一、重积分的深入理解一、重积分的深入理解二、习题难点解答二、习题难点解答三、典型例题三、典型例题四、课堂提问与解答四、课堂提问与解答1.二重积分的换元法(,)fxyxOy定理设在平面的闭区域(,),(,)Dxxuvyyuv==上连续,变换将'uOvDxOy面上的闭区域一对一映到面,(,),(,)'DxuvyuvD上的闭区域函数在上'D具有一阶连续偏导数,且在上雅可比行列式(,)0(,)xxxyuvJuvyyuv∂∂∂∂∂==≠∂∂∂∂∂,'(,)(,)([,],[,]).(,)DDxyfxydfxuvyuvdudvuvσ∂=∂∫∫∫∫则有xyOuvO1(),DIxydxdyD=+∫∫计算:是圆域:22.xyxy+≤+xOy22221111()()222xyxyxy+≤+−+−=解:由得,22111':222xuyvDDuv−=−=+=令,,则变为,'11(,)()22(,)DxyIuvdudvuv∂=+++∂∫∫所以'(1)[]Duvdudv=++∫∫由对称性'2Ddudvπ==∫∫uOv22'D3cossin2404(cossin)drdrπθθπθθθ+−+∫∫解2:原式=34441(cossin)3dππθθθ−=+∫34411cos4(12sin2)32dππθθθ−−=++∫2π=222,2,,xyxyyx===例2求曲线:0.511.520.511.52D22,xyuvyx==解:作变换,0.511.520.511.5222yx=所围成的平面图形的面积。2222(,)111(,)(,)32(,)2xyuvuvxyxyyxxyxy∂===∂∂−∂−。'13DDSdxdydudv==∫∫∫∫所以,'12,12Duv≤≤≤≤其中为:。221111.33Sdudv==∫∫所以(,)(,)1(,)(,)(,)(,)xxuvxyuvyyuvuvxy=⎧∂⎨∂=∂⎩∂补例.设,证明:=。(,)(,),,(,)(,)xxuvxyJuvyyuv=⎧∂=⎨∂=⎩证明:令设,,uvxyx视为的函数,两边同时对求导得1,;0uxvxvuxxuxvxxuxvyyuvyuyvJJ=+⎧⇒==−⎨=+⎩,.vuyyxxuvJJ=−=同理可得:故22(,)111(,)vvuvuuuvyxxxuvyxyyxyJJJ−∂===−∂.(,)1.(,)(,)(,)xyuvuvxy∂∂∂∂即:=()sin(),Dxyxydxdy+−∫∫例3计算,uxyvxy=−=+解:令,则(,)111(,)11(,)2(,)11xyuvuvxy∂===∂∂∂−,'0,0Duvππ≤≤≤≤为::0,0.Dxyxyππ≤−≤≤+≤其中:()sin()Dxyxydxdy+−∫∫所以'1sin2Dvududv=∫∫001sin2dvvuduππ=∫∫22π=,yxyxDIedxdyD−+=∫∫例4计算:其中是以点(0,0),(1,0),(0,1)为顶点的三角形内部。,(0,0),(1,0),(0,1)yxuyxv−=⎧⎨+=⎩解:令则三点(0,0),(1,1),(1,1).−变为xyODxyO'D(,)111(,)11(,)2(,)11xyuvuvxy∂===−∂−∂∂。'12yxuyxvDDedxdyedudv−+=⋅∫∫∫∫所以1012uvvvdvedu−=⋅∫∫14ee−−=xyO'D2222522zxyzxy=+=−−例求由椭圆抛物面与所围成立体的体积。2222(2)(2)DVxyxydxdy=−−−+∫∫解:22(223).Dxydxdy=−−∫∫223{(,)|1}.2Dxyxy=+≤2cos,sin,3xryrθθ==作变换:'{(,)|01,02}.DDrrθθπ=≤≤≤≤变为(,)2.(,)3xyJruv∂==∂2'22(1)3DVrrdrdθ=−∫∫2120026(1)3drrdrπθ=−∫∫63π=22Question6Find(),where{(,)|1}.DIxydxdyDxyxyxy=+=+≤++∫∫2211Solve:Disacircle()()2.22xy−+−≤2211Let,,'isacircle2.22xuyvDuv=+=++≤'()(1)DDIxydxdyuvdudv=+=++∫∫∫∫22002(cossin)drrrdrππθθθ=++∫∫822.3π=+222()()7()()xyRaxbyIdxyϕϕσϕϕ+≤+=+∫∫例计算二重积分:.222()()()()xyRaxbyIdxyϕϕσϕϕ+≤+=+∫∫解:2222221()()()()[]2()()()()xyRxyRaxbybxayddxyxyϕϕϕϕσσϕϕϕϕ+≤+≤+++++∫∫∫∫=2221()()()()2()()xyRaxbybxaydxyϕϕϕϕσϕϕ+≤+++=+∫∫2221()2xyRabdσ+≤=+∫∫22abRπ+=2.三重积分的换元法(,,)fxyzV定理设在闭区域上连续,变换:(,,),(,,),(,,),xxuvwyyuvwzzuvw===(,,)'VxyzV建立了区域中的点与区域中(,,)uvw的点之间的一一对应关系,函数(,,),(,,),(,,)xxuvwyyuvwzzuvw==='V在区域中有一阶连续偏导数,且雅可比行列式。(,,)0(,,)xxxuvwxyzyyyJuvwuvwzzzuvw∂∂∂∂∂∂∂∂∂∂==≠∂∂∂∂∂∂∂∂∂∂,(,,)VfxyzdV∫∫∫则有:'[(,,),(,,),(,,)]||VfxuvwyuvwzuvwJdudvdw=∫∫∫12(),VIxyzdV=++∫∫∫例计算三重积分222(222)VIxyzxyyzzxdV=+++++∫∫∫解:2220.VVVxydVyzdVzxdV===∫∫∫∫∫∫∫∫∫由对称性得:222123().VIxyzdVIII=++=++∫∫∫所以222222:1.xyzVabc++≤其中3sincos,sinsin,cosxaIybzcρϕθρϕθρϕ=⎧⎪=⎨⎪=⎩求作变换222222:1xyzVabc++≤它把区域,2(,,)sin(,,)xyzJabcρϕρϕθ∂==∂':01,0,02.Vρϕπθπ≤≤≤≤≤≤变为区域23VIzdV=∫∫∫于是21324000cossinabcdddππθϕϕϕρρ=∫∫∫232001cossin5abcddππθϕϕϕ=∫∫3312425315abcabcππ=⋅⋅=2222'cossinVcabcdddρϕρϕρϕθ=∫∫∫331244,.1515IabcIabcππ==同理可得:2224().15Iabcabcπ=++所以:22222,xyLIedsLxya+=+=∫1计算其中为圆周yxx=与直线和轴在第一象限内扇形的边界。xOxy=y1L2L3La1:0,0;Lyxa=≤≤解:2:cos,sin,0;4Lxayaπθθθ==≤≤32:,0.2Lyxx=≤≤22123xyLLLIeds+=++∫∫∫0axedx=∫40aeadπθ+⋅∫22202axedx+⋅∫(2)2.4aeaπ=+−2(),()[,]fxgxab设在上连续,证明:2,[()()]0,batfxtgxdx∀+≥∫证明1:有即222(())(2()())()0bbbaaagxdxtfxgxdxtfxdx++≥∫∫∫,()2222()()4()()0bbbaaafxgxdxgxdxfxdxΔ=−≤∫∫∫,()222()()()().bbbaaafxgxdxfxdxgxdx≤∫∫∫即:()222()()()()bbbaaafxgxdxfxdxgxdx≤∫∫∫.()1gx=特别地,令,有()22()()()bbaafxdxbafxdx≤−∫∫()22222()()2()()bbbaaafxdxgxdxfxgxdx−∫∫∫证明:2222()()()()bbbbaaaafxdxgydyfydygxdx=+∫∫∫∫2222()()()()bbbbaaaafxgydxdyfygxdxdy=+∫∫∫∫2()()()()bbaafxgxdxfygydy−∫∫2()()()()bbaafxgxfygydxdy−∫∫2(()()()())0bbaafxgyfygxdxdy=−≥∫∫,()22()()()().bbbaaafxgxdxfxdxgxdx⇒≤∫∫∫222,3(0)0,yzayazx⎧+=⎨=⎩求圆周绕轴旋yOzya2P2ay−(2)yay−Q转一周所得到的旋转面的方程,并求此旋转面所围成的立体的体积。yOzya2P2ay−(2)yay−QyOzya2P2ay−(2)yay−Q(,0)(2)QyPQyay=−解:设点坐标为,则,PQz所以绕轴旋转一周所得圆柱面的面积为2(2)2(2),SyyaydVyyaydyππ=−−,体积微元为=2022(2)aVyyaydyπ=−∫所以22204()[cos]ayayadyyaaπθ=−−=+∫令04(cos)sin(sin)aaaadππθθθθ=+−∫0324(1cos)sinadππθθθ=−+∫3234[0]22aaπππ=+=yOzya2P2ay−(2)yay−Q41:xy+=求平面包含在椭球面2221236xyz++=内的那部分面积。Ozx2445zxx=+−p=linspace(0,pi,200);t=linspace(0,2*pi,200);[ph,th]=meshgrid(p,t);x=sqrt(2)*sin(ph).*cos(th);y=sqrt(3)*sin(ph).*sin(th);z=sqrt(6)*cos(ph);mesh(x,y,z);daspect([111]);axistight;gridon;camlight;lightinggouraud;alpha(.5)holdont=linspace(-1.5,1.5,200);s=linspace(-2.5,2.5,200);[x,z]=meshgrid(t,s);mesh(x,1-x,z)holdonmesh(x,z,z*0)axisequalxlabel('x');ylabel('y');zlabel('z');colormaphsv;xOz解:把图形投影到坐标面,由方程2221,326,xyyxyz+=⎧⎨++=⎩消去得方程:2222544445xxzzxx−+==+−,即,2204450,zxx≥+−≥由可得22622655x−+≤≤所以有,2445.zzxx≥=+−取0的一支图象可得1yx=−因为所求平面方程为,'2'212.xzdSyy=++=所以Ozx2445zxx=+−得平面的面积2226445522605222xzxxDSdxdzdxdz++−−==∫∫∫∫222625226522445xxdx+−=+−∫226252265242225[()]255xdx+−=−−∫241025π=Ozy2445zxx=+−222225,0LxyzaIxdsLxyz⎧++==⎨++=⎩∫求其中为圆周:.222LLLxdsydsxds==∫∫∫解:由对称性得:,2221()3LIxyzds=++∫213Lads=∫2312233aaaππ=⋅=Acrobat22222262,.LxyzaIyzdsLyx⎧++==+⎨=⎩∫求其中为圆周:yx=解:由,则222LIyzds=+∫222Lyyzds=++∫2Lads=∫2aaπ=⋅Lads=∫22aπ=222Lxyzds=++∫2222227(),:SIxydSSxyza=+++=∫∫计算:其中.222,SSSxdSydSzdS==∫∫∫∫∫∫解:由对称性得:22242()3SIxyzdS=++∫∫所以有:222233SSadSadS==∫∫∫∫22428433aaaππ=⋅=8计算曲面所围成几何体的体积:22223()(0)xyzaza++=3cos,aρϕ=解:其球面坐标方程为:0202πθπϕ≤≤≤≤其中,.32cos22000sinaVdddππϕθϕρϕρ=∫∫∫392012cossin3a