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1、13.7 Triple Integrals in Cartesian CoordinatesProblem IntroductionHow to extend single and double integrals to tripleintegrals for three variables?Triple Integrals Triple Integrals in Cartesian CoordinatesIntroduction:Quality of a Space ObjectWe have a space object with space bounded closed area ,it
2、s volume density at,is a continuous function ,.Find its quality.(1)Divide into n small pieces.Any small piece(volume)is approximately equal to a homogeneous object.(2)(3)(4)iMM M=(,)iiiifv 1(,)niiiiifv =1(,)niiiiifv =0limDefinitionFor a geometry with space bounded closed area ,the integral of ,with
3、area is called the triple integral,that is,(,)f x y z dv If ,0,the physical significance of is the quality of an nonhomogeneous object with a volume density ,.Definition(,)f x y z dv Triple Integrals in Cartesian CoordinatesIn Cartesian Coordinates,we use the planes which are parallel to the coordin
4、ate planes to divide .Then,the small cuboid .ijklvxyz=So,.dd d dvx y z=That is,the triple integral .(,)d(,)d d df x y zvf x y zx y z=DefinitionProjection MethodIdea:Change the triple integral into three integrals.As shown in figure,the projection of the closed area in xoy-plane is the closed area.11
5、:(,),Szz x y=22:(,),Szzx y=The line through passes from 1to 2.(,)x yDxyzO Dab)(1xyy=)(2xyy=1S1(,)zz x y=2S2(,)zzx y=),(yx1z2zTriple Integrals in Cartesian CoordinatesdD21(,)(,)(,)(,)dzx yzx yF x yf x y zz=Projection MethodLet,be fixed numbers and let ,be a function of,that is,Then,find the double in
6、tegrals of ,in the closed area,12:()(),D y xyy xaxbX-Type21(,)(,)(,)d zx yzx yf x y zz(,)dDF x y=(,)df x y zv=21(,)(,)(,)dzx yzx yf x y zz21()()dyxyxydbaxDefinitionProjection Method2211()(,)()(,)(,)ddd(,)dbyxzx yayxzx yf x y zvxyf x y zz=Note:There are at most two intersections between the line and
7、the boundary surface of the closed area .If is Y-Type,then2211()(,)()(,)(,)dd(,)ddxyzx ycxyzx yf x y zvdyxf x y zz=The line is parallel to the-axis and passes through the closed area .DefinitionProjection MethodAlso,we can get the projection of the closed area in plane or.So,triple integrals can be
8、converted into three integrals of six different orders.Thus,we should select appropriate three integrals according to the specific ,and .DefinitionExample 1Evaluate the iterated integral 532204xxydzdydx+532204xxydzdydx+()532204xxydz dydx+=532204xxyzdydx+=()5320448xxydydx=+53202428xxyyydx=+()522624xx
9、 dx=+14=130dxxExample 2Evaluate where is the box34cos d d d,VIx yz x y z=(,)01,01,0.2Vx y zxyz=The upper and lower limits of three integrals are constants.1 1114 520=I=140dyyxyzO2 cos dz z022222112112dd(,)dxxxxyIxyf x y zz+=Example 3(,)d d dIf x y zx y z=222zxy=+22zx=22:1D xy+22222zxyzx=+=That is,pr
10、ojective areaSo,：2211xyx11x z222xy+22xxyzO22xz=222yxz+=Changeinto three integrals.isbounded byandMethod of SectionsProjection:from the area to the axis(eg:-axis).Projective interval:12,c c 1,2,cut by using a plane which goes through the-axis and is parallel to the plane.xzoy 1c2czzDThen,we get(the r
11、ed part).Calculate(,)d dzDf x y zx yThe result is a function of,that is,Calculate 21()d.ccF zzStep1Step2Step3Step4When the integrand is only related to the variable and the cross section is easy to find,we use the formula above.Method of SectionsThat is,21(,)dd(,)d dzDcf x y zvzf x y zx yc=21()dcF z
12、zc=There are two other formulas for the method of section.Note:Example 4Evaluate ,is bounded by three coordinateplanes and .d d dz x y z1xyz+=Method of Sections d d dz x y z=d dzDx y(,)|1zDx yxyz=+21d d(1)2zDx yz=10dz z111xyzO1xyz+=zD12011d d d(1)d.224z x y zzzz=Example 4Projection Method111000dddzy
13、 zz zyx=1100d(1)dzz zyzy=dz xdyzD1yz0d d dz x y z=1201(1)d2zzz=1.24=111xyzO1xyz+=d d dz x y z10ddxyx yDz z=Example 5Ellipsoid:,the volume density at,is,find the mass.2222221xyzabc+222222xyzabc+222222dVxyzMvabc=+222222dddVVVxyzvvvabc=+As is equal to the area of Example 5222222222222ddddVVVVxyzxyzMvvv
14、vabcabc=+=+And 22dVxva=d dxDy z222222111xxxbcbcaaa=22daaxxad dy zxD2222221yzxbca+=Example 5Similarly,415abc=22224dd15VVyzvvabcbc=Thus,4.5Mabc=So,22dVxva=22daaxxad dxDy z)1(dd22axbczyxD=222202(1)dabcxxxaa=Example 6Evaluate which is bounded by 222224 and zaxyzxy=+xyzOIntersecting lines:=+=22222ayxazPr
15、ojection:xyD2222xya+dVVv=222224ddxyaxyxyDz+=22222(4)dxyDaxyxy=+222200d(4)aad=polar coordinates38(22).3a=Summary of the Infinite Sequences(,)df x y zv=dbax21()()dyxyxy21(,)(,)(,)dzx yzx yf x y zzxyzO Dab)(2xyy=1S1(,)zz x y=2S2(,)zzx y=),(yx1z2zNote:There are at most two intersections between the line and the boundary surface of the closed area .If is Y-style,then2211()(,)()(,)(,)dd(,)ddxyzx ycxyzx yf x y zvdyxf x y zz=Triple Integrals in Cartesian Coordinates