Question

计算由曲面z=x^2+y^2,三个坐标面及平面x+y=1所围立体的体积,答案是1/6,

Answer 1

求由x=0 y=0 x+y=1围成的三棱柱的体积

下底为z=0 上底为z=x^2+y^2(圆锥)

=∫(0,1)dx∫(0,1-x)dy∫(0,x^2+y^2)dz

=∫(0,1)dx∫(0,1-x)[z](0,x^2+y^2)dy

=∫(0,1)dx∫(0,1-x)[x^2+y^2]dy

=∫(0,1)[x^2y+y^3/3](0,1-x)dx

=∫(0,1)[x^2(1-x)+(1-x)^3/3]dx

=[x^2/3-x^/2-(1-x)^4/12](0,1)

=1/6