求半椭圆x^2/9 + y^2/4=1(y≥0)绕x轴一周所得到的旋转体的体积

Question

如题所述

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.