求由星形线x^(2/3)+y^(2/3)=a^(2/3)所围成的图形绕x轴旋转而成的旋转体的体积 答案:(32/105)πa^3

Question

如题所述

Answer 1

y=±[a^（2/3）-x^(2/3)]^(3/2),
星形线分成上下两个半支，考虑X轴对称关系，只求上半支即可，
从-a至a以Y轴左右对称，可求从0至a积分,再乘以2，
V=2π∫[0,a]{[a^(2/3)-x^(2/3)]^(3/2)}^2dx
=2π∫[0，a][a^2-3a^(4/3)x^(2/3)+3a^(2/3)x^(4/3)-x^2)dx
=2π[0,a][a^2x-3a^(4/3)x^(2/3+1)/(2/3+1)+3a^(2/3)x^(4/3+1)/(4/3+1)-x^3/3]
=2π[a^3-9a^3/5+9a^3/7-a^3/3)
=32πa^3/105.
用到差的立方公式，(a-b)^3=a^3-3a^2b+3ab^2-b^3.