Question

求第一类曲线积分∬z ds，其中锥面z=√(x^2+y^2 )在柱体x^2+y^2≤2x内的部分

Answer 1

{ z = √(x^2 + y^2)
{ x^2 + y^2 = 2x ==> (x - 1)^2 + y^2 = 1 <-- Σ在xy面下的投影域D
积分域关于x轴对称
考虑在xy面上的积分。
z'x = x/√(x^2 + y^2)、z'y = y/√(x^2 + y^2)
∫∫Σ z dS
= ∫∫D √(x^2 + y^2) * √[1 + x^2/(x^2 + y^2) + y^2/(x^2 + y^2)] dxdy
= ∫∫D √(x^2 + y^2) * √2 dxdy
= √2∫(- π/2→π/2) dθ ∫(0→2cosθ) r^2 dr
= 2√2∫(0→π/2) (1/3)r^3：(0→2cosθ) dθ
= (2√2)/3 * ∫(0→π/2) 8cos^3(θ) dθ
= (16√2)/3 * 2/(3 * 1)
= (32√2)/9