Question

求摆线x=a(t-sint)y=a(1-cost)的一拱与x轴围成的图形绕y轴旋转所成旋转体的体积 求过程

Answer 1

垂直于x轴的微面积ydx绕y轴旋转的微体积等于лx²(ydx)；
则 V=∫лx²(ydx)=∫a³л (t-sint)²(1-cost) [(1-cost)dt]；积分区间t[0,2л]；
V=a³л∫[(t-sint)*(1-cost)]²dt=a³л∫(t-sint-tcost+sintcost)²dt
=a³л∫[(t²+sin²t+t²cos²t+sin²tcos²t)-(tsint+t²cost-tsintcost)+(tsintcost-sin²tcost)-tsintcos²t]dt
=a³л[t²+(1-cos2t)/2+t²(1+cos2t)/2+(sin2t)²/4-(tsint+t²cost)+t(sin2t)-sin²tcost-tsint[(1+cos2t)/2]dt
=a³л[t³/3+t/2+(t³/6+tcos2t/4)+t/8-(-tcost+2tcost)-t(cos2t)/2-0+tcost/2+tcos(3t/2)/6-tcos(t/2)/2] {0,2л}；
=a³л[8л³/2+2л*5/8+2л*(1/4-1-1/2+1/2)+2л*(-1)-2л*(-1)]
=a³л(4л³+л/4)；
积分运算较繁琐，结果供参考；