《高数双语》课件section 10.2.pptx

1、Section 10.21Riemann,Bernhard2Geometric meaning of the double integralcylindrical body in threedimensional space.Volume of a Cylindrical body01()(,)lim(,)nkkkdkf x y df then it can be think of aSuppose that(,)0,(,)(),zf x yx y 3Volume of a Cylindrical body01()(,)lim(,)nkkkdkf x y df Volume=4Calculat

2、ing Double Integrals Over Rectangles(,)f x yis continuous defined on a rectangular region()x y A():,.axbcyd Then we make a network of lines parallel These lines divide()into to x-and y-axes.We number.Ax y small pieces of area 12,.nAAA these in some order then(,)kkxyChoose a point,kA in each piece 1(

3、,).nnkkkkSf xyA ()(,)f x y d given by 5go to zero,the sums approach a limit called theCalculating Double Integrals Over RectanglesIf f is continuous throughout (),then as we refined the mesh width to y make both x andThe notation for it isdouble integral of f over ().()(,).f x y dxdy or01()(,)lim(,)

4、nkkkAkf x y df xyA Thus,()(,),f x y d As with functions of a single variable,the sums approach this limit nothat determine()are partitioned,c dmatter how the intervals,a bandas long as the norms of the partitions both go to zero.()(,)f x y dA 6Fubinis Theorem for Calculating Double IntegralsSuppose

5、that we wish to calculate the volume under the plane4zxyon the xy-plane.():02,01xy over the rectangular region 10()(4),yyA xxy dy If we denote the area of the cross-section at x asFor each value of x,we may calculate()A x,then the volume is()A xas the integral20().xxA x dx ()A x4zxywhich is the area

6、 under the curve in the plane of the cross section at x.421xyzxO4zxy()(4)Vxy d 7Fubinis Theorem for Calculating Double Integrals,x is held fixed and the integration takes place with()A xIn calculating Then the volume of the entire solid is respect to y.221000Volume()(4).xxA x dxxy dydx 12220007422yx

7、xxxyyyxydxx dx cubic units.2207522xxIf we just want to write instructions for calculating the volume,without carrying out any of the integrations,we could have written 221000Volume()(4)xxyxxyA x dxxy dy dx8()A yyFubinis Theorem for Calculating Double IntegralsWhat would have happened if we had calcu

8、lated the volume by slicingyzxO4zxy421with planes perpendicular to the yaxis?As a function of y,the typical cross-section area is()A y20(4)xxxy dx 22042xxxxxy 62.yTherefore10Volume()yyA y dy 10(62)yyy dy 12065.yy9Fubinis Theorem for Calculating Double Integralsis continuous throughout the rectangula

9、r(,)f x yIf():,axb,cydthen()(,)(,)(,).dbbdcaacf x y df x y dxdyf x y dydx region Fubinis theorem says that double integrals over rectangles can beThus,we can evaluatecalculated as iterated integrals累次积分累次积分.a double integral by integrating with respect to one variable at a time.10Evaluating a Double

10、 IntegralExample for()(,)f x y d Calculate():02,11.xy and2(,)16f x yx ySolutionBy Fubinis theorem,123012xxxx ydy 122()10(,)(16)f x y dx y dxdy 4 .11(216)y dy 12128yy Reversing the order of integration gives the same answer:21212221010(16)3yyx y dydxyx ydx 2220(13)(13)xxdx 202dx 4 .11Double Integrals

11、 over Bounded Nonrectangular Regions()(,)f x y d If ()is a region like the one shown in the xy-planebounded“above”and“below”by thein the left figure,and on the sides1()ygx curves 2()ygx and,xb we may again,xa by the linescalculate the volume by the methodWe first calculate the cross-section areaof s

12、licing.21()()()(,)y gxy gxA xf x y dy (,)f x yis continuous defined on a closed region().12Double Integrals over Bounded Nonrectangular Regionsfromto get the volume as an xb Then the integrate()A xxa toiterated integral21()()()(,).bbgxaagxVA x dxf x y dydx 13Double Integrals over Bounded Nonrectangu

13、lar Regions,then theSimilarly,if()is a region like the one shownin the right figure,bounded by theand1()xh y curves 2()xhy andyd the linesyc andvolume calculated by slicing is givenby the iterated integralVolume21()()(,).dhychyf x y dxdy 14Double Integrals over Bounded Nonrectangular Regionsbe conti

14、nuous on a region().(,)f x yLet with h1 and h2 continuous12,()(),cyd h yxhy2.If()is defined by,then ,c don21()()()(,)(,).dhychyf x y dAf x y dx dy 12,()(),axb gxygx1.If()is defined by with g1 and g2 continuous,then ,a bon21()()()(,)(,).bgxagxf x y dAf x y dy dx y type region x type region Double Int

15、egrals over Bounded Nonrectangular Regions15Example Findwhere()is the region bounded by the line x=1,y=0 and the parabola22()()Ixyd 2.yx 2yx x=1 Oxy21222200()()()xIxydxydy dx 61403xxdx 26105 Double Integrals over Bounded Nonrectangular Regions16Example Findwhere()is the region bounded by the line x=

16、1,y=0 and the parabola22()()Ixyd 2.yx 2yx x=1 Oxy11122220()()()yIxydxydx dy 3251220133yyydy 26105 17Finding VolumeExample Find the volume of the prism whose and1x andyx whose top lies in the plane(,)3.zf x yxybase is the triangle in the xy plane bounded bythe x axis and the linesSolution For any x b

17、etween 0 and 1,y may Hence,.yx may vary from 0y to210032y xyyyxydx 100(3)xVxy dydx 210332xxdx 1230322xxxx cubic unit.1 18Finding VolumeSolution(continued)When the order of integration is reversed,theintegral for the volume is121032xxyxxxydy 110(3)yVxy dxdy cubic unit.1 12053422yydy 13205222yyyyy 212

18、013322yyyydy The two integrals are equal,as they should be.19Double Integrals over Bounded Nonrectangular RegionsNoteabxyOcdx0y0If the region()is of both x-type and y-type,then from the Fubinis Theorem(Stronger From)we know2211()()()()()(,)(,)(,)dhybgxchyagxf x y dAf x y dx dyf x y dy dx 20Reversing

19、 The Order of IntegrationExample Sketch the region of integration for the integral 2220(42)xxxdydx and write an equivalent integral with the order of integration reversed.Solution the region bounded by the curves and2yx 2():2,02.xyxxTherefore,xOy(2,4)2yx 2yx 2yx between 2.x 0 x andTo find the integr

20、ating in the reverse order,we imagine a horizontal line passing from left to right through the region.21Reversing The Order of IntegrationSolution(continued)and leaves at The common value of these integrals is 8.4.y To include all such lines,we let y xOy(2,4)2yx 2yx 2yx The line enters at.xy The int

21、egral isrun from0y to40/2(42)yyxdxdy 22Finding AreaExample Find the area of the region()enclosed by the parabola2.yxand the line 2yx Solution xyO(1,1)(2,4)If we divide()into the regions 1 and 2 shown in the right figure,we may calculate the area as12AdAdA412.yydxdy 10yydxdy 2yx 2yx1 2 23Finding Area

22、Solution(continued)xyO(1,1)(2,4)On the other hand,reversing the orderof integration gives2221.xxAdydx 2yx 2yxThis result is simpler and is the onlyone we would bother to write down inThe area ispractice.2221xxAydx 221(2)xxdx square units.92 1 2 Example Find the area of the region()enclosed by the pa

23、rabola2.yxand the line 2yx 24Integrals in Polar Coordinatesis defined over a region()that is bounded(,)f Suppose that a function and by the continuous curves by the rays and2()g 1().g for every valueand120()()ggaSuppose also that Then()lies in a fan-shaped region Q definedof between and.andby the in

24、equalities.0a approach a limit as we refine the grid to make Integrals in Polar Coordinatesbe the center of the polar rectangle whose area is.kA We let(,)kk By“center”,we mean the point that lies halfway between the circular We then form the sumarc on the ray that bisects the arcs.1(,).nnkkkkSfA If

25、f is continuous throughout,this sum will 2k k Large sectorSmall sectorgo to zero.andThe areas of the circular sectors subtended by these arcs at the originare2k 21.22k Outer:21,22k Inner:26Integrals in Polar CoordinatesTherefore,kA area of large section-area of small section.k 22222kk 22k Then,1(,).

26、nnkkkkSf By the Fubinis theorem now says that the limit approached by thesesums can be evaluated by repeated single integrations with respect to and as21()()()(,)(,).ggfdAfd d 27How to Integrate in Polar Coordinateswhere()is the region determinedExample Evaluateby the inequalities2222.axyb22()(),xyd

27、ydx xyOab():2222.axyb,02.ab22220()()baxy dydxdd 44()2ba 28Finding Limits of IntegrationExample Find the limits of integration for integrating(,)f over and outside the1cos the region()that lies inside the cardioid circle 1.Solution right figure.Step 1:sketch.Show in theStep 3:the limits of integratio

28、nStep 2:the limits of integrationThen the integral is/21 cos/21(,).fd d 29Changing Cartesian Integrals into Polar IntegralsExample Convert the iterated integralto an iterated integral in polar coordinates.2112201()xxIdxf xydy Oyx1yx21yx11（）:211,01.xyxx1,1.sincos 11,0.sincos2 21122011210sincos()()xxI

29、dxf xydydfd 30Evaluating Integrals Using Polar Coordinateswhere()is the semicircular region222100 xyedydxed d Example Evaluatebounded by the x axis and the curve 21.yxSolution In Cartesian coordinates,the integral in question is a nonelementary integral and there is no direct way to integrate 22xye

30、with respect to either x or y.Substituting cos,sinxyand replacing d d enables us to evaluatedydxby the integral as22,xyedydx (1).2e 31Integrals in Polar CoordinatesThe boundary of the region can also be divided into two monodromy branches w.r.t.on the interval a,b.The boundary equations are abNote1()2()1()2().and21()()()()(,)(cos,sin)(cos,sin).baf x y dxdyfd dfdd The Cartesian integral then becomes