《高数双语》课件section 5-5.pptx

1、Applications of Definite IntegralsMethod of elements for setting up integral representationsWhat kind of quantities can be calculated by definite integrals?How can we set up the integral representations?2QuestionsMany quantities we want to know in science and technology can be calculated by definite

2、 integrals.Method of elements for setting up integral representations3We had seen that the area of trapezoid with curved top,the mass of a stick and the displacement of a movement along a straight line can all be expressed by definite integrals.They both have the following properties.1)They are all

3、distributed non-uniformly but continuously on an interval a,b2)These quantities are all additive,that is,the total quantity on the interval a,b equals the sum of all those local quantities distributed on the subintervals of a,b.In general,a quantity with these two characteristics may be calculated b

4、y a definite integral.Method of elements for setting up integral representations4In order to calculate the whole area,we will first find the approximate value of every local area.To do this,we need two steps:(1)“partition”Divide a,b into many very small subintervals;(2)“homogenization”Regard the loc

5、al area distributed on each small interval x,x+x as a rectangle with height f(x).Using multiplication for the uniform distribution,we have()Af xxMethod of elements for setting up integral representations5After we found the approximation value of every local area,we can easily obtain the precise valu

6、e of the whole area A by the other two steps:(3)“summation”()Af xx (4)“precision”0lim()()bxaAf xxf x dx Because the function f(x)is continuous on a,b,the limit of the sum is just the definite integral.Method of elements for setting up integral representations6So the local area is just the increment

7、of this function.Since A By the geometric meaning of the definite integral with varying upper limit,we known that the area of the definite integral with the curve y=f(x)and on the Interval a,x is expressed by ,namely()xaf t dt()()xaA xf t dt ()(),xadAdf t dtf xdxdx the approximate value of the local

8、 area ,is actually the differentialA()f xx(1)of the function(1).Method of elements for setting up integral representations7In general,the procedures may be simplified into the following two steps:(1)Find the element Find the approximate value of the local required quantity,dQ,on the subinterval,x xd

9、x()QdQf x dx(2)Setup the integration For the differential ,write down the corresponding definite integral over the interval ,we obtain()f x dx,a b()baQf x dx ()f x dxInfinitesimal element of the integral or element of the integralThe procedure summarized by the above two steps is called the element

10、method of integration.The Area of a Plane Region8Find the area A of the region enclosed by the parabolas21yxand2.7yxSolutionFrom the system2217yxyx we can easily obtain the abscissae of the intersections of the two parabolas,2.x It is easy to see that the area A is distributed continuously but non-u

11、niform on the interval-2,2,and is additive.The Area of a Plane Region9Solution(continued)By the method of element,we calculate A by two steps:(1)Find the element222(7)(1)2(4).MNxxxPartition the interval-2,2 and consider the subinterval x,x+dx.On this subinterval,the area distributed non-uniformly ma

12、y be regarded approximately as uniform,that is a rectangle with heightFind the area A of the region enclosed by the parabolas21yxand2.7yxThe Area of a Plane Region10Solution(continued)222(7)(1)2(4).MNxxxThus,we obtain the element of area:22(4).dAx dxFind the area A of the region enclosed by the para

13、bolas21yxand2.7yx(1)Find the element(2)Setup the integralThe Area of a Plane Region11Solution(continued)22(4)dAx dxover the interval-2,2.222220642(4)4(4).3Ax dxx dx The whole area A is just the integral of the elementSoFinish.Find the area A of the region enclosed by the parabolas21yxand2.7yxThe Are

14、a of a Plane Region12Find the area A of the region enclosed by the parabolayx and the straight lines,1.yx y Solution This area A may be regarded as a quantity which is distributed non-uniformly on the interval 0,1 on the y-axis.To find the element of the area ,partition the interval 0,1,and consider

15、 the subinterval,regarding as the area of a rectangle with widthA()MNyy The Area of a Plane Region13Solution(continued)Hence,the total area is107().6Ayy dy Then the element of the area is()dAyy dyFinish.Find the area A of the region enclosed by the parabolayx and the straight lines,1.yx y The Area o

16、f a Plane Region14Find the area A of the region enclosed by the cardioidSolution(1 cos),(0).aaBy the symmetry of the graph of the cardioid,it is enough to calculate the area located in the upper half-plane.Since the equation of the cardioid is expressed in polar coordinates,the area may be regarded

17、as a distribution on the interval 0,and the distribution is non-uniform.The Area of a Plane Region15Solution(continued)To calculate the area,we partitionthe interval 0,and consider thesubinterval ,d On this interval,we regard asA constant .Thus the elementof area is()22211()(1cos).22dAdadFind the ar

18、ea A of the region enclosed by the cardioid(1 cos),(0).aaThe Area of a Plane Region22211()(1cos).22dAdad16Solution(continued)Thus,the total area is2012()2Ad 220(1cos)ad 23.2a Finish.Find the area A of the region enclosed by the cardioid(1 cos),(0).aaThe Area of a Plane Region22404sin2tadt 17yxOFind

19、the area A of the region enclosed by the curve(sin),0,02;(1 cos)xa ttatyat and the x-axis.Solution(2)(0)()xxAf x dx 20()()y t x t dt 20(1cos)(1cos)atat dt 22424420028sin8sinsintuauduauduudu 242016sinaudu 23 116()4 2 2a 2.3 a 21|()|()ttAy tx t dt 18The Arc Length of a Plane CurveTo study motion along

20、 a space curve,we need to have a measurable length along the curve.(),xbyf x a xyOM0Mba0?M M 19The Arc Length of a Plane Curve0?M M 11iiiiiMMMML 011niiiM MMM xyOMi-1Mi01max011liminniiiLiiM MMML(1)Find the element of the arc lengthxdx ydy 22221()()()()iiiLMMxydxdy 22()()dsdxdy221()1()dydxdxdxdsfx 20T

21、he Arc Length of a Plane Curve(2)Set up the integrationxyOMi-1Mixdx ydy 221()1()dydxdxdxdsfx 21()badxsfx 21The Arc Length of a Plane CurveArc Length for a smooth curve in the case of the rectangular coordinates(),yf x axb1)If the equation of the plane curve is then the length of the arc is 2211().bb

22、aadysdxfxdxdx(),xg y cyd2)If the equation of the plane curve is then the length of the arc is 2211().ddccdxsdyg ydydy 22()()dsdxdyThe Arc Length of a Plane Curve22 Find the arc length of the plane curve2(0,02);2xypxpp2211bdacdydxsdx or sdydxdy23The Arc Length of a Plane Curve Find the length of an a

23、rc of the plane curve211ln,(1).42xyyye2211bdacdydxsdx or sdydxdy24The Arc Length of a Plane CurveArc Length for a Smooth Curve in the case of the parametric form(),()(),xttyt If the equation of the plane curve is then the length of the arc is 22()().sttdt 2222()()()()dsdxdyt dtt dt where(t)and(t)are

24、 continuously derivable,and(t),(t)0,25Example Find the length of an arc of the cycloidThe Arc Length of a Plane Curve(sin),(1cos);(02).xa ttyatt 22()()sttdt (sin),(1cos),(02)xa ttyatt 26The Arc Length of a Plane CurveArc Length for a Smooth Curve in the case of the polar coordinates22()()dsdxdyIf is

25、 expressed in polar coordinates,(),then the length of the arc is 2222()()dxdysdddd ()cos()()sin()xy 27The Arc Length of a Plane CurveExample Find the length of an arc of the plane curve(1cos);().awhole arc 22()()Ld The Volume of a Solid28Consider a solid like the one shown in the following figure.At

26、each the cross section of the solid is a region whose area is a knowncontinuous function .Express the volume of this solid by an integration.,xa b()A xSolutionWe partition a,b so that the solidcan be cut into many slices by perpendicular planes through thePoints of the partition.Consider the slice c

27、orresponding tothe subinterval .We havethe element of volume,x xdx()dVA x dx(1)Find the element of the volume.The Volume of a Solid29Solution(continued)By the method of elements we obtain the volume of the given solid as follows()baVA x dx (2)Setup the integration.Consider a solid like the one shown

28、 in the following figure.Ateach the cross section of the solid is a region whose area is a knowncontinuous function .Express the volume of this solid by an integration.,xa b()A xThe Volume of a Solid30How to Find Volumes by the Method of SlicingStep 1.Sketch the solid and a typical cross section.Ste

29、p 2.Find a formula for A(x).Step 3.Find the limits of integration.Step 4.Integrate A(x)to find the volume.The Volume of a Solid31A curved wedge is cut from a cylinder of radius 3 by two planes.One plane is perpendicular to the axis of the cylinder.The second plane crossesthe first plane at a 45o ang

30、le at the center of the cylinder.Find the volume of thewedge.SolutionWe draw the wedge and sketch a typical cross section perpendicular to the x-axis.The cross section at x is a rectangle of area2()(height)(width)()(2 9)A xxx(1)A sketch.(2)The formula for A(x).229 units squared.xxThe Volume of a Sol

31、id32Solution(continued)(4)Integrate to find the volume.320()29baVA x dxxx dxThe rectangles run from to 0 x 3.x (3)The limits of integration.323/23/2022(9)0(9)33x 18 unit cubed.Finish.A curved wedge is cut from a cylinder of radius 3 by two planes.One plane is perpendicular to the axis of the cylinde

32、r.The second plane crossesthe first plane at a 45o angle at the center of the cylinder.Find the volume of thewedge.The Volume of a Solid33The most common application of the method of slicing is to solids of revolution旋转体旋转体.Solids of revolution are solids whose shapes can be generated by revolving p

33、lane Regions about axes.The only thing that changes when the cross sections are circular is the formula for the area A(x).The Volume of a Solid34The typical cross section of the solid perpendicular to the axis of revolution is a disk of radius R(x)and area22()(radius)().A xR xFor this reason,the met

34、hod is often called the disk method.The Volume of a Solid35The region between the curve ,and thex-axis is revolved about the x-axis to generate a solid.Find its volume.,04yxxThe Volume of a Solid36SolutionWe draw figures showing the region,a typical radius,and the generated solid.The volume is4220()

35、baVR xdxxdx40 xdx 4220(4)22x8 units cubed.Finish.The region between the curve ,and thex-axis is revolved about the x-axis to generate a solid.Find its volume.,04yxxThe Volume of a Solid37Find the volume of the solid generated by revolving a regionbounded by and the lines ,about the line .yx 1y 4x 1y

36、 The Volume of a Solid38SolutionWe draw figures showing the region,a typical radius,and the generated solid.The volume is2421()1baVR xdxxdx423/212223xxx 7 units cubed.6 Finish.Find the volume of the solid generated by revolving a regionbounded by and the lines ,about the line .yx 1y 4x 1y The Volume

37、 of a Solid39How to Find Volumes for Circular Cross Sections(Disk Method)Step 1.Draw the region and identify the radius function R(x).Step 2.Square R(x)and multiply by.Step 3.Integrate to find the volume.22()(radius)().A xR xThe Volume of a Solid40Find the volume of the solid generated by revolving

38、a regionbetween the y-axis and the curve ,about the y-axis.2/xy 14yThe Volume of a Solid41SolutionWe draw figures showing the region,a typical radius,and the generated solid.The volume isFinish.24212()baVR ydydyy4214dyy 4113444y 3 units cubed.Find the volume of the solid generated by revolving a reg

39、ionbetween the y-axis and the curve ,about the y-axis.2/xy 14yThe Volume of a Solid42If the region we revolve to generate a solid does not border on or cross the axisof revolution,the solid has a hole in it.The cross sections perpendicular to theaxis of revolution are washers instead of disks.The di

40、mensions of a typical washer areOuter radius:()R xInner radius:()r xThe washers area is 2222()()()()().A xR xr xR xr xThe Volume of a Solid43How to Find Volumes for Washer Cross SectionsStep 1.Draw the region and sketch a line segment across it perpendicular to the axis of revolution.When the region

41、 is revolved,this segment will generate a typical washer cross section of the generate solid.Step 2.Find the limits of integration.Step 3.Find the outer and inner radii of the washer swept out by the line segment.Step 4.Integrate to find the volume.The Volume of a Solid44The region bounded by the cu

42、rve and the line21yx3yx is revolved about the x-axis to generate a solid.Find thevolume of the solid.Solution Step 1.Draw the region and sketch a l i ne s e g m e n t a c r o s s i t perpendicular to the axis of revolution.The Volume of a Solid45Solution(continued)12x 21x Step 2.Find the limits of i

43、ntegrationby finding the x-coordinates of the intersection points of the curve and the line in right figure.The region bounded by the curve and the line21yx3yx is revolved about the x-axis to generate a solid.Find thevolume of the solid.The Volume of a Solid46Solution(continued)Outer radius:()3R xx

44、Inner radius:2()1r xxStep 3.Find the outer and inner radii of the washer that would be swept out by the line segment if it were revolved about the x-axis along with theregion.The region bounded by the curve and the line21yx3yx is revolved about the x-axis to generate a solid.Find thevolume of the so

45、lid.The Volume of a Solid47Solution(continued)Step 4.Evaluate the volume integral.22()()baVR xr xdx 1222231xxdx 124286xxxdx 125228335xxxx 117 units cubed.5 Finish.The region bounded by the curve and the line21yx3yx is revolved about the x-axis to generate a solid.Find thevolume of the solid.The appl

46、ications of the definite integral in physics48Pumping Liquids from ContainersHow much work does it take to pump all or part of the liquid from a container?To find out,we imagine lifting the liquid out one thin horizontal slab at a time and applying the equation W=Fd to each slab,where F is the force

47、 and d is the distance of the object along with the direction of F.We then evaluate the integral this lead to as the slabs become thinner and more numerous.The integral we get each time depends on the weight of the liquid and the dimensions of the container,but the way we find the integral is always

48、 the same.The applications of the definite integral in physics49How much work does it take to pump the water from a fullupright circular cylindrical tank of radius 5m and height 10m to a level of 4m above the top of the tank?Solution We draw the tank as right figure,add coordinate axes,and imagine t

49、he waterdivided into thin horizontal slabs by plansperpendicular to the y-axis at the points ofa partition P of the interval 0,10.The applications of the definite integral in physics50Solution(continued)The typical slab between the planes at y andyy has a volume of2(radius)(thickness)V The force F(y

50、)required to lift the slab is equalto its weight,()9800F yV23(5)25myy 9800(25)245,00N0.yyHow much work does it take to pump the water from a fullupright circular cylindrical tank of radius 5m and height 10m to a level of 4m above the top of the tank?The applications of the definite integral in physi