Question

高数题求解~

Answer 1

答：6πa⁴

x² + y² + z² = 2az
x² + y² + (z - a)² = a²，a ≤ z ≤ 2a
运用广义球坐标：
x = a sinφ cosθ
y = a sinφ sinθ
z = a + a cosφ
dS = a² sinφ dφdθ
Σ --> Σ'：r = a，0 ≤ φ ≤ π/2
∫∫_(Σ) (x² + y² + z²) dS
= ∫∫_(Σ) 2az dS
= 2a∫∫_(Σ) z dS
= 2a∫(0,2π) dθ ∫(0,π/2) a² sinφ * (a + a cosφ) dφ
= 2a(2π)(a³誉局)∫(0,π/2) (sinφ + sinφcosφ) dφ
= 4πa⁴ * 3/2
= 6πa⁴

一般方法派虚和：
z = a + √(a² - x² - y²)
z'x = - x/√(a²-x²-y²)，z'y = - y/√(a²-x²-y²)
dS = √[ 1 + (x²+y²)/(a²-x²-y²) ] = a/√(a²-x²-y²)
∫∫_(Σ) (x²+y²+z²) dS
= ∫尘盯∫_(Σ) 2az dS
= 2a∫∫_(D) [a+√(a²-x²-y²)]*a/√(a²-x²-y²) dxdy
= 2a²∫∫_(D) [a/√(a²-x²-y²)+1] dxdy
= 2a²∫(0,2π) dθ ∫(0,a) [a/√(a²-r²)+1]*r dr
= 2a²(2π)*3a²/2
= 6πa⁴