求大神I=∫∫∑√(x^2+y^2)dsx^2+y^2+z^2=R^2（z>=0）求解

Question

求大神I=∫∫∑√(x^2+y^2)ds x^2+y^2+z^2=R^2（z>=0）求解

Answer 1

x² + y² + z² = R²在z ≥ 0的部分
取Σ：z = √(R² - x² - y²)
∂z/∂x = - x/√(R² - x² - y²)、∂z/∂y = - y/√(R² - x² - y²)
∫∫Σ √(x² + y²) dS
= ∫∫D √(x² + y²) * √[1 + x²/(R² - x² - y²) + y²/(R² - x² - y²)] dxdy、D：x² + y² ≤ R²
= ∫∫D √(x² + y²) * R/√(R² - x² - y²) dxdy
= R∫(0,2π) dθ ∫(0,R) r * 1/√(R² - r²) * r dr、r = Rsinz、dr = Rcosz dz
= R * 2π * ∫(0,π/2) R²sin²z * 1/(Rcosz) * (Rcosz) dz
= 2πR³∫(0,π/2) (1 - cos(2z))/2 dz
= πR³ * [z - (1/2)sin(2z)]：(0,π/2)
= (π²R³)/2