大一高等数学用二重积分求重心 第四题
1个回答
展开全部
由对称性, 质心纵坐标 q = 0。
质心横坐标 p = m<x>/m
m = ∫∫<D>dxdy = (π/4)(b^2-a^2)
m<x> = ∫∫<D>xdxdy
= ∫<-π/2,π/2>dt∫<acost, bcost>rcost rdr
= (1/3)(b^3-a^3)∫<-π/2,π/2>(cost)^4dt
= (1/12)(b^3-a^3)∫<-π/2,π/2>(1+cos2t)^2dt
= (1/12)(b^3-a^3)∫<-π/2,π/2>[1+2cos2t+(cos2t)^2)]dt
= (1/12)(b^3-a^3)∫<-π/2,π/2>[3/2+2cos2t+(1/2)cos4t]dt
= (1/12)(b^3-a^3)[3t/2+sin2t+(1/8)sin4t]<-π/2,π/2>
= (π/8)(b^3-a^3)
p = m<x>/m = (a^2+ab+b^2)/[2(a+b)]
质心横坐标 p = m<x>/m
m = ∫∫<D>dxdy = (π/4)(b^2-a^2)
m<x> = ∫∫<D>xdxdy
= ∫<-π/2,π/2>dt∫<acost, bcost>rcost rdr
= (1/3)(b^3-a^3)∫<-π/2,π/2>(cost)^4dt
= (1/12)(b^3-a^3)∫<-π/2,π/2>(1+cos2t)^2dt
= (1/12)(b^3-a^3)∫<-π/2,π/2>[1+2cos2t+(cos2t)^2)]dt
= (1/12)(b^3-a^3)∫<-π/2,π/2>[3/2+2cos2t+(1/2)cos4t]dt
= (1/12)(b^3-a^3)[3t/2+sin2t+(1/8)sin4t]<-π/2,π/2>
= (π/8)(b^3-a^3)
p = m<x>/m = (a^2+ab+b^2)/[2(a+b)]
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询