北大版高数第七章习题解答

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5pŒ˘61SK)SSSKKK7.13.…Œf(x,y)3k.4«DºY,g(x,y)3DK,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)∈Df(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«DºY,K,RRDf(x,y)dxdy=0.y†f(x,y)=0,(x,y)∈D.y.ˇǑfK,ef??Ǒ,Kf3,:P∈D?u0.qˇfºY,ˇd3PSfu12f(P).u·RRDf(x,y)dxdy0,gæ.SSSKKK7.2Oe›¨'.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−sin4.110.RRDy2√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.|^4IOe\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%9r=a(1+cosθ)–r=a(a0)⁄„4:«.I=Rπ2−π2dθRa(1+cosθ)arrdr=Rπ2−π2dθa3(cosθ+cos2θ+13cos3θ)=(229+π2)a3.19.|^›¨'A¿´y†:dθ=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.%9r=a(1+cosθ)(a0,0≤θ2π)⁄«¡¨.).S=R2π0dθRa(1+cosθ)0rdr=R2π0dθ12a2(1+cosθ)2=32πa2.Oe›¨'.21.RRD(2x2−xy−y2)dxdy,¥Ddy=−2x+4,y=−2x+7,y=x−2,y=x+1⁄.2).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=xy=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,¥DǑx2+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′Ǒu2+v2≤12,I=RRD′(v+12)dudv=0+12·π2=π4.24.RRD(x2+y2)dxdy,¥DǑx2a2+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(Jlima→+∞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.SSSKKK7.3Oen›¨'.1.RRRΩ(z+z2)dV,¥ΩǑ¥x2+y2+z2≤1.I=RRRΩzdV+RRRΩz2dV=0+R2π0dθRπ0dϕR10drr2sinϕ(rcosϕ)2=4π15.32.RRRΩx2y2zdV,¥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−y2z=px2+y2⁄.I=R2π0dθRπ40dϕRR0r2sinϕdrr2sin2ϕ=2π(215−√212)R5.12.RRRΩpx2+y2+z2dV,Ωdz=x2+y2+z2=z⁄.4I=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=2a2az=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,Ω:(x−x0)2+(y−y0)2+(z−z0)2≤a2.x=x0+u,y=y0+v,z=z0+w,Ω′:u2+v2+w2≤a2.KD(x,

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