce S insurfaces unit normal vector en equals the integral of F nover surface S.nFrFCSde dS Curl IntegralCounterclockwise circulationNote If two different oriented surface 2Shave the same boundary 1SandC,their curl integral are same.Stokes,George(1819-1903),Irish mathematician and physicist()()CSRQPR

28、QPPdxQdyRdzdydzdzdxdxdyyzzxxy 27Greens Theorem versus Stokes TheoremIf C is a curve in the xy plane,oriented counterclockwise,and R isanddSdxdy is the region in the xy plane bounded by C,then Fr.CQPddxdyxy Under these conditions,Stokes equation becomes FnFk.QPxy Greens TheoremStokes:Green:28Find Lin

29、e Integral by Stokes Theorem()dydzdzdxdxdy Example Findby the three coordinates plane,and its positive direction 1xyzhas right hand relation with the normal up vector.Solution I)By Stokes Theorem,we havezdxxdyydz zdxxdyydz ()33.2xyd By the symmetry,we havezdxxdyydz ()dydzdzdxdxdyxyzzxy xyzwhere is t

30、he boundary of the triangle cuts from the planeFind Line Integral by Stokes Theorem29111(1,1,1)(,)333dS Solution II)zdxxdyydz ()33.2xydxdy Then(,)Fz x y(1,1,1)ijkFxyzzxyxyznFrFCSddSe By Stokes Theorem,we have111(,)333ne zdxxdyydz ()()333xydSdxdy 30ReviewGauss Formula and DivergenceStokes Formula and Curl