Question

高数题求解答

设平面曲线L:x²+y²=1，y≥0，则∫L arctan(e的x²＋y²次方)ds=

Answer 1

L : x²+y² = 1，y = √(1-x^2), y' = -x/√(1-x^2)
ds = √(1+y'^2)dx = dx/√(1-x^2)
∫<L> arctan[e^(x²＋y²)]ds = ∫<-1, 1> arctan(e^1)dx/√(1-x^2)
= ∫<-1, 1> arctan(e^1)dx/√(1-x^2)
= 2arctane∫<0, 1> dx/√(1-x^2) 令 x = sint
= 2arctane∫<0, π/2> costdt/cost = πarctane

Answer 2

见下图：