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计算二重积分∫∫sin√x^2+y^2DxDy=?,D:π^2≤x^2+y^2≤...

答案在图片上,希望得到采纳,谢谢。愿您学业进步☆⌒_⌒☆

用分步积分法 ∫ r sinr dr =-∫ r dcosr =-rcosr +∫ cosrdr =(-rcosr+sinr) 会了吧

∫∫ e^(x²+y²) dxdy =∫∫ e^(r²)*r drdθ =∫[0→2π]dθ∫[0→2] e^(r²)*r dr =2π∫[0→2] e^(r²)*r dr =π∫[0→2] e^(r²) d(r²) =πe^(r²) |[0→2] =π(e^4-1)

答:(3π-4)a³/9 D为x²+y²≤ax,配方得 (x-a/2)²+y²≤(a/2)² 极坐标化简得0≤r≤a*cosθ 整个积分区域D都黏在y轴右边,故-π/2≤θ≤π/2 ∫∫_(D) √(a²-x²-y²) dxdy = ∫(-π/2,π/2) dθ ∫(0,a*cosθ) √(a²-r...

设x=rcost y=rsint -π/2

x^2+y^2=2x 极坐标 ρ=2cosθ,θ∈[-π/2,π/2] ∫∫D y²/x² dxdy =∫[-π/2,π/2]tan²θdθ∫[0,2cosθ]ρdρ =∫[-π/2,π/2]tan²θ * 2cos²θdθ =4∫[0,π/2]sin²θ dθ =2∫[0,π/2] (1-cos2θ) dθ =π

解:原式=∫dθ∫ln(r^2)rdr (作极坐标变换) =4π∫r*lnrdr =4π[(ln2-1)/8] (应用分部积分法计算) =π(ln2-1)/2。

解:∵∫x²sinxdx=(-x²cosx+2xsinx+2cosx)│ (应用分部积分法) =π²-2-2 =π²-4 ∫sin³xdx=∫(1-cos²x)sinxdx =∫(cos²x-1)d(cosx) =(cos³x/3-cosx)│ =-1/3+1-1/3+1 =4/3 ∴∫∫(x²-y²)dxdy=∫dx∫(x²...

化为极坐标 x=pcosa,y=psina p∈[0,2] a∈[0,π] ∫∫ D √(x^2+y^2) dxdy =∫[0,π]∫ [0,2] p*pdpda =∫[0,π]da∫ [0,2] p*pdp =8π/3

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