V=∫(-a,a) S(x) dx
截面：y^2/[(1-a^2/x^2)b^2] + z^2/[(1-a^2/x^2)c^2]=1
因此,截面积S(x)=bc(1-x^2/a^2)π
那么,
V
=∫(-a,a) S(x) dx
=∫(-a,a) bc(1-x^2/a^2)π dx
=bcπ∫(-a,a) 1-x^2/a^2 dx
=bcπ(x-x^3/3a^2) | (-a.a)
=[abcπ-abcπ/3]*2
=(4/3)abcπ
有不懂欢迎追问