下册 8.1 二重积分 第41题

\section*{解题过程：} （1）方法 1：因 故 $$ \begin{aligned} \iint_{D} f(1-y) f(x) \mathrm{d} x \mathrm{~d} y & =\int_{0}^{1} f(x) \mathrm{d} x \int_{0}^{1-x} f(1-y) \mathrm{d} y=\int_{0}^{1} f(x) \mathrm{d} x \int_{x}^{1} f(t) \mathrm{d} t \\ & =\int_{0}^{1} f(t) \mathrm{d} t \int_{0}^{t} f(x) \mathrm{d} x=\int_{0}^{1} f(x) \mathrm{d} x \int_{0}^{x} f(t) \mathrm{d} t \\ \iint_{D} f(1-y) f(x) \mathrm{d} x \mathrm{~d} y & =\frac{1}{2}\left(\int_{0}^{1} f(x) \mathrm{d} x \int_{x}^{1} f(t) \mathrm{d} t+\int_{0}^{1} f(x) \mathrm{d} x \int_{0}^{x} f(t) \mathrm{d} t\right) \\ & =\frac{1}{2} \int_{0}^{1} f(x) \mathrm{d} x \int_{0}^{1} f(t) \mathrm{d} t=\frac{1}{2}\left(\int_{0}^{1} f(x) \mathrm{d} x\right)^{2} \end{aligned} $$ 方法 2：设 $1-y=u, x=v$ ，则 $$ \Delta_{u}=\{(u, v) \mid 0 \leqslant u \leqslant 1,0 \leqslant v \leqslant u\} \text { 或 } \Delta_{v}=\{(u, v) \mid 0 \leqslant v \leqslant 1, v \leqslant u \leqslant 1\}, J=\frac{\partial(x, y)}{\partial(u, v)}=1 . $$ 于是 $$ \begin{aligned} & \iint_{D} f(1-y) f(x) \mathrm{d} x \mathrm{~d} y=\iint_{\Delta_{u}} f(u) f(v) \mathrm{d} u \mathrm{~d} v=\int_{0}^{1} f(u) \mathrm{d} u \int_{0}^{u} f(v) \mathrm{d} v=\int_{0}^{1} f(x) \mathrm{d} x \int_{0}^{x} f(y) \mathrm{d} y . \\ & \iint_{D} f(1-y) f(x) \mathrm{d} x \mathrm{~d} y=\iint_{\Delta_{v}} f(u) f(v) \mathrm{d} u \mathrm{~d} v=\int_{0}^{1} f(v) \mathrm{d} v \int_{v}^{1} f(u) \mathrm{d} u=\int_{0}^{1} f(x) \mathrm{d} x \int_{x}^{1} f(y) \mathrm{d} y . \end{aligned} $$ 从而 $$ \begin{aligned} \iint_{D} f(1-y) f(x) \mathrm{d} x \mathrm{~d} y & =\frac{1}{2}\left(\int_{0}^{1} f(x) \mathrm{d} x \int_{x}^{1} f(y) \mathrm{d} y+\int_{0}^{1} f(x) \mathrm{d} x \int_{0}^{x} f(y) \mathrm{d} y\right) \\ & =\frac{1}{2} \int_{0}^{1} f(x) \mathrm{d} x \int_{0}^{1} f(y) \mathrm{d} y=\frac{1}{2}\left(\int_{0}^{1} f(x) \mathrm{d} x\right)^{2} . \end{aligned} $$ （2）左边被积函数为 $f(x-y)$ ，考虑变换 $u=x-y$ 。为此令 $x-y=u, x+y=v$ ，则 $\displaystyle \frac{D(u, v)}{D(x, y)}=2$ ，积分范围变为 $D_{1}:|u+v| \leqslant 2 a,|u-v| \leqslant 2 a$ ．于是 $$ \iint_{D} f(x-y) \mathrm{d} x \mathrm{~d} y=\iint_{D_{1}} \frac{1}{2} f(u) \mathrm{d} u \mathrm{~d} v $$ 又 $f(u)$ 为连续偶函数，故 $\iint_{D} f(x-y) \mathrm{d} x \mathrm{~d} y=\iint_{D_{2}} f(u) \mathrm{d} u \mathrm{~d} v$ ，其中 $D_{2}$ 为 $D_{1}$ 的右半部分，即 $$ D_{2}=\{(u, v)|u \geqslant 0,|u+v| \leqslant 2 a,|u-v| \leqslant 2 a\} . $$ 从而 $$ \iint_{D} f(x-y) \mathrm{d} x \mathrm{~d} y=\iint_{D_{2}} f(u) \mathrm{d} u \mathrm{~d} v=\int_{0}^{2 a} \mathrm{~d} u \int_{u-2 a}^{2 a-u} f(u) \mathrm{d} v=2 \int_{0}^{2 a}(2 a-u) f(u) \mathrm{d} u $$ （3）设 $x-y=u, y=v$ ，即 $x=u+v, y=v$ ，则 $\displaystyle J=\frac{\partial(x, y)}{\partial(u, v)}=1$ ， $$ D=\{(x, y) \mid-a \leqslant y \leqslant a, 0 \leqslant x \leqslant a\} \text { 变为 } \Delta=\{(u, v) \mid-a \leqslant v \leqslant a,-v \leqslant u \leqslant a-v\} \text {. } $$ 于是 $$ \begin{aligned} \int_{0}^{a} \mathrm{~d} x \int_{-a}^{a} f(x-y) \mathrm{d} y & =\iint_{\Delta} f(u) \mathrm{d} u \mathrm{~d} v=\int_{-a}^{0} f(u) \mathrm{d} u \int_{-u}^{a} \mathrm{~d} v+\int_{0}^{a} f(u) \mathrm{d} u \int_{-u}^{a-u} \mathrm{~d} v+\int_{a}^{2 a} f(u) \mathrm{d} u \int_{-a}^{a-u} \mathrm{~d} v \\ & =\int_{-a}^{0}(a+u) f(u) \mathrm{d} u+\int_{0}^{a} a f(u) \mathrm{d} u+\int_{a}^{2 a}(2 a-u) f(u) \mathrm{d} u \\ & =\int_{a}^{0}(a-t) f(-t)(-\mathrm{d} t)+\int_{0}^{a} a f(u) \mathrm{d} u+\int_{a}^{2 a}(2 a-u) f(u) \mathrm{d} u \end{aligned} $$ $$ =\int_{0}^{a}(a-u) f(u) \mathrm{d} u+\int_{0}^{a} a f(u) \mathrm{d} u+\int_{a}^{2 a}(2 a-u) f(u) \mathrm{d} u=\int_{0}^{2 a}(2 a-u) f(u) \mathrm{d} u . $$ （4）记 $D=\{(x, y) \mid 0 \leqslant x \leqslant 1,0 \leqslant y \leqslant x\}$ ，则 $$ \int_{0}^{1} \mathrm{~d} x \int_{0}^{x} f(1-x, 1-y) \mathrm{d} y=\iint_{S} f(1-x, 1-y) \mathrm{d} x \mathrm{~d} y $$ 令 $x=1-u, y=1-v$ ，则 $0 \leqslant v \leqslant 1,0 \leqslant u \leqslant v,|J|=1$ 。于是 $$ \int_{0}^{1} \mathrm{~d} x \int_{0}^{x} f(1-x, 1-y) \mathrm{d} y=\int_{0}^{1} \mathrm{~d} v \int_{0}^{v} f(u, v) \mathrm{d} u=\int_{0}^{1} \mathrm{~d} v \int_{0}^{v} f(v, u) \mathrm{d} u=\int_{0}^{1} \mathrm{~d} x \int_{0}^{x} f(x, y) \mathrm{d} y . $$