三重积分z=√(5-x^2-y^2)及x^2+y^2=4z

问题描述:

三重积分z=√(5-x^2-y^2)及x^2+y^2=4z
答案是2(5√5-4)π/3

Ω2:{ z = √(5 - x² - y²) --> x² + y² + z² = 5、上球
Ω1:{ x² + y² = 4z、抛物面
5 - z² = 4z --> z = 1
圆环Dz:x² + y² = 4、r = 2
Ω = Ω1 + Ω2
体积∫∫∫ dV = ∫(0→1) dz ∫∫(Dz) dxdy + ∫(1→√5) ∫∫(Dz) dxdy
= ∫(0→1) π(4z) dz + ∫(1→√5) π(5 - z²) dz
= 4π • (1/2)[ z² ] |(0→1) + π • [ 5z - (1/3)z³ ] |(1→√5)
= 2π + π • [ (5√5 - (1/3) • 5√5) - (5 - 1/3) ]
= (2/3)(5√5 - 4)π
体积∫∫∫ dV
= ∫(0→2π) dθ ∫(0→2) r dr ∫(r²/4→1) dz + ∫(0→2π) dθ ∫(0→2) r dr ∫(1→√(5 - r²)) dz
= 2π∫(0→2) r • (1 - r²/4) dr + 2π∫(0→2) r • [√(5 - r²) - 1] dr
= 2π∫(0→2) [ r - r³/4 + r√(5 - r²) - r ] dr
= 2π • (5√5 - 4)/3
= (2/3)(5√5 - 4)π