一道高数题 ∫∫(x²+y²)dxdy,其中D={(x,y)|0≤x≤1,√x≤y≤2√x}

原式=∫(0,1)dx∫(√x,2√x)(x^2+y^2)dy=∫(0,1)dx*[x^2*y+(1/3)*y^3]|(√x,2√x)=∫(0,1)[x^(5/2)+(7/3)*x^(3/2)]dx=[(2/7)*x^(7/2)+(14/15)*x^(5/2)]|(0,1)=128/105

∫∫D√(x²-y²)dxdy D是积分区域 范围:0≤x≤1;0≤y≤x

∫∫D√(x²-y²)dxdy= ∫[0,1] dx ∫[0,x] √(x²-y²) dy= ∫[0,1] dx {1/2[y√(x²-y²)+y^2arcsin(y/x)]|[0,x]= ∫[0,1] πx^2/4 dx= π/4 { x^3/3 |[0,1] }= π/12

计算二重积分I=∫∫√(y-x²)dxdy,其中D={(x,y)| |x|≤1,x²≤y≤1}

解:原式=∫<-1,1>dx∫<x²,1>√(y-x²)dy =(2/3)∫<-1,1>(1-x²)^(3/2)dx =(2/3)∫<-π/2,π/2>(cost)^4dt (令x=sint) =(2/3)∫<-π/2,π/2>[3/8+cos(2t)/2+cos(4t)/8]dt =(2/3)[3t/8+sin(2t)/4+cos(4t)/32]│<-π/2,π/2> =(2/3)(3π/16+3π/16) =π/4.

二重积分∫∫√x²+y²dxdy D:x²+y²≤a²

将直角坐标换为极坐标 x=rcost,y=rsint 那么 ∫∫√x²+y²dxdy D:x²+y²≤a² =∫∫r√(rcost)²+(rsint)²drdt D:0≤r≤a,0≤t≤2π=∫∫r*rdrdt=∫∫r²drdt=∫r²dr *∫dt D:0≤r≤a,0≤t≤2π=r³/3(从0到a)* t(从0到2π)=a³/3*2π=2πa³/3

高等数学二重积分 ∫∫(x+y)dxdy D:y=x y=x^2 D

D∫∫(x+y)dxdy=∫{0,1}dx∫{x²,x}(x+y)dy=∫{0,1}dx∫{x²,x}(xy+y²/2)=∫{0,1}[3x²/2-x³-(x^4)/2]dx=(x³/2-x^4/4-x^5/10) |{0,1}=1/2-1/4-1/10=3/20;

∫∫D|1-x²-y²|dxdy,其中D={(x,y)|x²+y²≤x,y≥0}

解:∵在区域D={(x,y)|x²+y²≤x,y≥0}中,1-x²-y²≥0 ∴∫∫<D>|1-x²-y²|dxdy=∫∫<D>(1-x²-y²)dxdy =∫<0,π/2>dθ∫<0,cosθ>(1-r²)rdr (作极坐标变换)=∫<0,π/2>[cos²x(2-cos²x)/2]dθ =(1/32)∫<0,π/2>[5+4cos(2θ)-cos(4θ)]dθ (应用倍角公式) =(1/32)[5(π/2)+2sin(2(π/2))-sin(4(π/2))/4] =(1/32)(5π/2) =5π/64.

∫∫(x²-y²)e∧(x+y)²dxdy,D={(x,y)|y≤x≤√(1-y2),0≤y≤√2

计算二重积分∫∫-{D} x²/y² dxdy,其中D:1≤x≤2,1/x≤y≤x?

令u=x-1,v=y-1,则d={(x,y)|x2+y2≤2x+2y}={(u,v)|u2+v2≤2}, 从而, ∬ d (x+y2)dxdy= ∬ u2+v2≤2 (u+v2+2v+2)dudv. 因为d关于u,v轴均对称,u为关于u的奇函数,2v为关于v的.

求∫∫'D(x²+y²)d6,其中D={(x,y),0≤y≤√2ax-x²,0≤x≤2a}的积分

d:y=1-x原式=∫ (0到1)dx ∫ (0到(1-x)) x+2y dy =∫ (0到1) (xy+y²)|(y=0到y=1-x)dx =∫ (0到1) (x(1-x)+(1-x)²)dx =∫ (0到1) (-2x²+x+1)dx =((-2/3)x³+(1/2)x²+x)|(从x=0到x=1)=-2.

∫∫x²y²dxdy 其中d={x.y}0≤x≤1 0≤y≤2?