Question

高数三重积分

Answer 1

记Ω1为球 x^2+y^2+z^2≤R^2, Ω2为圆柱 x^2+y^2=Rx 被球截取的部分。则
I =∫∫∫<Ω>z^2dv =∫∫∫<Ω1>z^2dv - ∫∫∫<Ω2>z^2dv
=∫<0,π>dφ∫<0,2π>dθ∫<0, R>(rcosφ)^2*r^2sinφdr
-∫<-π/2,π/2>dθ∫<0, Rcosθ>rdr∫<-√(R^2-r^2),√(R^2-r^2)> z^2dz
=(2πR^5/5)∫<0,π>(cosφ)^2sinφdφ
-(2/3)∫<-π/2,π/2>dθ∫<0, Rcosθ>(R^2-r^2)^(3/2)rdr
=(-2πR^5/5)∫<0,π>(cosφ)^2dcosφ
-(1/3)∫<-π/2,π/2>dθ∫<0, Rcosθ>-(R^2-r^2)^(3/2)d(R^2-r^2)
= 4πR^5/15 - (2R^5/15)∫<-π/2,π/2>[1-(sinθ)^5]dθ
= 4πR^5/15 - 2πR^5/15 = 2πR^5/15 .