Question

计算三重积分∫∫∫zdxdydz,Ω由x^2+y^2+z^2=4与z=1/3(x^2+y^2)所围的闭区域

Answer 1

联立解 x^2+y^2+z^2 = 4 与 z = (1/3)(x^2+y^2) 得交线 x^2+y^2 = 3， z = 1.
取球坐标，x = rsinφcosθ, x = rsinφsinθ, z = rcosφ,
则 x^2+y^2+z^2 = 4 变为 r = 2 ；
z = (1/3)(x^2+y^2) 变为 r = 3cosφ/(sinφ)^2， 则
∫∫∫zdxdydz
= ∫<0, π/6>dφ∫<0, 2π>dθ∫<0, 2>rcosφ r^2sinφ dr
+ ∫<π/6, π/2>dφ∫<0, 2π>dθ∫<0, 3cosφ/(sinφ)^2>rcosφ r^2sinφ dr
= 2π∫<0, π/6>cosφsinφdφ∫<0, 2>r^3dr
+ 2π∫<π/6, π/2>cosφsinφdφ∫<0, 3cosφ/(sinφ)^2>r^3dr
= (π/2)∫<0, π/6>16cosφsinφdφ + (π/2)∫<π/6, π/2>81[(cosφ)^5/(sinφ)^7]dφ
= 4π[(sinφ)^2]<0, π/6> + (81π/2)∫<π/6, π/2>[(cosφ)^4/(sinφ)^7]dsinφ
= π + (81π/2)∫<π/6, π/2>[(cosφ)^4/(sinφ)^7]dsinφ
= π + (81π/2)∫<π/6, π/2>[1-(sinφ)^2]^2/(sinφ)^7]dsinφ
= π + (81π/2)∫<π/6, π/2>[1/(sinφ)^7-2/(sinφ)^5+1/(sinφ)^3]dsinφ
= π + (81π/2)[-1/{6(sinφ)^6} + 1/{2(sinφ)^4} - 1/{2(sinφ)^2}]<π/6, π/2>
= π + (81π/2)[-1/6 + 1/2 - 1/2 + 32/3 - 8 + 2 ]<π/6, π/2>
= π + 729π/4 = 733π/4