y^3–y=x и x=0
y^3-y=0
y(y^2-1)=0
y=0; y= ± 1
V_(Оу)=π ∫ ^(b)_(a)g^2(y)dy
[b]g(y)=y^3-y[/b]
Так как области симметричны
V_(Оу)=2π ∫ ^(1)_(0)(y^3-y)^2dy=
=2π∫ ^(1)_(0)(y^6-2y^4+y^2)dy=
=2π*((y^7/7)-2(y^5/5)+(y^3/3))|^(1)_(0)=
=2π*((1/7)-(2/5)+(1/3))=16π/108= [b]4π/27[/b]