求摆线x=a(t-sint),y=a(1-cost),0≤t≤2π.与x轴所围成图形绕y轴旋转所的旋转体的体积

2025-01-24 11:32:52
推荐回答(1个)
回答1:

首先取体积微元,在x=a(t-sint)处,x变化量为dx,形成的圆环面积为:
dS=2πxdx,
圆环所在柱面体积:dV=ydS=2πxydx
又dx=d[a(t-sint)]=a(1-cost)dt
将x,y参数方程代入得:
dV=2π[a(t-sint)][a(1-cost)][a(1-cost)dt]=2πa3(t-sint)(1-cost)2dt
V=

a3(t?sint)(1?cost)2dt
作变换u=t-π,则 t=u+π,dt=du,
原积分变为:
V=
a3[(u+π)?sin(u+π)]?[1?cos(u+π)]2du
 
=a3
[π+(u+sinu)](1+cosu)2du
=2π2a3
(1+cosu)2du+
a3
(u+sinu)(1+cosu)2du

上式积分的第二部分被积函数 (u+sinu)(1+cosu)2为奇函数,因此在[-π,π]上,积分为0
V=2π2a3
(1+cosu)2du
=2π2a3
(1+2cosu+cos2u)du

=4π2a3+4π2a3
cosudu
+π2a3
(1+cos2u)du

=4π2a3?4π2a3sinu
|
+2π2a3?
1
2
π2
a3sin2u
|

=6π2a3