1个回答
展开全部
x^2+y^2 = 2x, (x-1)^2+y^2 = 1
x = 1+cost, y = sint, x'<t> = -sint, y'<穗族t>庆御 = cost
ds = √[(-cost)^2+(sint)^2]dt = dt
I = ∮<L>(x^2+y^2)ds = ∫<0, 2π>[(1+cost)^2+(sint)^2]dt
= ∫<誉族岩0, 2π>(2+2cost)dt = [2t+2sint]<0, 2π> = 4π
x = 1+cost, y = sint, x'<t> = -sint, y'<穗族t>庆御 = cost
ds = √[(-cost)^2+(sint)^2]dt = dt
I = ∮<L>(x^2+y^2)ds = ∫<0, 2π>[(1+cost)^2+(sint)^2]dt
= ∫<誉族岩0, 2π>(2+2cost)dt = [2t+2sint]<0, 2π> = 4π
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
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
