L是上半圆周y=(1-x^2)^1/2,则∫y^3ds=()(a)0(b)2∫(0~1)y^3dy(c)∫(-1~1)(1-x^2)^3/2dx(d)2∫(0~1)(1-x^2)dx

Question

L是上半圆周y=(1-x^2)^1/2,则 ∫ y^3ds=()
(a)0 (b)2∫(0~1) y^3dy
(c) ∫(-1~1)(1-x^2)^3/2dx (d)2∫(0~1)(1-x^2)dx

Answer 1

∵y=√(1-x²),则y'=-x/√(1-x²)
∴ds=√(1+y'²)dx
=√(1+(-x/√(1-x²))²)dx
=dx/√(1-x²)
于是,∫ y³ds=∫(√(1-x²))³dx/√(1-x²)
=∫(1-x²)dx
=∫(1-x²)dx+∫(1-x²)dx
=-∫(1-x²)dx+∫(1-x²)dx (在第一个积分中,用-x代换x)
=∫(1-x²)dx+∫(1-x²)dx (在第一个积分中,交换上下限)
=2∫(1-x²)dx
故 应该选择答案(d)2∫(1-x²)dx.