Question

求曲面1-z=√（x∧2+y∧2）,x=z,x=0所围成的立体体积

Answer 1

简单计算一下即可，答案如图所示

Answer 2

V=∫∫∫(Ω)dV，其中Ω是所求立体
=∫∫(D)dxdy*∫(x,1-√(x^2+y^2))dz，其中D是Ω在xoy平面上的投影

=∫∫(D)[1-x-√(x^2+y^2)]dxdy
接下来求D的范围：将x=z代入1-z=√(x^2+y^2)
(1-x)^2=x^2+y^2
1-2x=y^2
所以D={(x,y)|0<=x<=(1-y^2)/2}
V=∫(-1,1)dy∫(0,(1-y^2)/2)[1-x-√(x^2+y^2)]dx
=∫(-1,1)dy*{x-(x^2)/2-(x/2)*√(x^2+y^2)-[(y^2)/2]*ln|x+√(x^2+y^2)|}|(0,(1-y^2)/2)
=∫(-1,1)dy*{(1-y^2)/2-(1-2y^2+y^4)/8-[(1-y^2)/4]*[(1+y^2)/2]+[(y^2)/2]*ln|y|}
=∫(-1,1){(1-y^2)/4+[(y^2)/2]*ln|y|}dy
=∫(0,1)[(1-y^2)/2+y^2*lny]dy
=∫(0,1)(1-y^2)/2dy+∫(0,1)y^2*lnydy
=[y/2-(y^3)/6]|(0,1)+(1/3)*∫(0,1)lnyd(y^3)
=1/3+(1/3)*[lny*y^3|(0,1)-∫(0,1)y^2dy]
=1/3-(1/9)*(y^3)|(0,1)
=2/9