2025年考研数二第20题解析：二重积分的化简与计算

一、题目

二、解析

首先，根据题目所给的积分区域的表达式 $D=\left\{ \left( x,y \right) \mid x^{2}+y^{2} \leqslant 4x, x^{2}+y^{2} \leqslant 4y \right\}$ 可知：

$$
\begin{aligned}
& \ x^{2}+y^{2} \leqslant 4x \\
\textcolor{lightgreen}{ \leadsto } & \ \left( x-2 \right)^{2} + y^{2} \leqslant 2^{2} \\ \\
& \ x^{2}+y^{2} \leqslant 4y \\
\textcolor{lightgreen}{ \leadsto } & \ x^{2} + \left( y-2 \right)^{2} \leqslant 2^{2}
\end{aligned}
$$

于是可知，积分区域 $D$ 实际上就是圆心为 $\left( 2,0 \right)$, 半径为 $2$ 的圆和圆心为 $\left( 0, 2 \right)$, 半径为 $2$ 的圆重叠的区域，如图 01 所示：

图 01.

由于积分区域 $D$ 关于 $y=x$ 对称，我们将 $y=x$ 两侧的积分区域 $D$ 分别记为 $D_{1}$ 和 $D_{2}$.

设 $D_{1}$ 为区域 $D$ 位于 $y = x$ 下方的部分，则：

$$
\begin{aligned}
& \ D_{1} = \left\{ \left( x,y \right) \mid x^{2} + \left( y-2 \right)^{2} \leqslant 2^{2} \right\} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{gray}{x = r \cos \theta, y = r \sin \theta} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{gray}{r^{2} \cos^{2} \theta + \left( r \sin \theta – 2 \right)^{2} \leqslant 4} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ D_{1} = \left\{ \left( r,\theta \right) \mid 0 \leqslant \theta \leqslant \frac{\pi}{4}, 0 \leqslant r \leqslant 4 \sin \theta \right\}
\end{aligned}
$$

于是：

$$
\begin{aligned}
& \ \iint_{D}\left( x-y \right)^{2} \mathrm{~d}x \mathrm{d} y \\ \\
= & \ 2 \iint_{D_{1}}\left( x-y \right)^{2} \mathrm{~d} x \mathrm{d} y \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{gray}{x = r \cos \theta, y = r \sin \theta} \\ \\
= & \ 2 \int_{0}^{\frac{\pi}{4}} \mathrm{d} \theta \int_{0}^{4 \sin \theta} \left( r\cos \theta-r\sin \theta \right)^{2} \cdot r \mathrm{~d} r \\ \\
= & \ 2 \int_{0}^{\frac{\pi}{4}} \left( \cos \theta-\sin \theta \right)^{2} \mathrm{~d} \theta \int_{0}^{4\sin \theta}r^{3}\mathrm{~d} r \\ \\
= & \ 2 \int_{0}^{\frac{\pi}{4}} \left( 1-2\sin \theta \cos \theta \right) \cdot \frac{1}{4} \left( 4 \sin \theta \right)^{4} \mathrm{~d} \theta \\ \\
= & \ 32 \int_{0}^{\frac{\pi}{4}} \left( 1-\sin 2\theta \right) \cdot \left( 2 \sin^{2} \theta \right)^{2} \mathrm{~d} \theta \\ \\
= & \ 32 \int_{0}^{\textcolor{lightgreen}{\frac{\pi}{4}}} \left( 1-\sin 2\theta \right) \cdot \left( 1-\cos 2\theta \right)^{2}\mathrm{~d} \theta \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{gray}{t=2\theta} \\ \\
= & \ 16 \int_{0}^{\textcolor{orange}{\frac{\pi}{2}}} \left( 1-\sin t \right) \cdot \left( 1-\cos t \right)^{2} \mathrm{~d} t \\ \\
= & \ 16 \int_{0}^{\frac{\pi}{2}} \left( 1-\cos t \right)^{2} \mathrm{~d} t – 16 \int_{0}^{\frac{\pi}{2}} \sin t \cdot \left( 1-\cos t \right)^{2} \mathrm{~d} t \\ \\
= & \ 16\int_{0}^{\frac{\pi}{2}}\left( 1-2\cos t+\cos^{2}t \right)\mathrm{~d}t-16\int_{0}^{\frac{\pi}{2}}\left( 1-\cos t \right)^{2}\mathrm{~d}\left( 1-\cos t \right) \\ \\
= & \ \left. 16 \left( \frac{\pi}{2} – 2 + \frac{1}{2} \cdot \frac{\pi}{2} \right)-\frac{16}{3}\left( 1-\cos t \right)^{3}\right|_{0}^{\frac{\pi}{2}} \\ \\
= & \ \textcolor{lightgreen}{ 12 \pi – \frac{112}{3} }
\end{aligned}
$$

$\iint_{D}\left( x-y \right)^{2} \mathrm{~d}x \mathrm{d} y$ 的另一种计算方式：

$$
\begin{aligned}
& \ \iint_{D} \left(x-y\right)^{2} \mathrm{~d}x \mathrm{d}y \\ \\
= & \ 2 \iint_{D_{1}} \left(x^{2}+y^{2}-2xy\right) \mathrm{~d}x \mathrm{d}y \\ \\
= & \ 2 \int_{0}^{\frac{\pi}{4}} \mathrm{d}\theta \int_{0}^{4\sin\theta} \left(r^{2}-2r^{2}\cos\theta\sin\theta\right) r \mathrm{~d}r \\ \\
= & \ 128 \int_{0}^{\frac{\pi}{4}} \left(\sin^{4}\theta – 2\cos\theta\sin^{5}\theta\right) \mathrm{~d}\theta \\ \\
= & \ 128 \left[ \left. \int_{0}^{\frac{\pi}{4}} \left(\frac{1-\cos 2\theta}{2}\right)^{2} \mathrm{~d}\theta – \frac{1}{3}\sin^{6}\theta \right|_{0}^{\frac{\pi}{4}}\right] \\ \\
= & \ 128 \left[\frac{1}{4}\int_{0}^{\frac{\pi}{4}} \left(1-2\cos 2\theta+\cos^{2}2\theta\right) \mathrm{~d}\theta – \frac{1}{24}\right] \\ \\
= & \ 128 \left[ \left. \frac{1}{4}\left(\frac{\pi}{4}-\sin 2\theta\right) \right|_{0}^{\frac{\pi}{4}} + \frac{1}{2}\int_{0}^{\frac{\pi}{2}}\cos^{2}t \mathrm{~d}t – \frac{1}{24}\right] \\ \\
= & \ 128 \left(\frac{3\pi}{32}-\frac{7}{24}\right) \\ \\
= & \ \textcolor{lightgreen}{ 12 \pi – \frac{112}{3} }
\end{aligned}
$$

---