I=∫∫（D）ydσ
=∫(-2->1) ydy ∫(y^2->2-y) dx
=-9/4
2
∫(0->∞) e^(-x^2)dx=∫(0->∞) e^(-y^2)dy
所以∫(0->∞) e^(-x^2)dx
= √[∫(0->∞) e^(-x^2)dx * ∫(0->∞) e^(-y^2)dy]
=√[∫(0->∞)∫(0->∞)e^(-x^2-y^2)dxdy]
下面用极坐标计算∫(0->∞)∫(0->∞)e^(-x^2-y^2)dxdy
∫(0->∞)∫(0->∞)e^(-x^2-y^2)dxdy
=∫(0->π/2) ∫(0->+∞)e^(-r^2)rdrdθ
=(1/2)∫(0->π/2)dθ ∫(0->+∞) e^(-r^2)dr^2
=(π/4)∫(0->+∞) e^(-r^2)dr^2
=-(π/4)e^(-r^2) |∫(0->+∞)
=π/4
所以∫(0->∞) e^(-x^2)dx=√[∫(0->∞)∫(0->∞)e^(-x^2-y^2)dxdy]=(√π)/2
匿名回答于2024-05-17 19:24:50
