2. 考研
  3. 设D＝{(r，θ)｜2≤r≤2(1＋cosθ)，一π/2≤θ≤π/2}，f(x，y)在R2上连续，且满足f(x，y)＝x＋yf(x，y)dxdy，求f(x，y)及I＝xf(x，y)dxdy

设D＝{(r，θ)｜2≤r≤2(1＋cosθ)，一π/2≤θ≤π/2}，f(x，y)在R2上连续，且满足f(x，y)＝x＋yf(x，y)dxdy，求f(x，y)及I＝xf(x，y)dxdy

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问题 设D＝{(r，θ)｜2≤r≤2(1＋cosθ)，一π/2≤θ≤π/2}，f(x，y)在R2上连续，且满足f(x，y)＝x＋yf(x，y)dxdy，求f(x，y)及I＝xf(x，y)dxdy
选项
答案如图所示，由D关于x轴对称，知[*]ydxdy＝0 记A＝[*]f(x，y)dxdy([*]f(x，y)dxdy是一个常数)，则已知等式可化为f(x，y)＝x＋Ay，故 A＝[*]f(x，y)dxdy＝[*]dxdy＋A[*]ydxdy ＝∫－π/2π/2dθ∫22(1＋cosθ)rcosθ·rdr＋0 ＝8/3∫－π/2π/2[(1＋cosθ)－1]cosθdθ ＝16/3∫0π/2(3cos3θ＋3cos3θ＋cos4θ)dθ ＝16/3(2＋15π)＝32/3＋5π 于是 f(x＋y)＝x＋(32/3＋5π)y I＝[*]xf(x，y)dxdy＝[*][x2＋(32/3＋5π)xy]dxdy 由x2＋y2≤1关于y＝x对称，xy关于x是奇函数，可知 I＝[*]x2dxdy＝1/2[*](x2＋y2]dxdy ＝1/2∫02πdθ∫01r2·rdr＝1/2×2π×1/4＝π/4 [*]
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考研数学二

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