这道积分题目怎么用万能公式法做?高等数学
1个回答
2017-09-12
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标准球坐标
x²+y²+(z-a)² = a²
x²+y²+z² = 2az
x = r sinφ cosθ
y = r sinφ sinθ
z = r cosφ
dV = r²sinφ drdφdθ
Ω方程变为:r = 2acosφ
由于整个球面在xOy面上,所以0 ≤ φ ≤ π/2
∫_(Ω) (x²+y²+z²) dV
= ∫(0,2π) dθ ∫(0,π/2) sinφ dφ ∫(0,2acosφ) r² * r² dr
= (2π)∫(0,π/2) sinφ * (1/5)(32a⁵cos⁵φ) dφ
= (2π)(1/5)(32a⁵)(- 1)∫(0,π/2) cos⁵φ d(cosφ)
= (2π)(1/5)(32a⁵)(- 1)(1/6)[ cos⁶φ ]|(0,π/2)
= (2π)(1/5)(32a⁵)(- 1)(1/6)(0 - 1)
= 32πa⁵/15
x²+y²+(z-a)² = a²
x²+y²+z² = 2az
x = r sinφ cosθ
y = r sinφ sinθ
z = r cosφ
dV = r²sinφ drdφdθ
Ω方程变为:r = 2acosφ
由于整个球面在xOy面上,所以0 ≤ φ ≤ π/2
∫_(Ω) (x²+y²+z²) dV
= ∫(0,2π) dθ ∫(0,π/2) sinφ dφ ∫(0,2acosφ) r² * r² dr
= (2π)∫(0,π/2) sinφ * (1/5)(32a⁵cos⁵φ) dφ
= (2π)(1/5)(32a⁵)(- 1)∫(0,π/2) cos⁵φ d(cosφ)
= (2π)(1/5)(32a⁵)(- 1)(1/6)[ cos⁶φ ]|(0,π/2)
= (2π)(1/5)(32a⁵)(- 1)(1/6)(0 - 1)
= 32πa⁵/15
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