Question

用球坐标计算三重积分I=∫∫∫z^2dv 其中图形是由x^2+y^2+z^2<=R^2和x^2+y^2=z^2<=2Rz的公共部分 求解题步

Answer 1

解：∵方程组x²+y²+z²=R²与x²+y²+z²=2Rz的解是x²+y²=(√3R/2)²
∴两球体公共部分在xoy平面上的投影是S：x²+y²=(√3R/2)²
故 原式=∫∫<S>dxdy∫<R-√(R²-x²-y²),√(R²-x²-y²)>z²dz
=(1/3)∫∫<S>{[√(R²-x²-y²)]³-[R-√(R²-x²-y²)]³}dxdy
=(1/3)∫<0,2π>dθ∫<0,√3R/2>{[√(R²-r²)]³-[R-√(R²-r²)]³}rdr (做极坐标变换)
=(2π/3)∫<0,√3R/2>{2[√(R²-r²)]³-3R[√(R²-r²)]²+3R²[√(R²-r²)]-R³}rdr
=(-π/3)∫<0,√3R/2>{2[√(R²-r²)]³-3R[√(R²-r²)]²+3R²[√(R²-r²)]-R³}d(R²-r²)
=(-π/3)[(4/5)(R²-r²)^(5/2)-(3R/2)(R²-r²)²+2R²(R²-r²)^(3/2)-R³(R²-r²)]│<0,√3R/2>
=(-π/3)(-11R^5/160)
=11πR^5/480。

Answer 2

好久没有做数学题了，看着有点陌生了。