20楼: Originally posted by feixiaolin at 2013-12-03 11:05:19
依公式
面积=2*pi*x对sqrt(delta_x^2+ delta_y^2)在上的积分，
而sqrt(delta_x^2+ delta_y^2)= sqrt((delta_x/delta_y)^2+ 1)*dy
= sqrt((1/(dy/dx)^2+ 1)*dy
故，
Integrate
= Integrate
= pi/n* Integrate ...

20楼: Originally posted by feixiaolin at 2013-12-03 11:05:19
依公式
面积=2*pi*x对sqrt(delta_x^2+ delta_y^2)在上的积分，
而sqrt(delta_x^2+ delta_y^2)= sqrt((delta_x/delta_y)^2+ 1)*dy
= sqrt((1/(dy/dx)^2+ 1)*dy
故，
Integrate
= Integrate
= pi/n* Integrate ...

22楼: Originally posted by douhaosuda at 2013-12-03 15:06:08
没写完吗？还是？？？？...

23楼: Originally posted by feixiaolin at 2013-12-03 15:21:47
可积分，照那个思路做就可以了。

如果是绕x轴旋转，积分公式是面积=2*pi*y*sqrt((1+(dy/dx)^2)*dx
在x的某一区间上的积分，...