求解一道三重积分题
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用球面坐标:
I=∫(0,2π)dθ∫(0,π/2)dφ∫(0,2Rcosφ)r(r²sin²φsin²θ+r²cos²φ)dr
=∫(0,2π)dθ∫(0,π/2)(R^4/4)(sin²φsin²θ(cosφ)^4+(cosφ)^6)dφ
=(R^4/4)[∫(0,2π)sin²θdθ∫(0,π/2)((1-cos²φ )(cosφ)^4+∫(0,2π)dθ∫(0,π/2)(cosφ)^6)dφ]
=(R^4/4)[π((3/4)(1/2)(π/2)-(5/6)(3/4)(1/2)(π/2)+(2π)(5/6)(3/4)(1/2)(π/2)]
=11π²R^4/128
I=∫(0,2π)dθ∫(0,π/2)dφ∫(0,2Rcosφ)r(r²sin²φsin²θ+r²cos²φ)dr
=∫(0,2π)dθ∫(0,π/2)(R^4/4)(sin²φsin²θ(cosφ)^4+(cosφ)^6)dφ
=(R^4/4)[∫(0,2π)sin²θdθ∫(0,π/2)((1-cos²φ )(cosφ)^4+∫(0,2π)dθ∫(0,π/2)(cosφ)^6)dφ]
=(R^4/4)[π((3/4)(1/2)(π/2)-(5/6)(3/4)(1/2)(π/2)+(2π)(5/6)(3/4)(1/2)(π/2)]
=11π²R^4/128
追问
可是答案不是这个,其实一看你的答案就不太可能,最后的结果必有R的5次方的
追答
啊,把r²sinφ掉了,冤枉作这么久。
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