设D={(x,y)|(x2+y2)2≤4(x2-y2)},求x2dxdy。

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问题 设D={(x,y)|(x2+y2)2≤4(x2-y2)},求x2dxdy。

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答案设区域D位于第一象限的区域为D1, [*] =16∫0π/4cos2θcos22θdθ=8∫0π/4(1+cos2θ)cos22θdθ =4∫0π/4(1+cos2θ)cos22θd(2θ), 令2θ=t,4∫0π/4(1+cos2θ)cos22θd(2θ)=4∫0π/4(1+cost)cos2tdt =4∫0π/4(cos2t+cos3t)dt=4(π/4+2/3)=π+8/3。

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