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5p Œ˘61 SK) SSSKKK7.13. …Œf(x,y)3k.4« D ºY,g(x,y)3D K, g(x,y) f(x,y)g(x,y)3D ¨.y†:3D¥ 3 :(x0,y0)RRDf(x,y)g(x,y)dσ=f(x0,y0)RRDg(x,y)dσ.y. m,MǑf3D , .Kmg(x,y)≤f(x,y)g(x,y)≤Mg(x,y).ˇdRRDmg(x,y)dσ≤RRDf(x,y)g(x,y)dσ≤RRDMg(x,y)dσ.eRRDg(x,y)dσ=0,KRRDf(x,y)g(x,y)dσ=0, ? :(x0,y0)∈D•K⁄Æ.˜Kkm≤RRDf(x,y)g(x,y)dσRRDg(x,y)dσ≤M.d0 ‰n, 3(x0,y0)∈D f(x0,y0)=RRDf(x,y)g(x,y)dσRRDg(x,y)dσ,=RRDf(x,y)g(x,y)dσ=f(x0,y0)RRDg(x,y)dσ.4. …Œf(x,y)3k.4« D ºY, K, RRDf(x,y)dxdy=0.y†f(x,y)=0, (x,y)∈D .y.ˇǑf K,ef ??Ǒ,Kf3,:P∈D? u0.qˇfºY,ˇd3P Sf u12f(P).u·RRDf(x,y)dxdy0,gæ.SSSKKK7.2O e ›¨'.3.RRDydxdy, ¥Ddy=09y=sinx(0≤x≤π)⁄ .I=Rπ0dxRsinx0dyy=Rπ0dxsin2x2=π4.4.RRDxy2dxdy, ¥Ddx=1,y2=4x⁄ .I=R2−2dyR1y2/4dxxy2=R2−2dy12(1−y416)y2=3221.5.RRDexydxdy, ¥Ddy2=x,x=0,y=1⁄ .I=R10dyRy20dxexy=R10dyyey=1.6.R10dyR1y13√1−x4dx=R10dxRx30dy√1−x4=R10dxx3√1−x4=16.7.RRD(x2+y)dxdy, ¥Ddy=x2,x=y2⁄ .I=R10dxR√xx2dy(x2+y)=R10dx(12x+x52−32x4)=33140.8.Rπ0dxRπxsinyydy=Rπ0dyRy0dxsinyy=Rπ0dysiny=2.9.R20dxR2x2ysin(xy)dy=R20dyR20dx2ysin(xy)=R20dy2(1−cos2y)=4−sin4.1课后答案网√1−x2dxdy,D={(x,y)|x2+y2≤1}.I=4R10dxR√1−x20dyy2√1−x2=4R10dx13(1−x2)2=3245.11.RRD(|x|+y)dxdy,D={(x,y)||x|+|y|≤1}.I=RRD|x|dxdy+RRDydxdy=4R10dxR1−x0dyx+0=4R10dxx(1−x)=23.12.RRD(x+y)dxdy, ¥DǑdx2+y2=1,x2+y2=2y⁄ « ¥m ‹.I=RRDxdxdy+RRDydxdy=0+2R√320dxR√1−x21−√1−x2ydy=2R√320dx(√1−x2−12)=π3−√34.|^4 IO e \g¨'‰ ›¨'.13.R10dxR√1−x20(x2+y2)dy=Rπ20dθR10r2rdr=π8.14.R0−1dxR0−√1−x221+√x2+y2dy=R3π2πdθR1021+rrdr=π(1−ln2).15.R20dxR√1−(x−1)203xydy=Rπ20dθR2cosθ03rcosθrsinθrdr=Rπ20dθ12cos5θsinθ=2.16.RR0dxR√R2−x20ln(1+x2+y2)dy=Rπ20dθRR0ln(1+r2)rdr=π4[(1+R2)ln(1+R2)−R2].17.RRD1x2dxdy,D·dy=αx,y=βx(π2βα0),x2+y2=a2,x2+y2=b2(ba0)⁄ 31 '.I=RarctanβarctanαdθRba1(rcosθ)2rdr=(β−α)lnba.18.RRDrdσ, ¥D·d%9 r=a(1+cosθ) –r=a(a0)⁄ „4: « .I=Rπ2−π2dθRa(1+cosθ)arrdr=Rπ2−π2dθa3(cosθ+cos2θ+13cos3θ)=(229+π2)a3.19.|^ ›¨' A ¿´y†:d θ=α,r=β › r=r(θ)(α≤θ≤β)⁄ « D ¡¨ L«⁄12Rβα[r(θ)2]dθ.y.S=RRDdxdy=RβαdθRr(θ)0rdr=12Rβα[r(θ)2]dθ.20.%9 r=a(1+cosθ)(a0,0≤θ2π)⁄ « ¡¨.).S=R2π0dθRa(1+cosθ)0rdr=R2π0dθ12a2(1+cosθ)2=32πa2.O e ›¨'.21.RRD(2x2−xy−y2)dxdy, ¥Ddy=−2x+4,y=−2x+7,y=x−2,y=x+1⁄ .2课后答案网). u=2x+y,v=x−y.Kx=u+v3,y=u−2v3,D(x,y)D(u,v)=−13.ˇdI=R74duR2−1(uv)13dv=334.22.RRD(pyx+√xy)dxdy, ¥Ddxy=1,xy=9,y=x y=4x⁄ .). u=pyx,v=√xy.Kx=vu,y=uv,D(x,y)D(u,v)=−2vu.ˇdI=R21duR31(u+v)2vudv=8+523ln2.23.RRDydxdy, ¥DǑ x2+y2≤x+y.)1. x=12+rcosθ,y=12+rsinθ.KD(x,y)D(u,v)=r.ˇdI=R1√20drR2π0(12+rsinθ)rdθ=π4.)2. x=u+12,y=v+12.KD(x,y)D(u,v)=1. D′Ǒ u2+v2≤12, I=RRD′(v+12)dudv=0+12·π2=π4.24.RRD(x2+y2)dxdy, ¥DǑ x2a2+y2b2≤1.). x=arcosθ,y=brsinθ.KD(x,y)D(u,v)=abr.ˇdI=R10drR2π0(a2r2cos2θ+b2r2sin2θ)abrdθ=R10drπ(a2r2+b2r2)abr=π4(a2+b2)ab.26. a0,¿-I(a)=Ra0e−x2dx,J(a)=RRDae−x2−y2dxdy, ¥Da={(x,y)|x2+y2≤a2,x≥0,y≥0}.y†(1)[I(a)]2=RRRae−x2−y2dxdy, ¥Ra={(x,y)|0≤x≤a,0≤y≤a};(2)J(a)≤[I(a)]2≤J(√2a);(3)|^ !~10 (J lima→+∞Ra0e−x2dx=√π2.y.(1)[I(a)]2=Ra0e−x2dxRa0e−y2dy=RRRae−x2−y2dxdy.(2)Da⊂Ra⊂D√2a,⁄–RRDae−x2−y2dxdy≤RRRae−x2−y2dxdy≤RRD√2ae−x2−y2dxdy.(3)J(a)=Rπ20dθRa0e−r2rdr=π4(1−e−a2).ˇdlima→+∞J(a)=lima→+∞J(√2a)=π4.dY%‰n, lima→+∞I(a)=√π2.SSSKKK7.3O e n›¨'.1.RRRΩ(z+z2)dV, ¥ΩǑ ¥x2+y2+z2≤1.I=RRRΩzdV+RRRΩz2dV=0+R2π0dθRπ0dϕR10drr2sinϕ(rcosϕ)2=4π15.3课后答案网 ¥Ω·d2z=x2+y2,z=2⁄ ⁄ « .I=R2π0dθR20rdrR2r22dz(rcosθ)2(rsinθ)2z=R2π0(sin2θcos2θ)dθ·R20r5(2−r48)dr=π4·12815=32π15.3.RRRΩx2sinxdxdydz, ¥ΩǑd†¡z=0,y+z=19˛¡y=x2⁄ « .« Ω’uOyz†¡Ø¡, ¨…Œ·’ux …Œ,ˇd¨'Ǒ0.4.RRRΩzdxdydz, ¥Ωdx2+y2=4,z=x2+y29z=0⁄ .I=R2π0dθR20rdrRr20dzz=2π·163=32π3.5.RRRΩ(x2−y2−z2)dV,Ω:x2+y2+z2≤a2.RRRΩz2dV=R2π0dθRπ0dϕRa0r2sinϕdr(rcosϕ)2=4π15a5. nRRRΩx2dV=RRRΩy2dV=4π15a5.ˇdI=−4π15a5.6.RRRΩ(x2+y2)dV,Ω:3px2+y2≤z≤3.I=R2π0dθR10rdrR33rdzr2=2πR103(1−r)r3dr=3π10.7.RRRΩ(y2+z2)dV,Ω:0≤a2≤x2+y2+z2≤b2. ¥ IXx=rcosϕ,y=rsinϕcosθ,z=rsinϕsinθ. I=R2π0dθRπ0dϕRbar2sinϕdr(rsinϕ)2=2π·43·b5−a55=8π15(b5−a5).8.RRRΩ(x2+z2)dV,Ω:x2+y2≤z≤1.I=R2π0dθR10rdrR1r2dz(r2cos2θ+z2)=π12+π4=π3.9.RRRΩz2dV,Ω:x2+y2+z2≤R2,x2+y2≤Rx.I=4Rπ20dθRRcosθ02rdrR√R2−r20dzz2=4Rπ20dθ115R5(1−sin5θ)=215(π−1615)R5.10.RRRΩ(1+xy+yz+zx)dV, ¥ΩǑd›¡x2+y2=2z9x2+y2+z2=8⁄ z≥0 '.dØ¡5I=RRRΩ1dV=R2π0dθR20rdrR√8−r2r22dz=2π16√2−143.11.RRRΩ(x2+y2)dV,Ωdz=pR2−x2−y2 z=px2+y2⁄ .I=R2π0dθRπ40dϕRR0r2sinϕdrr2sin2ϕ=2π(215−√212)R5.12.RRRΩpx2+y2+z2dV,Ωdz=x2+y2+z2=z⁄ .4课后答案网=R2π0dθRπ20dϕRcosϕ0r2sinϕdrr=π10.13.RRRΩz2dV,Ω:p3(x2+y2)≤z≤p1−x2−y2.I=R2π0dθRπ60dϕR10r2sinϕdrr2cos2ϕ=2π15(1−3√38).14.RRRΩzdV√x2+y2+z2,Ωdx2+y2+z2=2az⁄ .I=R2π0dθRπ20dϕR2acosϕ0r2sinϕdrcosϕ=16π15.15.RRRΩ2xy+1x2+y2+z2dV,ΩǑdx2+y2+z2=2a2 az=x2+y2⁄ z≥0 '.dØ¡5I=RRRΩ1x2+y2+z2dV=R2π0dθRπ40dϕR√2a0r2sinϕdr1r2+R2π0dθRπ2π4dϕRacosϕsin2ϕ0r2sinϕdr1r2=2πa(√2−1+ln√2).16.RRRΩdV√x2+y2+(z−2)2,Ω:x2+y2+z2≤1.I=R2π0dθR10drRπ0dϕr2sinϕ1√r2−4rcosϕ+4=2πR10drr2|r+2|−|r−2|r=23π.17.RRRΩ(x3+siny+z)dV,Ωdx2+y2+z2≤2az,px2+y2≤z⁄ .dØ¡5I=RRRΩzdV=R2π0dθRπ40dϕR2acosϕ0r2sinϕdrrcosϕ=76πa4.18.RRRΩ(x2y+3xyz)dV,Ω:1≤x≤2,0≤xy≤2,0≤z≤1. u=x,v=xy,w=z.Kx=u,y=vu,z=w,D(x,y,z)D(u,v,w)=1u.I=R21duR20dvR10dw(uv+3vw)1ududvdw=2+3ln2.19.RRRΩ(x+1)(y+1)dV,Ω:x2a2+y2b2+z2c2≤1.dØ¡5I=RRRΩ(xy+x+y+1)dV=RRRΩ1dV=R2π0dθRπ0dϕR10abcr2sinϕdr=43πabc.20.RRRΩ(x+y+z)dV