题号:1845    题型:解答题    来源:2011年全国硕士研究生招生考试试题
已知函数 $f(x, y)$ 具有二阶连续偏导数, 且 $f(1, y)=0, f(x, 1)=0, \iint_{D} f(x, y) \mathrm{d} x \mathrm{~d} y=a$, 其中 $D=\{(x, y) \mid 0 \leqslant x \leqslant 1,0 \leqslant y \leqslant 1\}$, 计算二重积分 $I=\iint_{D} x y f_{x y}^{\prime \prime}(x, y) \mathrm{d} x \mathrm{~d} y$.
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答案:
解 由题设条件知积分区域 $D$ 可表示为: $D=\{(x, y) \mid 0 \leqslant x \leqslant 1,0 \leqslant y \leqslant 1\}$,
于是有
$$
\begin{aligned}
I &=\iint_{D} x y f_{x y}^{\prime \prime}(x, y) \mathrm{d} x \mathrm{~d} y \\
&=\int_{0}^{1} x \mathrm{~d} x \int_{0}^{1} y f_{x y}^{\prime \prime}(x, y) \mathrm{d} y=\int_{0}^{1} x \mathrm{~d} x\left[\int_{0}^{1} y \mathrm{~d} f_{x}^{\prime}(x, y)\right]
\end{aligned}
$$

\begin{aligned}
&=\int_{0}^{1} x \mathrm{~d} x\left[\left.\left(y f_{x}^{\prime}(x, y)\right)\right|_{0} ^{1}-\int_{0}^{1} f_{x}^{\prime}(x, y) \mathrm{d} y\right] \\
&=\int_{0}^{1} x \mathrm{~d} x\left[f_{x}^{\prime}(x, 1)-\int_{0}^{1} f_{x}^{\prime}(x, y) \mathrm{d} y\right] \\
&=\int_{0}^{1} x f_{x}^{\prime}(x, 1) \mathrm{d} x-\int_{0}^{1} x \mathrm{~d} x \int_{0}^{1} f_{x}^{\prime}(x, y) \mathrm{d} y \\
&=\int_{0}^{1} x \mathrm{~d} f(x, 1)-\int_{0}^{1} \mathrm{~d} y \int_{0}^{1} x f_{x}^{\prime}(x, y) \mathrm{d} x \\
&=\left.(x f(x, 1))\right|_{0} ^{1}-\int_{0}^{1} f(x, 1) \mathrm{d} x-\int_{0}^{1} \mathrm{~d} y \int_{0}^{1} x \mathrm{~d} f(x, y) \\
&=-\int_{0}^{1} \mathrm{~d} y \int_{0}^{1} x \mathrm{~d} f(x, y) \quad(\because f(x, 1)=0) \\
&=-\int_{0}^{1} \mathrm{~d} y\left[\left.(x f(x, y))\right|_{0} ^{1}-\int_{0}^{1} f(x, y) \mathrm{d} x\right] \\
&=-\int_{0}^{1}\left[f(1, y)-\int_{0}^{1} f(x, y) \mathrm{d} x\right] \mathrm{d} y \\
&=\int_{0}^{1} \mathrm{~d} y \int_{0}^{1} f(x, y) \mathrm{d} x \quad(\because f(1, y)=0) \\
&=\iint_{D} f(x, y) \mathrm{d} x \mathrm{~d} y=a, \\
&\text { 故 } \quad I=\iint_{D} x y f_{x y}^{\prime \prime}(x, y) \mathrm{d} x \mathrm{~d} y=a .
\end{aligned}
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