求三重积分∫∫∫ z㏑(x^2+y^2+z^2+1)/(x^2+y^2+z^2+1)dv,其中v是上半个球x^2+y^2
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求三重积分∫∫∫ z㏑(x^2+y^2+z^2+1)/(x^2+y^2+z^2+1)dv,其中v是上半个球x^2+y^2+z^2≤1且z>=0.
用球坐标:
∫∫∫ z㏑(x^2+y^2+z^2+1)/(x^2+y^2+z^2+1)dv
=∫∫∫ [rcosφln(r²+1)/(r²+1)]*r²sinφdrdφdθ
=∫ [0--->2π] dθ∫ [0--->π/2] sinφcosφdφ ∫[0---->1] r³ln(r²+1)/(r²+1)dr
=2π∫ [0--->π/2] sinφd(sinφ) ∫[0---->1] r³ln(r²+1)/(r²+1)dr
=πsin²φ∫[0---->1] r³ln(r²+1)/(r²+1)dr φ:[0--->π/2]
=π∫[0---->1] r³ln(r²+1)/(r²+1)dr
=(π/2)∫[0---->1] r²ln(r²+1)/(r²+1)d(r²)
令r²=u,u:0--->1
=(π/2)∫[0---->1] uln(u+1)/(u+1)du
=(π/2)∫[0---->1] (u+1-1)ln(u+1)/(u+1)du
=(π/2)∫[0---->1] ln(u+1)du-(π/2)∫[0---->1] ln(u+1)/(u+1)du
=(π/2)uln(u+1)-(π/2)∫[0---->1] u/(u+1)du-(π/2)∫[0---->1] ln(u+1) d[ln(u+1)]
=(π/2)uln(u+1)-(π/2)∫[0---->1] (u+1-1)/(u+1)du-(π/4)ln²(u+1)
=(π/2)uln(u+1)-(π/2)∫[0---->1] 1du+(π/2)∫[0---->1] 1/(u+1)du-(π/4)ln²(u+1)
=(π/2)uln(u+1)-(π/2)u+(π/2)ln(u+1)-(π/4)ln²(u+1) |[0---->1]
=(π/2)ln2-(π/2)+(π/2)ln2-(π/4)ln²2
=πln2-(π/2)-(π/4)ln²2
∫∫∫ z㏑(x^2+y^2+z^2+1)/(x^2+y^2+z^2+1)dv
=∫∫∫ [rcosφln(r²+1)/(r²+1)]*r²sinφdrdφdθ
=∫ [0--->2π] dθ∫ [0--->π/2] sinφcosφdφ ∫[0---->1] r³ln(r²+1)/(r²+1)dr
=2π∫ [0--->π/2] sinφd(sinφ) ∫[0---->1] r³ln(r²+1)/(r²+1)dr
=πsin²φ∫[0---->1] r³ln(r²+1)/(r²+1)dr φ:[0--->π/2]
=π∫[0---->1] r³ln(r²+1)/(r²+1)dr
=(π/2)∫[0---->1] r²ln(r²+1)/(r²+1)d(r²)
令r²=u,u:0--->1
=(π/2)∫[0---->1] uln(u+1)/(u+1)du
=(π/2)∫[0---->1] (u+1-1)ln(u+1)/(u+1)du
=(π/2)∫[0---->1] ln(u+1)du-(π/2)∫[0---->1] ln(u+1)/(u+1)du
=(π/2)uln(u+1)-(π/2)∫[0---->1] u/(u+1)du-(π/2)∫[0---->1] ln(u+1) d[ln(u+1)]
=(π/2)uln(u+1)-(π/2)∫[0---->1] (u+1-1)/(u+1)du-(π/4)ln²(u+1)
=(π/2)uln(u+1)-(π/2)∫[0---->1] 1du+(π/2)∫[0---->1] 1/(u+1)du-(π/4)ln²(u+1)
=(π/2)uln(u+1)-(π/2)u+(π/2)ln(u+1)-(π/4)ln²(u+1) |[0---->1]
=(π/2)ln2-(π/2)+(π/2)ln2-(π/4)ln²2
=πln2-(π/2)-(π/4)ln²2
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