《高数双语》课件section 11.3.pptx

1、Section 11.312Surface AreaThe right figure shows a surface(S)lyingin a plane().above its“shadow”regionThe surface is defined by beneath it.(,).f x y zc If the surface a equationis smooth,we can define and calculate its area as a double integral over().into small rectangles()The first step is to part

2、ition the region(),S()we may approximate itDirectly above()lies a patch of surfaceof a tangent plane.()with a portion 3Surface AreaIn the right figure gives a magnified view showing the gradient(),of()and(,)f x y z at P and a unit vectorvector pthat is normal to xOy plane.The angle()v lie along the

3、edges of the patchThus,both in the tangent plane.uv is.pbetween The vector u andandf f are normal to the tangent plane.andIt is easy to see that|(u)p|.v 4Surface Area,|cos|ororthen,we haveitself is the area|uv|Since,1|uv|p|cos|cos|Then the area is cos0.provided 00limlim|cos|dd ()1.|cos|d 5Surface Ar

4、eaMoreover,if the surface can be expressed as a equationNotice that(,1).xyzzthen,one of the normal vector of this surface is is the angle between this vector and the positive direction of z axis,then221cos.1xyzz So,()1|cos|xySd 22()1xyxyzzd 22()1.xyxyzzdxdy Similarly,if the surface can be expressed

5、as2(,),(,)()R,xyzf x yx y we have2(,),(,)()R,yzxf y zy z 2(,),(,)()R,xzyf x zx z oror22()1yzyzSxxdydz 22()1xzxzSyydxdz 6Surface AreaExample Find the area of the surface cut from the bottom of the by the plane 220 xyzparaboloid 4.z Solution 2223/2001(41)12rd 22()1xySzzdxdy 22221441xyxydxdy 2220041rrd

6、rd 17 171.6 Finish.77exists,then we say the function f is integrable over the surface(S),and the limit is called the surface integral with respect to surface area offunction f over that is is a portion of a smoothSurface integrals with respect to surface areasurface(S),f(x,y,z)is a function of three

7、(),kSS variables defined onon the subsurface.,kkk and kMis any point the subsurface ,If the limit of sum01()(,)lim(,).nkkkkdkSf x y z dSfS (),SSuppose ()kS 01lim(,),nkkkkdkfS xyOzkS Element of areaof the surfaceis the area of()kS 8Surface integrals with respect to surface areaJust as the arc length

8、can be found by evaluate the line integral of the first type,the surface area in space also can be found by evaluate().SdS Area01()(,)lim(,).nkkkkdkSf x y z dSfS This type of integral is called a surface integral with respect to surfacearea or surface integral of the first type.9Integrating Over a S

9、urfaceDefinition Integral of f over S and Surface IntegralIf(s)is the domain of a surface defined by the equation(,),zz x y is a continuous function defined at the points of(S),then(,)f x y zand the integral of f over S is the integral22()()(,)(,(,)1.xySf x y zf x y z x yzzd dSThe integral itself is

10、 called a surface integral.Surface integrals behave like other double integrals,the integral of thesum of two functions being the sum of their integrals and so on.The domain additivity property takes the form12.nSSSSfdfdfdfdSSSS()(,)Sf x y z dS10Integrating Over a SurfaceExample Integrate(,)f x y zx

11、yz over the surface of the cube cut and1.z from the first octant by the planes1,1xySolution We integrate xyz over each of thesix sides and add the results.on0 xyz Since the sides that lie in the coordinate planes,theintegral over the surface of the cube reduces toxyzdS Cube surface.xyzdS Side CxyzdS

12、 Side AxyzdS Side B11Integrating Over a SurfaceSolution(continued)over the square region(,)1f x y zzSide A is the surface():01,01,xyxy Forin the xy plane.this surface and region,1,xyxydSdddxdy(1),xyzxyxyand1.4 xyxyzdSxyzd Side ASide A()xyxydxdy 1100 xydxdy 102ydy 12Integrating Over a SurfaceSolution

13、(continued)Symmetry tell us that the integrals of xyz over sides B and C are also1.4Hence,1113.4444xyzdS Cube surfaceFinish.School of Science,BUPTIntegrating Over a SurfaceExample Evaluate,dSz is a sphere()where 2222xyza(0).zhhacut from the top by a plane222.zaxySolution isThe equation of()on the xO

14、y plane is,xyDThe projection of()2222:(,)|.xyDx yxyahSince 222221,xyazzaxywe have222xyDdSadxdyzaxy 2ln.aah 22xyDardrdar 2222200arrdrdar Finish.13School of Science,BUPTIntegrating Over a SurfaceExample Evaluate 4xyz dSis bounded by()where 0,0,0 xyzrespectively.1xyzSolution The surface can be divided

15、into fourWe denote the part onparts.0,0,0 xyzand 1.xyzand 3 by 12,are both zero,thenThen,xyzdS xyz dSSince the integrand on the surface 123,4,andandxyz dS1234.xyzxyzxyzxyzdSdSdSdS3(1),xyDxyxy dxdy 14School of Science,BUPT15Integrating Over a SurfaceSolution(continued)is bounded by the lines xyDthen0

16、,0 xywhere1,xyand3(1)xyDxyxy dxdy xyzdS 3.120 11003(1)xxdxyxy dy 1221003(1)23xyyxxdx 310(1)36xxdx 123403(33)6xxxxdx Finish.School of Science,BUPT16Surface Integrals with respect to CoordinatesOriented surface A closed surface has an inside andUsually,a surface has two sides.a surface which is not cl

17、osed may have front and back or an outside;This kind of surface is called two side surface.upper and right sides.In fact,if we choose any point on P the surface and a normal vector at P,then,we move the point in the surface arbitrarily and return to the sameposition of P,the direction of normal vect

18、or may not change its directionfor two side surface.If the normal vector changes direction,the surface is called one sidesurface.School of Science,BUPT17The Fluxe.nIt is easy to see that the flux will change sign,if the direction of flux changes to the opposite direction.The negative sign means that

19、 the flow is flowing to the opposite side of be a field of velocity of a incompressible3v()()R)MMGSuppose that is a unit normal vector of flow and(S)be a directed surface in(G).If()Sthen the1 and the density of the fluid is flux Q passing through(S)to the given sideisv e.nQSenSchool of Science,BUPT1

20、8Surface Integrals with respect to Coordinates Suppose that(S)is a rectifiable oriented surface and a sideof(S)is given.Let the vector filed A(M)is given.Arbitrarily partition(S)into n small parts and denote the k-th parts by(),1,2,.kSknand form the scalar product(),kkMS Select any point 1A()e().nkk

21、kkkMMS ine()nkMwherekMis a unit normal vector to the surface at the point the direction of the given sides.Form the sum:A()e()(1,2,),kkkkMMSknIf for any partition of the surface(S)and any selection of(),kkMS School of Science,BUPT19Surface Integrals with respect to Coordinates01()A()SlimA()e().nknkk

22、dkSMdMMS then the integral can be written aseS,ndSd If we denotewhere d is the maximum of the 0,d the limit of the sum exists as then this value of the limit is called(),1,2,kSkndiameters of allthe surface integral of the second type or the surface integral with respect to coordinates of the vector

23、field A(M)over the(S)and is denoted by 01()A()elimA()e().nnknkkdkSMdSMMS School of Science,BUPT20Surface Integrals with respect to CoordinatesThe of the vector field A(M)over the(S)and is denoted by 01()A()SlimA()e().nknkkdkSMdMMS 01()A()elimA()e().nnknkkdkSMdSMMS orwhere en is a unit normal vector

24、to the directed surface.School of Science,BUPT21Surface Integrals with respect to CoordinatesIf we denote A()(,),(,),(,),MP x y z Q x y z R x y z e()cos,cos,cos,nkkkkMM or ()()()A()S,cos,cos,cos,SSSMdP x y zQ x y zR x y zdSP x y z dydzQ x y z dzdxR x y z dxdy ,coskkkkkRS 01lim,cos,cosnkkkkkkkkkkdkPS

25、QS Then 01()A()SlimA()e()nknkkdkSMdMMS Sd School of Science,BUPT22Properties of Surface Integrals of the Second TypeLet us suppose that all the integrals of the second type exist.Property 1 If we change the side of the surface of integration then thesign of the integral will change,that is()()A dSA

26、dS.SS Property 2 If the surface(S)is divided into two pieces(S1)and(S2),then12()()()A dSA dSA dS,SSS12()()(),SSS where all surfaces of integration are on the same side.andSchool of Science,BUPT23Properties of Surface Integrals of the Second TypeProperty 3 Suppose the domain(V)in space bounded by the

27、 oriented(V1)and(V2)are denoted by(S1)and(S2)respectively.Then surface located inside the domain(V)and the boundaries of the domainsclosed surface(S)is divided into two domains(V1)and(V2)by another12()()()A dSA dSA dS,SSSwhere the surfaces in the above equality are both outside or both inside.School

28、 of Science,BUPT24Calculate Surface Integrals of the Second TypeSince ()()A()S,cos,cos,cosSSMdP x y zQ x y zR x y zdS (),cosSP x y z dS (),cosSQ x y z dS (),cos.SR x y z dSIf the surface (S)can be expressed by an then it is easy to(,),zz x y explicit function ()(),cos,.xyxySR x y zR x y z x yd dSand

29、 then we have ,xy obtain its projection School of Science,BUPT25Calculate Surface Integrals of the Second TypeSimilarly,if the equation of(S)has the form(,),yy x z and(,)xx y z we have respectively ()(),cos,.xzxzSQ x y zQ x y x zz d dS ()(),cos,yzyzSP x y zP x y zy z d dSandare the projection of(S)o

30、nto the yOz plane xz where yz and zOx plane.The sign of the integration depends on the angle between the normal vector and the positive direction of the axes.dS()()A()S,cos,cos,cosSSMdP x y zQ x y zR x y z (),xyxyR x y z x yd (),yzyzP x y zy z d (),.,xzxzQ x y x zz d School of Science,BUPT26Calculat

31、e Surface Integral of the Second TypeExample Calculate the integral 222,x dydzy dzdxz dxdy is the outside surface of the rectangle where ,0,0,0.x y zxaybzcSolution We divide the surface into six parts:6:0 0,0yxazcThe left 1:0,0zcxaybThe top 2:0 0,0zxaybThe bottom 3:0,0 xaybybThe front 4:0 0,0 xybybT

32、he back 5:0,0ybxazcThe right School of Science,BUPT27Calculate Surface Integral of the Second TypeSolution(continued)areas of the projection of the other 34,Exceptparts onto yOz plane are all zero,then2z dxdy and34222,x dydzx dydzx dydz2.c ab Similarly,we have222()()0 xyxyx dydza dydzdydz 2.a bc and

33、2y dzdx 2,b ac School of Science,BUPT28Calculate Surface Integral of the Second TypeSolution,where2 Divide into2211:1,zxyand1 2222:1,zxythen Example Find is the outside of a sphere,where0,0.xycut by 2221xyzxyzdxdy the unit normal vectors are shown as in right then we have12,xyzdxdyxyzdxdyxyzdxdyfigu

34、re,School of Science,BUPT29Calculate Surface Integral of the Second TypeSolution(continued)132200sin21drr dr 222211xyxyDDxyzdxdyxyxy dxdyxyxydxdy 2221xyDxyxy dxdy 222sincos1xyDrr rdrd 221.1515 School of Science,BUPT30Calculate Surface Integral of the Second Typeand the 0,(0).zzH H222xyRExample Evalu

35、ate()Swhereis outside 22()(),SIxydydzzdxdy of the surface of the solid bounded by the cylinderplanes xyzHRO1()S3()S2()SDivide(S)into three parts:the cylinder(S1),the base(S2)and the top(S3).Divide(S1)two parts:the front pieces(S11),the back pieces(S12).31(S2)(S1)OCalculate Surface Integral of the Se

36、cond TypeExample Find the flux of the vector field r=x,y,z at the(1)(S)is the outside of the sphere 2221.xyz,wheredirected surface(S)and the plane1.z (2)(S)is the outside of the surface of the solid bounded by the cone 22zxyO(S)32ReviewSurface integrals with respect to surface areaSurface integrals with respect to coordinatesThe relationship between two kinds of surface integrals