二重积分习题二重积分的定义定义:),(yxf设将区域D任意分成n个小区域任取一点若存在一个常数I,使可积,),(yxf则称),(yxfI为称在D上的二重积分.称为积分变量yx,积分和积分域被积函数积分表达式面积元素记作是定义在有界区域D上的有界函数,性质1性质2区域可加性(重积分与定积分有类似的性质)重积分的性质(二重&三重)线性性gdVfdVgdVf2121fdVfdVfdV性质3性质4性质5保序性fxgxfdVgdV()mfxMmVfdVMV绝对可积性fdVfdV设函数),(yxf在闭区域D上连续,为D的面积,则在D上至少存在一点),(使得性质6二重积分中值定理(数一)),(),(fdyxfD性质7二重积分的对称性(1)普通对称性:设积分区域D关于Y轴对称,则(2)轮换对称性:若积分区域D关于y=x对称,则yxfyxfyxfyxfdxdyyxfdxdyyxfDD,,,0,,,,,2,1dxdyxyfdxdyyxfDD,,a0xbzyx)(0xA),(yxfz)(1xy)(2xy.),()(),()()(21DbaxxbadyyxfdxdxxAdyxf.),(),()()(21Ddcyydxyxfdydyxf.)sin,cos(),(DDrdrdrrfdxdyyxf一些公式:2sin,1sin,22sin,221sin0202440xdxxdxxdxxdx;,0,,cos2cos;sin2sin;1,32231,,221231cossin2002002020为奇数为偶数的奇数为大于为正的偶数nnxdxxdxxdxxdxnnnnnnnnnnxdxxdxnnnnnn;,0,,sin4cossin202020为奇数为偶数nnxdxxdxxdxnnn.sinsin2sin;cossin20002020dxxfdxxfdxxxfdxxfdxxf一、基本概念及性质例1.(2005,8题,4分)设,dyxID221cosdyxID)cos(222,dyxID2223)cos(其中}1),{(22yxyxD,则(A)(B)(C)(D)123III321III312III213III由于cosx在)2,0(上为单调减函数,于是22cos0yx)cos(22yx222)cos(yx因此dyxD22cosdyxD)cos(22dyxD222)cos(故应选(A)解:在区域}1),{(22yxyxD上,有1022yx从而有2212yx22yx0)(222yxdyxID221cosdyxID)cos(222dyxID2223)cos(}1),{(22yxyxD(2010,6题,4分)(A)(B)(C)(D)2211lim()()nnxijnninj12001(1)(1)xdxdyxy1001(1)(1)xdxdyxy11001(1)(1)dxdyxy112001(1)(1)dxdyxy例2.解:22221111.()()(1)1()limlimnnnnxxijijnnijninjnnnn1122002111111.(1)(1)11()limnnxijdxdyijnxynn例3.(2013,3题,4分)设kD是圆22{(,)|1}Dxyxy位于第k象限的部分,()kkDIyxdxdy1,2,3,4k则()(A)(B)(C)(D)10I30I20I40I故应选B.2/2/)1(2/2/)1(102/2/)1(cossin31cossin31cossinkkkkkkDkdrdrrrddxdyxyIk032,0,032,04321IIII二、二重积分的基本计算例4.(2012,16题,10分)计算二重积分,xDexydxdy其中D为由曲线1yxyx与以及y轴所围域.xDexydxdy解:110xxxxedxydy1122001111(1)0222xxxxedxexedx2111121(22)022222xeeexxexyO(1,1)D例5.(2013,17题,10分)设平面区域D由直线x=3y,y=3x及x+y=8围成,计算2Dxdxdy解:y=3x与x+y=8的交点为(2,6),x=3y与x+y=8的交点为(6,2)。2368222110233xxxxDxdxdydxxdydxxdy26220211416(3)(8)=333xxxdxxxxdyxyOD(2,6)(6,2)(2009,19题,10分)22,112,DxydxdyDxyxyyx求,其重积分中二例6.解:22(1)(1)22(sincos)xyr得由32(sincos)404()(cossin)Dxydxdydrrrdr2(sincos)3340432441(cossin)38(cossin)(sincos)(sincos)3rdd22,112,DxydxdyDxyxyyx求,其重积分中二3344333444448(cossin)(sincos)3881(sincos)(sincos)(sincos)33483dd(+05,7分)计算二重积分d)(31DyxxI,其中D为直线1yx,0x和0y所围成的平面区域.例7xyo11解d)(31DyxxIxyyxxx103110d)(d10103231d)(23xyxxx1031d)(23xxx.83xyo11总结:注意观察被积函数,选择合适的积分次序(2011,13题,4分)222_____.DDyxxyyyxyd设平面区域则二重积分由直线,圆及轴组成,例8.解:2sin2204cossinDxyddrrdr易得圆的极坐标方程为r=2sin,于是55224474sincos4sinsin12dd(2011,13题,4分)222_____.DDyxxyyyxyd设平面区域则二重积分由直线,圆及轴组成,例8.三、利用区域的对称性及函数的奇偶性计算积分例9.(2008,11题,4分)设,则22{(,)1}Dxyxy2()Dxydxdy______.解:22221()2DDDxydxdyxdxdyxydxdy利用函数奇偶性21200124drrdrDDDdxdyxyfyxfdxdyxyfdxdyyxf,,21,,若积分区域D关于y=x对称,则DDDdxdyyxdxdyydxdyx222221对于这道题,有三、利用区域的对称性及函数的奇偶性计算积分(2005,10题,4分)(A)(B)(C)(D)22{(,)4,0,0}(),()()()DfxDDxyxyxyafxbfydfyafbx,为上的正值连续函数,为常数,则设区域.ab().ab.2ab.2ab例10.解:()()()()()()()()DDafxbfyafybfxddfxfyfyfx由轮换对称性,有2()()()()1[]2()()()()12.2242.DDafxbfyafybfxdfxfyfyfxabbbdDaa应选(04,8分)求Dyyxd)(22,其中D是由圆422yx和1)1(22yx所围成的平面区域。例11xy2DO解由对称性知,0dDy.Dyxd22原式cos20223220220ddddrrrr2323dcos38316.)23(916xy2DO2222211202cos02cosxyxyxrrr四、分块函数积分的计算例12.(2005,17题,9分)计算二重积分dyxD122其中}10,10),{(yxyxD解:}),(,1),{(221DyxyxyxD}),(,1),{(222DyxyxyxD于是dyxD122=1)1(22Ddxdyyx2)1(22Ddxdyyx20210)1(rdrrd=Ddxdyyx)1(221)1(22Ddxdyyx8=20102210210)1()1(rdrrddyyxdx+.314=例12.(2005,17题,9分)计算二重积分dyxD122其中}10,10),{(yxyxD(08,11分)计算Dyxxydd)1,max(,其中}20,20|),(yxyxD.例13解Dyxxydd)1,max(20210d1dyxxyx10221d1d21221ddxyxyx2ln4152ln21.2ln419xy21DO22D213D计算二重积分Dyxfd),(,其中}2||||),{(yxyxD。例14(07,11分)设二元函数2||||1,11||||,),(222yxyxyxxyxfxy21DO-22-211-1-12D解xy21DO-22-211-1-12DDyxfd),(,d),(4d),(421DDyxfyxfxy21DO-22-211-1-12DDyxfd),(,d),(4d),(421DDyxfyxf11dd),(2DDxyxfxyxx10210dd102d)1(xxx;12122d1d),(22DDyxyxf22d1d),(22DDyxyxfcossin2cossin120d1drrr20dcossin120d)4csc(21x20|)4cot()4csc(|ln21xx,)12ln(2.)12ln(2431d),(Dyxf(07,4分)设函数),(yxf连续,则二次积分1sin2d),(dxyyxfx等于例15(A)yxyxfyarcsin10d),(d(B)yxyxfyarcsin10d),(d(C)yxyxfyarcsin210d),(d(D)yxyxfyarcsin210d),(dxyo2D五、交换积分次序或改变坐标系解【答案】应选(B).1sin,2:yxxDxyyDarcsin,10:xyo2DyPI-ar