高数,求第五题转动惯量和第六题二重积分的过程和答案
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5. I = ∫∫<D> y^2 dσ = ∫<0, π>dθ∫<2sinθ, 4sinθ> (ρsinθ)^2 ρdρ
= ∫<0, π>(sinθ)^2dθ∫<2sinθ, 4sinθ>ρ^3dρ
= ∫<0, π>(sinθ)^2dθ[(1/4)ρ^4]<2sinθ, 4sinθ>
= 60∫<0, π>(sinθ)^6dθ = (15/2)∫<0, π>(1-cos2θ)^3dθ
= (15/2)∫<0, π>[1-3cos2θ+3(cos2θ)^2-(cos2θ)^3]dθ
= (15/2)∫<0, π>[5/2-3cos2θ+(3/2)cos4θ]dθ - (15/4)∫<0, π>[1-(sin2θ)^2]dsin2θ
= (15/2)[5θ/2-(3/2)sin2θ+(3/8)sin4θ]<0, π> - (15/4)[sin2θ-(1/3)(sin2θ)^3]<0, π>
= (15/2)(5π/2) = (75/4)π
6. 等于积分域面积的 5 倍, 积分域看不清 ?
= ∫<0, π>(sinθ)^2dθ∫<2sinθ, 4sinθ>ρ^3dρ
= ∫<0, π>(sinθ)^2dθ[(1/4)ρ^4]<2sinθ, 4sinθ>
= 60∫<0, π>(sinθ)^6dθ = (15/2)∫<0, π>(1-cos2θ)^3dθ
= (15/2)∫<0, π>[1-3cos2θ+3(cos2θ)^2-(cos2θ)^3]dθ
= (15/2)∫<0, π>[5/2-3cos2θ+(3/2)cos4θ]dθ - (15/4)∫<0, π>[1-(sin2θ)^2]dsin2θ
= (15/2)[5θ/2-(3/2)sin2θ+(3/8)sin4θ]<0, π> - (15/4)[sin2θ-(1/3)(sin2θ)^3]<0, π>
= (15/2)(5π/2) = (75/4)π
6. 等于积分域面积的 5 倍, 积分域看不清 ?
追问
嗷嗷,谢谢呀
第六题积分区域是
(x-1)²+(y+2)²≤3
追答
6. I = 5(3π) = 15π
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