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.若D是平面区域{1≤x2+y2≤4},则∬Ddxdy=(  C  )
A.πB.4πC.3πD.15π
解:∬Ddxdy=SD=π(22−12)=3π

2.若D是平面区域−1≤x≤1,0≤y≤1,则∬D(x3+2y)dxdy=(  C  )
A.0B.1C.2D.3
解:∬D(x3+2y)dxdy=∫1−1dx∫10(x3+2y)dy=∫1−1[x3y+y2]10dx=∫1−1(x3+1)dx=[14x4+x]1−1=2

(二)填空题
1.交换二次积分次序,I=∫10dx∫x0f(x,y)dy=  ∫10dy∫1yf(x,y)dx  −−−−−−−−−−−−−−−−
解:I=∫10dx∫x0f(x,y)dy=∫10dy∫1yf(x,y)dx

2.交换二次积分次序,I=∫10dx∫1x√f(x,y)dy=  ∫10dy∫y20f(x,y)dx  −−−−−−−−−−−−−−−−−
解:已知D:0≤x≤1,x√≤y≤1;变换后,0≤y≤1,0≤x≤y2I=∫10dx∫1x√f(x,y)=∫10dy∫y20f(x,y)dx

3.交换二次积分次序,I=∫e1dx∫lnx0f(x,y)dy=  ∫10dy∫eeyf(x,y)dx  −−−−−−−−−−−−−−−−
解:已知D:1≤x≤e,0≤y≤lnx;变换后:0≤y≤1,ey≤x≤e;I=∫e1dx∫lnx0f(x,y)dy=∫10dy∫eeyf(x,y)dx

4.设区域D由x2+y2=2围成,则∬D(x+x3)dxdy=  0  −−−
解:D区域对称,x+x3是奇函数,积分结果为0

5.设f(x,y)为连续函数,则∬x2≤y≤1xyf(x2+y2)dσ=  0  −−−
解:y≥x2≤0,−1≤x≤1,D关于y轴对称F(x,y)=xyf(x2+y2)F(−x,y)=−xyf(x2+y2)=−f(x,y),关于x为奇函数积分等于0

(三)解答题
1.计算∬D(x2+y)dxdy,其中D是由曲线y=x2与y2=x围成的平面区域
解:解得方程y=x2,y2=x的交点(0,0),(1,1)D:0≤x≤1,x2≤y≤x√∬D(x2+y)dxdy=∫10dx∫x√x2(x2+y)dy=∫10[x2y+12y2]x√x2dx=∫10[x52+12x−32x4]dx=[27x72+14x2−310x5]10=33140

2.计算∬Dyxdxdy,其中D是有直线y=x,y=2x,x=2及x=4围成的平面区域
解:D:2≤x≤4,x≤y≤2x;∬Dyxdxdy=∫42dx∫2xxyxdy=∫42[y22x]2xxdx=∫423x2dx=[34x2]42=9

3.计算∬Dx2ydxdy,其中D是由曲线xy=1,y=x√,x=2所围成的平面区域
解:D:1≤x≤2,1x≤y≤x√;∬Dx2ydxdy=∫21dx∫x√1xx2ydy=∫21[12x2y2]x√1xdx=∫21[12x3−12]dx=[18x4−12x]21=118

4.计算∬D(x2+y2)dxdy,其中D是由曲线y=x2与y=x围成的平面区域
解:D:0≤x≤1,x2≤y≤x;∬D(x2+y2)dxdy=∫10dx∫xx2(x2+y2)dy=∫10[x2y+13y3]xx2dx=∫10[43x3−x4−13x6]dx=[13x4−15x5−121x7]10=335

5.计算∬Dx2ydxdy,其中D是由曲线x2+y2=2,y=x及y轴围成的在第一象限内的区域.
解:D的极坐标区域:π4≤θ≤π2,0≤r≤2√;∬Dx2ydxdy=∫π2π4dθ∫2√0(rcosθ)2rsinθrdr=∫π2π4cos2θsinθdθ∫2√0r4dr=∫π2π442√5cos2θsinθdθ=−42√5∫π2π4cos2θdcosθ=−42√5[13cos3θ]π2π4=215

6.计算二重积分I=∬Df(x,y)dxdy,其中D为x2+y2≤4,
f(x,y)={1,x2+y2≤1,xy2,1≤x2+y2≤4.
解:极坐标区域D1:0≤θ≤2π,0≤r≤1;极坐标区域D2:0≤θ≤2π,1≤r≤2;I=∬Df(x,y)dxdy=∫2π0dθ∫10rdr+∫2π0dθ∫21rcosθ(rsinθ)2rdr=∫2π012dθ+∫2π0cosθsin2θ[15r5]21dθ=π+315∫2π0sin2θdsinθ=π+315[13sin3θ]2π0=π

7.计算∬D(x−3y2)dxdy,其中D为0≤x≤2,1≤y≤2
解:∬D(x−3y2)dxdy=∫20dx∫21(x−3y2)dy=∫20[xy−y3]21dx=∫20[x−7]dx=[12x2−7x]20=−12

8.计算∬Dysinxydσ,其中D为1≤x≤2,0≤y≤π
解:∬Dysinxy=∫π0dy∫21ysinxydx=∫π0dy∫21d(−cosxy)=∫π0[−cosxy]21dy=∫π0[cosy−cos2y]dy=[siny−12sin2y]π0=0

9.计算∬De−(x2+y2)dσ,其中D为1≤x2+y2≤4
解:极坐标区域D:0≤θ≤2π,1≤r≤2;∬De−(x2+y2)dσ=∫2π0dθ∫21e−[(rcostheta)2+(rsinθ)2]rdr=∫2π0dθ∫21e−r2rdr=∫2π0[−12e−r2]21dθ=∫2π0[12e−1−12e−4]dθ=(e−1−e−4)π

10.计算∬Ddσ4−x2−y2−−−−−−−−−√,其中D为x2+y2≤1
解:极坐标区域D:0≤θ≤2π,0≤r≤1;∬Ddσ4−x2−y2−−−−−−−−−√=∫2π0dθ∫1014−[(rcosθ)2+(rsinθ)2]−−−−−−−−−−−−−−−−−−−−√rdr=∫2π0dθ∫1014−r2−−−−−√rdr=∫2π0dθ∫10[−12(4−r2)−12]d(4−r2)=∫2π0[−(4−r2)12]10dθ=[(2−3√)θ]2π0=2(2−3√)π

11.计算∬Dln(x2+y2)dxdy,其中D为1≤x2+y2≤2
解:极坐标区域D:0≤θ≤2π,1≤r≤2√;∬Dln(x2+y2)dxdy=∫2π0dθ∫2√1ln[(rcosθ)2+(rsinθ)2]⋅rdr=∫2π0dθ∫2√1(lnr)dr2=∫2π0{[r2lnr]2√1−∫2√1r2dlnr}dθ=∫2π0{[2ln2√−∫2√1rdr}dθ=∫2π0{[2ln2√−[12r2]2√1}dθ=∫2π0[2ln2√−12]dθ=[(2ln2√−12)θ]2π0=(ln4−1)π