微积分教学资料——chapter15

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1、Chapter 15 Multiple Integrals15.1 Double Integrals over Rectangles15.2 Iterated Integrals15.3 Double Integrals over General Regions15.4 Double Integrals in polar coordinates15.5*Applications of Double Integrals15.6*Surface Area15.1 Double Integrals over RectanglesVolumes and Double IntegralsA functi

2、on f of two variables defined on a closed rectangle2()Ra,bc,dx,yaxb,cydand we suppose that ()0f x,y The graph of f is a surface with equation().zf x,yLet S be the solid that lies above R and under the graphof f,that is,3()0()()sx,y.zzf x,y,x,yR(See Figure 1)Find the volume of SFigure 11)Partition:Th

3、e first step is to divide the rectangle R into subrectangles.21j-11j-1()ijiijiijRx,xy,yx,yxxx,yyyEach with areaAx y badcx,ymn 2)Approximation:ij()R*ijijx,yA thin rectangular box:Base:Height:ijR()*ijijf x,yWe can approximate byijVij()A*ijijVf x,y3)Sum:1111()Amnmn*ijijijijijVVf x,y 4)Limit:1111()Amnmn

4、*ijijijm,njiijVVlimf x,y A double Riemann sumDefinition The double integral of f over the rectangleR is11()()Amn*ijijm,nijRf x,y dAlimf x,y if this limit exists.The sufficient condition of integrability:f is continuous on Rfis integral on RTheorem1.Theorem2.f is bounded on R,and f is discontinuous o

5、nly on a finite number ofsmooth curvesfis integral on RNoteIf then the volume V of the solid that()0f x,y()RVf x,y dA.lies above that the surface is()zf x,yExample 1 If ()11 22Ra,bc,dx,yx,x,evaluate the integral21Rx dASolution22111422Rx dA15.2 Iterated Integralss a function of two variables that sco

6、ntinuous on the rectanglef iiPartial integration with respect to y defines a functionof x:()()dcA xf x,y dyWe integrate A with respect to x from x=a to x=b,we get()()bbdaacA x dxf x,y dy dx ()Ra,bc,dx,y axb,cyd()()bbdaacA x dxf x,y dy dx The integral on the right side is called an iterated integrala

7、nd is denoted by()bdacf x,y dydx Thus ()()()bdbdbdacacacf x,y dydxf x,y dy dxdxf x,y dy Similarly ()()()dbdbdbcacacaf x,y dxdyf x,y dx dydyf x,y dx Fubinis theorem If f is continuous on the rectangle()Ra,bc,dx,y axb,cydthen()()()bddbaccaRf x,y dAf x,y dydxf x,y dxdy More generally,this is true that

8、we assume that f is boundedon R,f is discontinuous only on a finite number of smooth curves,and the iterated integrals exist.The proof of Fubinis theorem is too difficult to includeIn our class.If f(x,y)0,then we can interpret the double integral()Rf x,y dAas the volume V of the solid S that lies ab

9、ove R and under the surface z=f(x,y).So()()()baRbdacf x,y dAVA x dxf x,y dy dx Or()()()dcRdbcaf x,y dAVA y dyf x,y dx dyExample22Evaluate the double integral(162)RxydA,02 02whereR(x,y)x,y22222200(162)(162)RxydAxydy dx 202032)3216(dxyyxy48)2380(202dxxSolutionExample SolutionExample Evaluate the doubl

10、e integralRxydxdy,()02 03where Rx,yx,y2300232300009Rxydxdyxydy dxxydy dxxdxydy SolutionSpecially If ()()()and R=a,b c,df x,yg x h yThen()()()bdacRf x,y dAg x dxh y dySome examples of type IA plane region D is said to be of type I if 12()()()Dx,y axb,g xyg xWhere and are continuous on a,b1()g x2()gxS

11、ome examples of type IIA plane region D is said to be of type II if 212()(y)(y)Dx,ycyd,hxhWhere and are continuous on c,d1(y)h2(y)hExample 4 Find the volume of the solid enclosedby the paraboloid and the planes223yxz.0,1,0zxyyxSolutionand above regionDdAyxV)3(22The solid lies under the paraboloid223

12、yxzSo the volume is1,10),(yxxyxDSuppose that D is a bounded region,the double integral of f over D is()Df x,y dA15.3 Double Integrals over General RegionsSuppose that D is a bounded region which can be enclosed in a rectangular region R.A new function F with domain R:()()()0()f x,yifx,y is in DF x,y

13、ifx,y is in R but not in D0D0DIf the integral of F exists over R,then we define the double integral of f over D by()()DRf x,y dAF x,y dASome examples of type IA plane region D is said to be of type I if 12()()()Dx,y axb,g xyg xWhere and are continuous on a,b1()g x2()gxEvaluate where D is a region of

14、 type I()Df x,y dA()()()0()f x,yifx,y is in DF x,yifx,y is in R but not in DA new function F with domain R:2()whereRx,yaxb,cydcontainsD2121()()()()()DRbdacbg(x)ag(x)bg(x)ag(x)f x,y dAF x,y dAF x,y dydxF x,y dydxf x,y dydx 1g(x)2g(x)If f is continuous on type I region D such that 212()()()Dx,yaxb,g x

15、yg xthen 2121()()()bg(x)ag(x)Dbg(x)ag(x)f x,y dAf x,y dydxdxf x,y dy Some examples of type IIA plane region D is said to be of type II if 212()(y)(y)Dx,ycyd,hxhWhere and are continuous on c,d1(y)h2(y)hIf f is continuous on type II region D such that then 2121()()()dh(y)ch(y)Ddh(y)ch(y)f x,y dAf x,y

16、dxdydyf x,y dx 212()(y)(y)Dx,ycyd,hxhExample 1 Evaluate ,where D is the region2()Dxy dAbounded by the parabolas 22andyxyx.2xy 2yx Solution),1,1(,)0,0(22 yxxy xyxx210Type I DdAyx)(2dxdyyxxx)(1022dxxxxxx)(21)(42102 14033 dxyyxxx2)21(10222xy 2yx Type II yxyy210DdAyx)(2dydxyxyy 1 0 22)(14033 Properties

17、of Double IntegralSuppose that functions f and g are continuous on a bounded closed region D.Property 1 The double integral of the sum(or difference)of two functions exists and is equal to the sum(or difference)of their double integrals,that is,Property 2 Property 3 where D is divided into two regio

18、ns D1 and D2 and the area of D1 D2 is 0.DDD f(x,y)g(x,y)dAf(x,y)dAg(x,y)dA.if is a constant.DDkf(x,y)dAkf(x,y)dA,k12DDDf(x,y)dAf(x,y)dAf(x,y)dA,Property 4 If f(x,y)0 for every(x,y)D,thenProperty 5 If f(x,y)g(x,y)for every(x,y)D,thenMoreover,since it follows from Property 5 that hence 0Df(x,y)dA.DDf(

19、x,y)dAg(x,y)dA.,),(,),(),(),(Dyxyxfyxfyxf everyfor DDDf(x,y)dAf(x,y)dAf(x,y)dA,DDf(x,y)dAf(x,y)dA.where S is the area of D.SdADProperty 6Property 7 Suppose that M and m are respectively the maximum and minimum values of function f on D,then where S is the area of D.Property 8(The Mean Value Theorem

20、for Double Integral)If f(x,y)is continuous on D,then there exists at least a point(,)in D such that where S is the area of D.f(,)is called the average Value of f on DDf(x,y)dAf(,)S,DmSf(x,y)dAMS,Example 2 Evaluate ,where D is the regionDxydAbounded by the parabolas 226andthe line 1yxyx-Solution22612

21、5 41yx(,),(,),yxType II 1312626xDxyx Type I215126xDxyx12DDDExample 3 Evaluate ,where D is the regionDsinxdAxbounded by the parabolas 2andthe line yxyxSolutionType Ixyo11201xD:xyx1sin1)sin(sin)(sinsinsin10102102dxxxxdxxxxxdyxxdxdAxxxxDyxyyD10:xyo1110sinsinyyDdxxxdydAxxType IIExample change the order

22、of integration yydxyxfdy2202),(xyo42yx2 2yx DyxyyD2202solution：xyxxD240 Dyydxdyyxfdxyxfdy),(),(2022 402),(xxdyyxfdxWe haveAn alternative description of D isxyo231yx 3yx2 yxy20,10 yxy 30,31xyxx 321,20 xyoyx2 xyoyx2 Example change the order of integration 12330010(,)(,)yydyf x y dxdyf x y dx solution：

23、Example Prove thatSolutionyxayD0,0:xy a0),(aaayxaxD,0:An alternative description of D isayaxbdxxfedy00)()(aaxbdxxfexa0)()()(where 0a We haveayaxbdxxfedy00)()(Daxbdxfe)()(aaxaxbdyxfedx0)()(aaxbdxxfexa0)()()(aaxaxbdydxxfe0)()(Soxy a0),(aaChapter 10 Parametric Equations andPolar Coordinates 10.3 Polar

24、coordinates10.3 Polar coordinates0Polar axisP(r,)rThe point o is called the poleThe point P is represented by theordered pair(r,)and r,are call-ed polar coordinates of Pis positive if measure in the cou-nterclockwise direction from the po-lar axis and negative in the clockwi-se directionThe connecti

25、on between polar and Cartesian coordinatesxrcosyr sinyrP(x,y)=p(r,)222yrxyarctanxExample Convert the point from polar to CartesiancoordinatesSolutionSince 2 andso321333the point is(1,3)in Cartesian coordinates.r,xrcoscos,yr sinso(2)3,Example Represent the point with Cartesian coordinatesSolution22Si

26、nce 1 and1 so21Because the point(1,1)lies in the fourth quadrant,7we can choose =or =.thus,one possible 447answer is(2,);another is(2,).44xy,yrxy,tanx-(1 1),in terms of polar coordinates.Example Identify the curve by finding a Cartesian equationfor the curve(1)2(2)r=3sinrSolution2222222222(1)2the Ca

27、rtesian equation is4(2)=3the Cartesian equation is3xysoxyyxysoxyxyyExample Find a polar equation for the curve representedby the given Cartesian equation 222(1)(2)4xyxyx Solution22(1)()the polar equation is (2)4the polar equation is4rcosr sinsorcsccotrrcossorcos D1212oxyxy 2422 ryx1122 ryx.21,40:rD2

28、22:ayxD xyaDarD020:xyxD2:22 xyD2ocos2cos22222rrrxyx.cos20,22:rDChapter 15 Multiple Integrals15.1 Double Integrals over Rectangles15.2 Iterated Integrals15.3 Double Integrals over General Regions15.4 Double Integrals in polar coordinates15.5*Applications of Double Integrals15.6*Surface Area15.4 Doubl

29、e Integrals in polar coordinatesA polar rectangle(r)R,arb,Evaluate the double integralRf(x,y)dA,(r)R,arb,whereThe“center”of the polar subrectangle 11(r)ijiijjR,rrr,has polar coordinates 1111()()22*iiijjjrrrThe area of is ijR22221111*111()2221()()2iiiiiiiiiiArrrrrrrrrri1111()()A(rr)r()mn*ijijm,nijRmn

30、*ijijim,nijbaf x,y dAlimf x,ylimfcos,sinrf r cos,r sinrdrd Change to polar coordinate in a double integralIf f is continuous on a polar rectangle R given by ,where ,then()()baRf x,y dAf r cos,r sinrdrd 0arb02SolutionExample 1 Evaluate ,where D is the region22()DxydAin the upper half plane bounded by

31、 the circles 22221 and4xyxy,the linesand0yxy.D1212oxyxy 0124R:,r.161512414)(4213404021322 rdrrddrdrdAyxD.)sin,cos()sin,cos()()()()(2121 rdrrrfddrdrrrf ADo)(1 r)(2 rDdAyxf),(1.If f is continuous on a polar region of the form then 12Dr,r AoD.)sin,cos()sin,cos()(0)(0 rdrrrfdrdrdrrf)(r2.If f is continuo

32、us on a polar region of the form then 0Dr,r DdAyxf),(.)sin,cos(.)sin,cos()(02020)(0 rdrrrfdrdrdrrfDoA)(r3.If f is continuous on a polar region of the form then 020Dr,r DdAyxf),(Solution：Example Find，。Dyxdxdye22222:ayxD xyaD020ra22222222000012()(1)02aaxyrrDraedxdyerdrdderdraee D is given bySoExample

33、Evaluate，Ddxdyyx22xyxD2:22 SolutionxyD2o:,02cos.22Dr2cos22220233222281632coscos339Dxy dxdyr drddd We haveImproper Integral(over the entire plane)2222222()()()limaxyxyxyaDRedAedydxedA where is the disk with radius and center the origin.aDa2222222()()()limaxyxyxyaSRedAedydxedA where is the square with

34、 vertics .aS(,)aa20Provethat.2xedx0022dxedxexx0022dyedxeyxdxdyeDyx222200rredrd 4DExample15.7 Triple Integrals f is defind on the rectanglar box B r,s()Ba,bc,dx,y,z axb,cyd,rzsDefinition The triple integral of f over the boxB is111()V(,z)lmn*ijkijkijkl,m,nijkBf x,y,z dlimf x,yV if this limit exists.f

35、 is continuous on Bfis integral on BTheorem1.Fubinis theorem If f is continuous on the rectanglar r,s()Ba,bc,dx,y,z axb,cyd,rzsthen()()sdbrcaBf x,y,z dVf x,y dxdydz box BExampleEvaluate the triple integral2Bxyz dV,Where B is the rectangular box given by()01 12 03Bx,y,zx,y,z SolutionThe triple integr

36、al over a general bounded region Ein three-dimensional space 12()()()()Ex,y,zx,yD,u x,yzu x,y21()()()()ux,yux,yEDf x,y,z dVf x,y,z dz dA A solid region E is said to be of type I ifthen1212()()(),()()Ex,y,z axb,g xyg x u x,yzu x,y2211()()()()()()bgxux,yagxux,yEf x,y,z dVf x,y,z dzdydx If the projecti

37、on D of E onto the xy-plane is a type I plane regionthenthenIf the projection D of E onto the xy-plane is a type II plane region1212()(y)(y),()()Ex,y,z cxd,hxhu x,yzu x,y2211(y)()(y)()()()dhux,ychux,yEf x,y,z dVf x,y,z dzdxdy ExampleEvaluate ,where E is the solid EzdVtetrahedron bounded by the four

38、planes 000 x,y,zand1xyz.tetrhi:drn Solution We have()01 0101Ex,y,zx,yx,zxyit is a type I region.15.8 Triple Integrals in Cylindrical and Spherical Coodinates1.Cylindrical Coodinates12()()()()Ex,y,zx,yD,u x,yzu x,yIfwhere D is given in polar coordinates by 12(r)()()D,r,hrh then2211()(rcos)()(rcos)()(

39、rcosrsin)rEhu,r sinhu,r sinf x,y,z dVf,zdzdrd ExampleA solid E lies within the cylinderbelow the plane2()020114Er,z,r,rzSolution221xy,4x,above the paraboloid221zxy.The density at any point is proportionalto its distance from the axis of the cylinder.Find the mass of E.We have22Bmkxy dV,Since the den

40、sity at any point is proportionalto its distance from the z-axis,the density function is22()=x,y,zkxy()x,y,zwhere k is the proportionality constant.Therefore,the mass isExample Evaluate the integral by changing to cylindrical coordinates.Solution222222311222211()xxyxxyxydzdydx The solid region has a

41、 much simpler description in cylindrical coordinates:222222()11112Ex,y,zx,xyx,xyzxy The solid region is22()02012Er,z,r,rzr2.Spherical Coodinateswhere E is a spherical wedge given by2()(cossin)Edbcaf x,y,z dVfsin,sin,cossinddd ()E,ab,cd If E is a general spherical region such as21(,)2()()(cossin)Edgc

42、g,f x,y,z dVfsin,sin,cossinddd 12()()()E,cd,g,g,thenExample Evaluate the integral by changing to spherical coordinates.Solution2222399222390zxxyxxyz dzdydx The solid region has a much simpler description in spherical coordinates:2222()339909Ex,y,zx,xyx,zxy The solid region is()020032Er,z,Example Use

43、 spherical coordinates to find the volumeSolution22zxyThe volume of E isThe solid is()02004Er,z,cosof the solid that lies above the cone andbelow the sphere222xyzz.EVdV15.9 Change of Variables in Mutiple IntegralsA transformation T from the uv-plane to the xy-plane()()T u,vx,ywhere()()xg u,vyh u,v()

44、()xx u,vyy u,vorT is a transformation,which means that and have continuous first-order partial derivatives.1CghA transformation T is a function whose doman and range are both subsets of If then the point is called the image of the point If no two points have the same image,T is called one-to-one.A t

45、ransformation T on a region S in the uv-plane.2.1111T()=(),u,vx,y11()x,y11().u,vT transforms S into a region R in the xy-plane called the image of S,consisting of the images of all points in S.If T is one-to-one transformation,then it has an inverse transformation from the xy-plane to the uv-plane.1

46、TExampleA transformation is defined by1sS is the triangular region with vertices0(01),vuSolution2xu,yv.(0 0),(1,1),(0,1).,The transformaton maps the boundary of S into the boundary of the image.the side is givenby whose image is0(01).yxThe side is given by ,whose image is1(01)uv2s1(01).xyFind the im

47、age of S.The side is given by ,whose image is(01)vuu3s2(01).yxyThe image of S is the region R bounded by the x-axis,the line and the parabola 1y 2.yxDefinition The Jacobian of the chansformation T given by()and()isxg u,vyh u,v()()xyx,yxyxyuuxyu,vuvv uvv Change of Variables in a Double IntegralSuppos

48、e that T is a transformation whose Jacobianis nonzero and that maps a region S in the uv-plane ontoa region R in the xy-plane.Suppose that f is continuouson R and that R and S are type I or type II plane regions.Suppose also that T is one-to-one,except perhaps on the boundary of S.then1C()()()()()RS

49、x,yf x,y dAf x u,v,y u,vdudvu,vExamplewhere R is the parallelogram encloseed by the linesSolution2024 31 38.xy,xy,xy,xythen the transformaton T from the uv-plane to xy-plane is23RxydA,xyLet 23uxyvxy12315555xuv,yuv Because the sides of R lie on the lines0841u,v,u,v2024 31 38.xy,xy,xy,xyand the image lines in the uv-plane areSo the region S is the rectangle with vertices(0,1)(4,1)(0,8)(4,8)and()04 18Su,vu,vThe Jacobine of T13()15521()555xyx,yuuxyu,vvv84102()3()15Rsxyux,ydAdudvxyvu,vududvv Thus