z^2=x^2+y^2, x^2+y^2+z^2 = 1, (z>0)
{z^2=x^2+y^2
{x^2+y^2+z^2=1
x^2+y^2+(x^2+y^2)=1
2x^2+2y^2=1
x^2+y^2=1/2 - окружность
V= ∫∫_(D)(sqrt(1-x^2-y^2)-sqrt(x^2+y^2))dxdy=
переходим к полярным координатам
= ∫_(0)^(2π ) [b]([/b][blue]∫_(0) ^(\sqrt(1/2))(sqrt(1- ρ ^2)-sqrt( ρ ^2)) ρ d ρ[/blue] [b])[/b]d θ =
=∫_(0)^(2π)[b]( [/b][blue](-1/2)∫_(0) ^(\sqrt(1/2))(sqrt(1-ρ ^2)*(-2ρ d ρ)-∫_(0) ^(\sqrt(1/2)) ρ ^2d [/blue] [b])[/b]d θ=
=∫_(0)^(2π )[b]( [/b](-1/2)*(1- ρ ^2)^(3/2)/(3/2)-( ρ ^3/3)[b])[/b]|_(0) ^(\sqrt(1/2))d θ =
=2π*((-1/3)*sqrt(1/8)+(1/3)sqrt(1)-(1/3)*(sqrt(1/2)^3)=...