Question

计算曲面积分∫根号下（x^2+y^2）ds，其中L：x^2+y^2=-2y，答案是8，求过程。。。

Answer 1

积分曲线x^2+(y+1)^2=1
所以参数方程是x=cost, y=-1+sint. t∈[0,2π]
ds=√[(x't)^2+(y't)^2]dt= dt
∫√（x^2+y^2）ds
=∫√(-2y)ds
=∫√[2(1-sint)] dt
=√2 ∫(0->2π) |sin(t/2)-cos(t/2)|dt
=2√2 ∫(0->2π) |sin(t/2)-cos(t/2)|d(t/2)
=2√2 ∫(0->π) |sinu-cosu|du
=2√2 [∫(0->π/4) (cosu-sinu)du + ∫(π/4->π) (sinu-cosu)du]
=2√2(√2+1)

追问

你答案好像算错了。。不过方法我懂啦，谢啦