Question

计算椭圆公式x^2/a^2+y^2/b^2=1(a>b)分别绕长轴和短轴旋转生成旋转体的体积

Answer 1

考虑对称性，只对第一象限的1/4图形旋转，再乘以2即可。
椭圆方程：y^2=b^2-b^2x^2/a^2, x^2=a^2-a^2y^2/b^2
绕X轴体积，V1=2π∫[0，a] (b^2-b^2x^2/a^2)dx
=2π(b^2x-b^2x^3/3)[0,a]
=2π[b^2a-b^2a^3/(3a^2)]
=2π(2ab^2)/3
=4πab^2/3,
同理绕Y轴体积：
V2=2π∫[0，b] (a^2-a^2y^2/b^2)dy
=2π[0，b][a^2y-a^2y^3/(3b^2)]
=2π[a^2b-a^2b^3/(3b^2)]
=2π(2a^2b/3)
=4πa^2b/3.

Answer 2

Vx=π∫y^2dx(上限是a，下限是-a）
=4/3πab^2
Vy=π∫x^2dy(上限是b，下限是-b)
=4/3πba^2