高数 07.11 多元函数微分学习题03B二重积分

二重积分 练习题
(一)单项选择题
1.D{1x2+y24},Ddxdy=(  C  )
A.πB.4πC.3πD.15π
:Ddxdy=SD=π(2212)=3π

2.D1x1,0y1,D(x3+2y)dxdy=(  C  )
A.0B.1C.2D.3
:D(x3+2y)dxdy=11dx10(x3+2y)dy=11[x3y+y2]10dx=11(x3+1)dx=[14x4+x]11=2

(二)填空题
1.,I=10dxx0f(x,y)dy=  10dy1yf(x,y)dx  
:I=10dxx0f(x,y)dy=10dy1yf(x,y)dx

2.,I=10dx1xf(x,y)dy=  10dyy20f(x,y)dx  
:D:0x1,xy1;,0y1,0xy2I=10dx1xf(x,y)=10dyy20f(x,y)dx

3.,I=e1dxlnx0f(x,y)dy=  10dyeeyf(x,y)dx  
:D:1xe,0ylnx;:0y1,eyxe;I=e1dxlnx0f(x,y)dy=10dyeeyf(x,y)dx

4.Dx2+y2=2,D(x+x3)dxdy=  0  
:D,x+x3,0

5.f(x,y),x2y1xyf(x2+y2)dσ=  0  
:yx20,1x1,DyF(x,y)=xyf(x2+y2)F(x,y)=xyf(x2+y2)=f(x,y),x0

(三)解答题
1.D(x2+y)dxdy,D线y=x2y2=x
:y=x2,y2=x(0,0),(1,1)D:0x1,x2yxD(x2+y)dxdy=10dxxx2(x2+y)dy=10[x2y+12y2]xx2dx=10[x52+12x32x4]dx=[27x72+14x2310x5]10=33140

2.Dyxdxdy,D线y=x,y=2x,x=2x=4
:D:2x4,xy2x;Dyxdxdy=42dx2xxyxdy=42[y22x]2xxdx=423x2dx=[34x2]42=9

3.Dx2ydxdy,D线xy=1,y=x,x=2
:D:1x2,1xyx;Dx2ydxdy=21dxx1xx2ydy=21[12x2y2]x1xdx=21[12x312]dx=[18x412x]21=118

4.D(x2+y2)dxdy,D线y=x2y=x
:D:0x1,x2yx;D(x2+y2)dxdy=10dxxx2(x2+y2)dy=10[x2y+13y3]xx2dx=10[43x3x413x6]dx=[13x415x5121x7]10=335

5.Dx2ydxdy,D线x2+y2=2,y=xy.
:D:π4θπ2,0r2;Dx2ydxdy=π2π4dθ20(rcosθ)2rsinθrdr=π2π4cos2θsinθdθ20r4dr=π2π4425cos2θsinθdθ=425π2π4cos2θdcosθ=425[13cos3θ]π2π4=215

6.I=Df(x,y)dxdy,Dx2+y24,
f(x,y)={1,x2+y21,xy2,1x2+y24.
:D1:0θ2π,0r1;D2:0θ2π,1r2;I=Df(x,y)dxdy=2π0dθ10rdr+2π0dθ21rcosθ(rsinθ)2rdr=2π012dθ+2π0cosθsin2θ[15r5]21dθ=π+3152π0sin2θdsinθ=π+315[13sin3θ]2π0=π

7.D(x3y2)dxdy,D0x2,1y2
:D(x3y2)dxdy=20dx21(x3y2)dy=20[xyy3]21dx=20[x7]dx=[12x27x]20=12

8.Dysinxydσ,D1x2,0yπ
:Dysinxy=π0dy21ysinxydx=π0dy21d(cosxy)=π0[cosxy]21dy=π0[cosycos2y]dy=[siny12sin2y]π0=0

9.De(x2+y2)dσ,D1x2+y24
:D:0θ2π,1r2;De(x2+y2)dσ=2π0dθ21e[(rcostheta)2+(rsinθ)2]rdr=2π0dθ21er2rdr=2π0[12er2]21dθ=2π0[12e112e4]dθ=(e1e4)π

10.Ddσ4x2y2,Dx2+y21
:D:0θ2π,0r1;Ddσ4x2y2=2π0dθ1014[(rcosθ)2+(rsinθ)2]rdr=2π0dθ1014r2rdr=2π0dθ10[12(4r2)12]d(4r2)=2π0[(4r2)12]10dθ=[(23)θ]2π0=2(23)π

11.Dln(x2+y2)dxdy,D1x2+y22
:D:0θ2π,1r2;Dln(x2+y2)dxdy=2π0dθ21ln[(rcosθ)2+(rsinθ)2]rdr=2π0dθ21(lnr)dr2=2π0{[r2lnr]2121r2dlnr}dθ=2π0{[2ln221rdr}dθ=2π0{[2ln2[12r2]21}dθ=2π0[2ln212]dθ=[(2ln212)θ]2π0=(ln41)π

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