二重积分典型习题

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1、第十章重积分一.将二重积分I=∫∫f(x,y)dσ化为累次积分(两种形式),其中D给定如下:D221.D:由y=8x与x=8y所围之区域.2.D:由x=3,x=5,x－2y+1=0及x－2y+7=0所围之区域.223.D:由x+y≤1,y≥x及x>0所围之区域.4.D:由

2、x

3、+

4、y

5、≤1所围之区域.28x4y解.1.I=f(x,y)dσ=dxf(x,y)dy=dyy2f(x,y)dx∫∫∫0∫x2∫0∫D8x+752.I=f(x,y)dσ=dx2f(x,y)dy∫∫∫3∫x+1D232y−15565=∫2dy∫3f(x,y)dx+∫3dy∫3f(x,y)dx+∫

6、5dy∫2y−7f(x,y)dx221−x23.I=∫∫f(x,y)dσ=∫dx∫f(x,y)dy0xD2y11−y22=∫0dy∫0f(x,y)dx+∫2dy∫0f(x,y)dx20x+111−x4.I=∫∫f(x,y)dσ=∫−1dx∫−x−1f(x,y)dy+∫0dx∫x−1f(x,y)dyD0y+111−y=∫−1dy∫−y−1f(x,y)dx+∫0dy∫y−1f(x,y)dx二.改变下列积分次序:2223−xaa−x1x321.∫dx∫a2−x2f(x,y)dy2.∫dx∫f(x,y)dy+∫dx∫f(x,y)dy000102a2202−x12−x3.∫

7、−1dx∫−xf(x,y)dy+∫0dx∫xf(x,y)dy22a2222aa−xa−yaa−y2解:1.dxa2−x2f(x,y)dy=dyf(x,y)dx+adyf(x,y)dx∫0∫∫0∫a2−2ay∫∫02a223−x1x313−2y2.∫dx∫f(x,y)dy+∫dx∫2f(x,y)dy=∫dy∫f(x,y)dx00100y2202−x12−x3.∫−1dx∫−xf(x,y)dy+∫0dx∫xf(x,y)dy11y22−y=∫0dy∫−yf(x,y)dx+∫0dy∫−2−yf(x,y)dx三.将二重积分I=∫∫f(x,y)dσ化为极坐标形式的累次积分,其

8、中:D1.D:a2≤x2+y2≤b2,y≥0,(b>a>0)2.D:x2+y2≤y,x≥03.D:0≤x+y≤1,0≤x≤1πb解.1.I=∫∫f(x,y)dσ=∫dθ∫f(ρcosθ,ρsinθ)ρdρ0aDπsinθ2.I=∫∫f(x,y)dσ=∫2dθ∫f(ρcosθ,ρsinθ)ρdρ00D103.I=f(x,y)dσ=dθcosθf(ρcosθ,ρsinθ)ρdρ∫∫∫−π∫0D4π1+∫2dθ∫cosθ+sinθf(ρcosθ,ρsinθ)ρdρ00四.求解下列二重积分:2xπx42πx1.∫1dx∫xsindy+∫2dx∫xsindy2y2y2y1x

9、−2.∫dx∫e2dy00y3.dxdy,D:由y=x4－x3的上凸弧段部分与x轴所形成的曲边梯形∫∫6xDxy4.dxdy,D:y≥x及1≤x2+y2≤2∫∫22x+yD解.2y22xπx42πx2yπx22πx1.∫1dx∫xsindy+∫2dx∫xsindy=∫1dy∫ysindx=−∫1ycosdy2y2y2yπ2yy22πy42πy=−ycosdy=−ydsinπ∫12π2∫1224πy42πy=−ysin+sindyπ22π2∫1212248πy4=−cos=(π+2)233ππ2π12222yyyy1x−1−11−1−2.dxe2dy=e2dydx=

10、e2dy−y2e2dy∫0∫0∫0∫y2∫0∫0222212yyyyy11−1−1−−1−−=∫e2dy+∫yde2=∫e2dy+ye2−∫e2dy=e200000y433.dxdy,D:由y=x−x的上凸弧段部分与x轴所形成的曲边梯形.∫∫6xD3221解.y'=4x−3x,y''=12x−6x=6x(2x−1)<0.解得0

11、解.使用极坐标变换xy5π42ρcosθρsinθdxdy=dθρdρ∫∫x2+y2∫π∫1ρ24D15π24=∫πsin2θdθ∫1ρdρ=024五.计算下列二重积分:2222⎛x⎞⎛y⎞xy1.1−⎜⎟−⎜⎟dxdy,D:+≤1.∫∫22D⎝a⎠⎝b⎠ab解.令x=aρcosθ,y=bρsinθ.雅可比行列式为∂(x,y)x'ρx'θacosθ−aρsinθ===abρ∂(ρ,θ)y'ρy'θbsinθbρcosθ2231⎛x⎞⎛y⎞2π121221−⎟−⎟dxdy=dθ1−ρabρdρ=−2πab(1−ρ)2=πab∫∫⎜a⎜b∫0∫033D⎝⎠⎝⎠0322

12、222+2