Question

三道高数题求解

求过程！

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Answer 1

1、u=u(x,y)
du=əu/əx*dx+əu/əy*dy
du/dx=əu/əx+əu/əy*dy/dx
满足u(x,y)|(y=x²)=1，则此时
du/dx=0=əu/əx+əu/əy*dy/dx
而əu/əx|(y=x²)=x，dy/dx|(y=x²)=2x
∴0=x+əu/əy*2x
∴əu/əy|(y=x²)=-1/2
选A
2、∫∫∫y²z²dv
=∫<0,1>y²dy∫<0,1>dx∫<-(x+y),(x+y)>z²dz
=∫<0,1>y²dy∫<0,1>dx[<-(x+y),(x+y)>z³/3]
=∫<0,1>y²dy∫<0,1>2(x+y)³/3dx
=2/3∫<0,1>y²dy∫<0,1>(x+y)³d(x+y)
=2/3∫<0,1>y²[<0,1>(x+y)^4/4]dy
=1/6∫<0,1>y²[(1+y)^4-y^4]dy
=1/6∫<0,1>y²[(1+4y+6y²+4y³+y^4)-y^4]dy
=1/6∫<0,1>(y²+4y³+6y^4+4y^5)dy
=1/6[<0,1>(y³/3+y^4+6y^5/5+4y^6/6)]
=1/6(1/3+1+6/5+4/6)
=1/6*16/5
=8/15
选C
3、I=∫∫√(4x²-y²)dxdy
=∫<0,1>dx∫<0,x>√(4x²-y²)dy
令y=2xsint，0≤t≤π/6，则
√(4x²-y²)=2xcost，dy=2xcostdt
∴I=∫<0,1>dx∫<0,π/6>2xcost*2xcostdt
=∫<0,1>x²dx∫<0,π/6>4cos²tdt
=∫<0,1>x²dx∫<0,π/6>2(cos2t+1)dt
=∫<0,1>x²dx∫<0,π/6>(cos2t+1)d2t
=∫<0,1>x²dx[<0,π/6>(sin2t+2t)
=(√3/2+π/3)∫<0,1>x²dx
=(√3/2+π/3)[<0,1>x³/3]
=(√3/2+π/3)*1/3
=√3/6+π/9
选B