北大版高数第七章习题解答

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1、5pêÆ61ÔÙS)S7.13.¼êf(x,y)3k.4«DþëY,g(x,y)3Dþ,g(x,y)f(x,y)g(x,y)3DþRRRRÈ.y²:3D¥3:(x0,y0)f(x,y)g(x,y)dσ=f(x0,y0)g(x,y)dσ.DDy.RRm,MǑf3DþRR,.mgRR(x,y)≤f(x,y)g(x,y)≤Mg(x,y).Ïdmg(x,y)dσ≤f(x,y)g(x,y)dσ≤Mg(x,y)dσ.RRDDRRDeg(x,y)dσ=0,f(x,y)g(x,y)dσ=0,?:(x0,y0)∈D·DRRDf(x

2、,y)g(x,y)dσDRR¤á.Äkm≤g(x,y)dσ≤M.d0½n,3(x0,y0)∈DRRDf(x,y)g(x,y)dσRRRRf(x,y)=DRR,00g(x,y)dσ=f(x,y)g(x,y)dσ=f(x0,y0)g(x,y)dσ.DDDRR4.¼êf(x,y)3k.4«DþëY,,f(x,y)dxdy=0.yD²f(x,y)=0,(x,y)∈D.y.ÏǑf,efØ?Ǒ",f3,RR:P∈D?u0.qÏfëY,Ïd1f(P).3PSfu2u´f(x,y)dxdy>0,gñ.DS7.2eÈ©.RR3.

3、ydxdy,Ù¥Ddy=09y=sinx(0≤x≤π)¤.DRπRsinxRπsin2xπI=dxdyy=dx=.00024RR4.xy2dxdy,2Ù¥Ddx=1,y=4x¤.DR2R12R21y4232I=−2dyy2/4dxxy=−2dy2(1−16)y=21.RRx5.eydxdy,2Ù¥Ddy=x,x=0,y=1¤.DR1Ry2xR1I=dydxey=dyyey=1.000R1R1√R1Rx3√R1√443416.dy11−xdx=dxdy1−x=dxx1−x=.0y30006RR7.(x2+y)dxdy,22Ù¥Ddy=x,x=

4、y¤.DRR√R1x21153433I=dx2dy(x+y)=dx(x+x2−x)=.0x022140RπRπsinyRπRysinyRπ8.dxdy=dydx=dysiny=2.0xy00y0R2R2R2R2R29.dx2ysin(xy)dy=dydx2ysin(xy)=dy2(1−cos2y)=4−sin4.0x0001RR√10.y21−x2dxdy,D={(x,y)

5、x2+y2≤1}.DR1R√1−x2√R1I=4dxdyy21−x2=4dx1(1−x2)2=32.000345RR11.(

6、x

7、+y)dxdy,D={(x,y)

8、

9、x

10、

11、+

12、y

13、≤1}.DRRRRR1R1−xR12I=

14、x

15、dxdy+ydxdy=4dxdyx+0=4dxx(1−x)=.0003DDRR12.(x+y)dxdy,2222Ù¥DǑdx+y=1,x+y=2y¤«¥m¬.D√√√RRRRR3R1−x2R3√I=xdxdy+ydxdy=0+22dx√ydy=22dx(1−x2−01−1−x20D√D1)=π−3.234

16、^4
egÈ©½È©.R1R√1−x2RπR113.dx(x2+y2)dy=2dθr2rdr=π.00008R0R0√√2R32πR1214.−1dx−1−x21+x2+y

17、2dy=πdθ01+rrdr=π(1−ln2).√R2R1−(x−1)2Rπ2R2cosθRπ2515.dx3xydy=dθ3rcosθrsinθrdr=dθ12cosθsinθ=000002.RRR√R2−x2RπRR16.dxln(1+x2+y2)dy=2dθln(1+r2)rdr=π[(1+R2)ln(1+00004R2)−R2].RR17.1dxdy,Dπ222x2´dy=αx,y=βx(2>β>α>0),x+y=a,Dx2+y2=b2(b>a>0)RR¤31Ü©.arctanβb1bI=arctanαdθa(rcosθ)2rd

18、r=(β−α)lna.RR18.rdσ,Ù¥D´d%9r=a(1+cosθ)±r=a(a>0)¤ØD¹4:«R.RRπa(1+cosθ)πI=2dθrrdr=2dθa3(cosθ+cos2θ+1cos3θ)=(22+π)a3.−πa−π3922219.

19、^È©AÛ¿Ây²:dθ=α,r=βr=r(θ)Rβ(α≤θ≤β)12¤«D¡È«¤2α[r(θ)]dθ.RRRβRr(θ)1Rβ2y.S=dxdy=αdθ0rdr=2α[r(θ)]dθ.D20.%9r=a(1+cosθ)(a>0,0≤θ<2π)¤«¡È.R2πRa(1+co

20、sθ)R2π12232).S=0dθ0rdr=0dθ2a(1+cosθ)=2πa.eÈ©.RR21.(2x2−xy−y2)dxdy,Ù¥Ddy=−2x+4,y=−2x+7,