高数讲座-重积分选讲.pdf

《高数讲座-重积分选讲.pdf》由会员分享，可在线阅读，更多相关《高数讲座-重积分选讲.pdf（7页珍藏版）》请在淘文阁 - 分享文档赚钱的网站上搜索。

1、 1 一一.关于重积分的概念关于重积分的概念 1.1.利用重积分利用重积分定义定义求极限求极限 重积分是用极限定义的,反过来也可以用它来求某种和的极限.回忆:1101limnniiff x dxnn,例如:22222111limlimnnnniinnninni 1221011lim141nniindxnx,口诀:“提出1n,凑in”,类似地,2110,10,11lim,nnnijijff x y dnn n,口诀:“提出21n,凑in和jn”,例如:114211110,10,10011limlim4nnnnnnijijijijxydxdxydynnn n.例例.32222211111limli

2、mnnnnnnijijnnnninjninj 112222110,10,100211ln2lim1141111nnnijddxdynxyijxynn.例例.3322221111111limlimnnnnnnnnijkijkknknn ninjn ninj 3221110,10,10,121limln281111nnnnijkkzndvnijxynn;口诀:“提出31n,凑in,jn和kn”.2.2.轮换不变性轮换不变性 设,x yy xDD,则,DDf x y dfy x d.例例.求 Daf xbfyIdf xfy,其中:1Dxy.解.2DDafybfxabIddabfyfx.2 例例.求2

3、2sincosDIxy d,其中22:1D xy.解.22221sinsin2DIxyxyd 22222211sinsinsin22DDyxyxdyxd 212001sin1 cos122dd.例例.证明:2211xyDede,其中22:1D xy.证.若 0fx,则 12DDDf xf xfydddfyfyf x,证毕.例例.设 fx连续,证明:221bbaafx dxf x dxba.证.2222,bbbaaaa ba ba ba bbafx dxdyfx dxfx dfy d 222,12ba ba ba ba bafxfydfx fy df x dx,证毕.例例.设 p x,fx,g

4、xa b,且 0p x,fx,g x单增,证明:,a ba ba ba bp x f x p y g y dp x fy p y g y d.证.左式-右式 ,a ba bp x p y g yf xfyd ,a ba bp y p x g xfyf xd ,102a ba bp x p yg yg xf xfyd,证毕.例例.设 fx连续,证明:21110012xdxf x fy dyfx dx.证.左式 1100,10,112ydyfy f x dxf x fy dxdy,即得,证毕.3 例例.设 fx连续,证明:31110013!yxxdx dyf x fy fz dzfx dx .证.

5、左式 11100,1,112zxxxxf x dx dzf z fy dyf x dxfy f z dydz 2211100111122xxxfxfy dydxfy dyfy dydx 1231011101126xxxfy dydfy dyfy dy,即得,证毕.注注.对于第一类的线面积分有类似的轮换不变性.二二.关于重积分的计算关于重积分的计算 例例.求ln 11DyxyxIdxy,其中D由1xy与两坐标轴围成.解一.ln 1ln 111211DDxyxyxxyyxyIddyxxy 2lnlnlnln1122111DDDxyxyxyxyxyxyxydddxyxyxy 12lnln11DDxy

6、xyxyxddIIxyxy,11111121000000lnlnlnln1111xuxxyxyuuuuuuIdxdydxdududxduxyuuu,111111222000000lnlnlnln1111xuxxyxuxuxuuuIdxdydxdududxduxyuuu,故12016151uIduu.解二.1cossin200cossinln 1tan1cossinIdd 4 12cossin21cossin00cossinln 1tan1cossintdd 2022201111cossinln 1tancossincossinttddt 02222222010ln 1tanln 1tan161

7、621tan1515cossin1tandtdtd.解三.令11uuxyxvyuvvyxv,2,1x yuu vv,则 112220000ln 1ln 116151111uvvuuIdudvdudvuuvv.例例.设 fx连续,证明:1213412151Dfxydu fudu,其中 22:1D xy.证.令534uxy,即345xyu,435xyv,则22221uvxy,于是1,1,x yu vu vx y,故34151DDfxydfududv 221112111512151uudufudvu fudu,证毕.例例.设,fx y连续,证明:222Dfxy dxf x dx,其中:1Dx,1y.

8、证.令22uvxuxyvxyvuy,1,2x yu v,则Df xy d 022222202211222uuuuduf udvduf udvuf u du,证毕.5 例例.证明:122210114arccosyxfxydxfx dxxfx dxx.证.左式 11cos44000144dfddfd ,而 1cos401dfd 224111arccos1arccos4dfdfd ,证毕.例例.求2Ix dv,其中由1z,22zxy,2212zxy围成.解一.2222222211222214121122xyxyxyxyIdxdyxydzdxdyxydz 2212112200002113224dddz

9、dddz .解二.222211222220004113224zzzxyzIdzxyddzdd.解三.1cos2arctan2arctan222232004413sinsintansec254Iddrr drd .例例.求2Ixyzdv,其中222222:xyzxy.解.是yOz面上的圆22:D yzy绕z轴旋转形成的立体,它关于 三个坐标面对称,在球坐标下,22:sinr,故 故1sin22222222260000488sinsin58Ixyzdvddrr drd .例例.求2222xyzx所围立体的体积.解.0 x,关于xOy面与xOz面对称,在球坐标下,3:sincosr,故3sincos

10、2222220000044sincossin33Vddr drdd .6 例例.求2cosIxyzxyzdxdydz,其中:01xy,01xz,01xyz.解.令uxyvxzwxyz,1,1,3x y zu v wu v wx y z,故 111220001coscossin16Iww dudvdwdu dv ww dw.例例.求111401xyIdx dy yz dz .解.111144400000111zzzzzxxxxIdx dz yz dydz dx yz dyz dz dx ydy 111223344340000111112 21226318zzxzzz dzdxz dzzz dz.

11、例例.求12000sin1yxzIdx dydzz .解.111122200000sinsinsin111xxxzzxzzzzIdx dzdydxdzdzxz dxzzz 1112200sin11sin1 cos12221zzxzxdzzdzz.三三.含重积分含重积分或累次积分或累次积分的极限的极限 例例.设,fx y连续,且0,01f,求(1)20130,limxxtxdtf t u duIx;(2)00220,limxx uxdtf t u duIx;(3)22200320,limxtx txdtf t u duIx.解.22000132000,0,01limlimlim3333xuxxx

12、xduf t u dtf t x dtfxfIxx;7 2222001,0,012limlim22Dxxf t u dtdufxfIxx(换顺序没用);2222,0322001,2limlim0,022tutx uxxf t u dtdufxIfxx.例例.设 fx连续,0lim1xf xx,求2222240limxyt xty fxydIt.解.arccos2 cos2222000044002sin2sinlimlimtttttdfddfdItt 222223000450002122248limlimlim55ttttttfdtfdfdfttttt .例例.设 fx连续,0lim1xf xx

13、,求2222222130limxyzttf xyzdvIt;2222222250limxyzttf xyzdvIt;2222222350limxyzt ztf xyzdvIt.解.2223130443lim003tftIft;222222200002554000sin444limlimlim55tttttddf rr drf rr drf ttIttt;arccos2 cos222222220000035500sin2sinlimlimrtttttddf rr drdrf rrdItt 222222223000560012122lim2 limtttttrf rrdrtf rr drf rr drttt 2222205400442322 lim2lim63015tttf rr drftttt.