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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
题目详情
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
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答案和解析
∵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=∫∫dxdy∫z²dz
=(1/3)∫∫{[√(R²-x²-y²)]³-[R-√(R²-x²-y²)]³}dxdy
=(1/3)∫dθ∫{[√(R²-r²)]³-[R-√(R²-r²)]³}rdr (做极坐标变换)
=(2π/3)∫{2[√(R²-r²)]³-3R[√(R²-r²)]²+3R²[√(R²-r²)]-R³}rdr
=(-π/3)∫{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)]│
=(-π/3)(-11R^5/160)
=11πR^5/480.
∴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=∫∫dxdy∫z²dz
=(1/3)∫∫{[√(R²-x²-y²)]³-[R-√(R²-x²-y²)]³}dxdy
=(1/3)∫dθ∫{[√(R²-r²)]³-[R-√(R²-r²)]³}rdr (做极坐标变换)
=(2π/3)∫{2[√(R²-r²)]³-3R[√(R²-r²)]²+3R²[√(R²-r²)]-R³}rdr
=(-π/3)∫{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)]│
=(-π/3)(-11R^5/160)
=11πR^5/480.
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