Question

数学第7题

Answer 1

1. 积分范围为 1/4 圆：x² + y² =1 在第 1 象限。换成极坐标，则 r =[0, 1]，θ = [0, π/2] 

   那么，这个积分就可以变换为：

   =∫∫√[(1-r²)/(1+r²)] * r * dr*dθ

   =∫dθ∫√[(1-r²)/(1+r²)] * r*dr

   =π/2 * ∫√[(1-r²)/(1+r²)] * 1/2 * d(r²)

   =π/4 * ∫√[(1-u)/(1+u)] * du                     注：u = r²，u 的积分范围 = [0, 1]

   =π/4 * ∫√(1-u²) /(1+u) * du

   =π/4 * ∫cosα * d(sinα)/(1+sinα)        注：设 u = sinα，则 du = d(sinα)，α = [0, π/2]

   =π/4 * ∫cos²α * dα/(1+sinα)

   =π/4 * ∫(1-sin²α)*dα/(1+sinα)

   =π/4 * ∫(1-sinα)*dα

   =π/4 * [∫dα - ∫sinα * dα]

   =π/4 * [α + cosα]|α=0→π/2

   =π/4 * [(π/2 -0) + cos(π/2) - cos0]

   =π/4 * (π/2 - 0 - 1)

   =π²/8 - π/4

2. 条件不全，无法计算。