Question

定积分,求椭球体x^2/a^2+y^2/b^2+z^2/c^2

Answer 1

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

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