求由曲线y=e^[-(x^2)(x>=0),x轴,y轴所围图形绕x轴和y轴旋转所成之旋转体的体积

Question

已知∫(0,+∞)e^[-(x^2)]dx=√π/2
答案是(1/2√2)*π^(3/2)；π
求详细解答，谢谢

Answer 1

基本体积公式：
围绕x轴：πy² = π∫ [f(x)]² dx
围绕y轴：πx² = π∫ [f(y)]² dy
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围绕x轴旋转：
V = π∫(0→+∞) [e^(-x²)]² dx
= π∫(0→+∞) e^(-2x²) dx
令u = 2x²，du = 4x dx
= π∫(0→+∞) e^(-u) • du/(4x)
= [π/(2√2)]∫(0→+∞) u^(-1/2) e^(-u) du
= [π/(2√2)] • Γ(1/2)
= [π/(2√2)] • √π
= [π^(3/2)]/(2√2)
围绕y轴旋转：
y = e^(-x²) => x² = -lny
V = π∫(0→1) -lny dy
= -π(ylny)|(0→1) + π∫(0→1) dy
= 0 + π•y|(0→1)
= π•(1 - 0)
= π