
1个回答
展开全部
∫∫∫<Ω>z^2dv = ∫<-c, c>z^2dz∫∫dxdy
= ∫<-c, c>z^2[πab(1-z^2/c^2)]dz = (4/15)πabc^3;
由对称轮换性 ∫∫∫<Ω>x^2dv = (4/15)πbca^3,
∫∫∫<Ω>y^2dv = (4/15)πacb^3
I = ∫∫∫<Ω>(x^2+y^2+z^2)dv = (4/15)πabc(a^2+b^2+c^2)
= ∫<-c, c>z^2[πab(1-z^2/c^2)]dz = (4/15)πabc^3;
由对称轮换性 ∫∫∫<Ω>x^2dv = (4/15)πbca^3,
∫∫∫<Ω>y^2dv = (4/15)πacb^3
I = ∫∫∫<Ω>(x^2+y^2+z^2)dv = (4/15)πabc(a^2+b^2+c^2)
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询