设D={(x,y)|x2+y2)≤π},计算积分I=sin(x2+y2)dxdy

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问题 设D={(x,y)|x2+y2)≤π},计算积分I=sin(x2+y2)dxdy

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答案依题意,有 I=eπ[*]464sin(x2+y2)dxdy=eπ0dθ[*]465sinr2dr =2πeπ[*]466sinr2dr[*]467πe4π0πe-tsintdt 记I1=∫0πe-tsintdt I1=-∫0πsintd(e-t)=-(e-tt|0π-∫0πe-tcostdt) =-∫0πcostd(e-t)=-(e-tt|0π+∫0πe-tsintdt) =e+1-I1 移项,得I1=1/2(1+e),故I=πeπI1=π/2(1+eπ)

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