Question

求下列曲线所围成的图形绕x轴旋转一周所形成的旋转体的体积:

Answer 1

1.
y^(2/3)=a^(2/3)-x^(2/3)
故-(a^(2/3)-x^(2/3))^(3/2)<=y<=(a^(2/3)-x^(2/3))^(3/2)
所以在点x处，旋转体的切面面积为
pai*((a^(2/3)-x^(2/3))^(3/2))^2=pai*(a^(2/3)-x^(2/3))^3
=pai*(a^2-3a^(4/3)x^(2/3)+3a^(2/3)x^(4/3)-x^2)
即对-a<=x<=a进行积分，即得旋转体体积
V=积分(-a,a)(pai*(a^2-3a^(4/3)x^(2/3)+3a^(2/3)x^(4/3)-x^2))
=pai*(a^2*2a-3a^(4/3)*6/5*a^(5/3)+3a^(2/3)*6/7*a^(3/7)-2/3*a^3)
=pai*(2-3*6/5+3*6/7-2/3)a^3=32/105*pai*a^3
2.
x的取值范围是[0,ln3]
对给定x，y的取值范围是[0,e^x-1],切面面积为pai*(e^x-1)^2=pai*(e^2x-2e^x+1)
体积V=积分(0,ln3)(pai*(e^2x-2e^x+1))
=pai*((e^(2ln3)-e^(2*0))/2-2(e^ln3-e^0)+(ln3-0))
=pai*(8-4+ln3)=(4+ln3)*pai
补充:
pai是圆周率
建议自己在草稿纸上画出平面上大致图像，便于理解