2026年考研数二第17题解析：二重积分、直角坐标系转极坐标系

一、题目

二、解析

根据题目所给的信息，我们知道，积分区域 $D$ 为：

$$
\begin{aligned}
& x \in (-1, 1) \\
& y \in (|x|, \sqrt{2-x^{2}})
\end{aligned}
$$

其中，$y = |x|$ 可以拆分为 $y = x$ 和 $y = -x$, $y = \sqrt{2-x^{2}}$ 实际上就是圆心位于 $(0,0)$ 点处的圆 $x^{2} + y^{2} = \sqrt{2}$ 在区间 $(-1, 1)$ 对应的第一、二象限的图象，于是，我们可以将积分区域 $D$ 绘制如下（绿色阴影部分）：

解法 1

首先，根据直角坐标系转极坐标系的方法，令 $x = r \cos \theta$, $y = r \sin \theta$, 则：

$$
\sin \sqrt{x^{2} + y^{2}} = \sin \sqrt{r^{2} \cos^{2} x + r^{2} \sin^{2} x}
$$

于是：

$$
\begin{aligned}
I & = \int_{\frac{\pi}{4}}^{\frac{3}{4}\pi} \mathrm{~d} \theta \int_{0}^{\sqrt{2}} r \sin \theta \sin r \cdot r \mathrm{~d} r \\ \\
& = \int_{\frac{\pi}{4}}^{\frac{3}{4}\pi} \sin \theta \mathrm{~d} \theta \int_{0}^{\sqrt{2}} r^{2} \cdot \sin r \mathrm{~d} r
\end{aligned}
$$

又因为：

$$
\begin{aligned}
\int_{\frac{\pi}{4}}^{\frac{3}{4}\pi} \sin \theta \mathrm{~d} \theta & = -\cos \theta \Bigg|_{\frac{\pi}{4}}^{\frac{3}{4} \pi} \\ \\
& = – \left( \cos \frac{3}{4} \pi – \cos \frac{\pi}{4} \right) \\ \\
& = – \left[ \cos \left( \frac{- \pi}{4} \right) – \cos \frac{\pi}{4} \right] \\ \\
& = – \left( – \frac{\sqrt{2}}{2} – \frac{\sqrt{2}}{2} \right) \\ \\
& = \sqrt{2} \\ \\ \\
\int_{0}^{\sqrt{2}} r^{2} \sin r \mathrm{~d} r & = \int_{0}^{\sqrt{2}} r^{2} \mathrm{~d}(-\cos r) \\ \\
& = -r^{2} \cos r \Bigg|_{0}^{\sqrt{2}} + \int_{0}^{\sqrt{2}} 2r \cdot \cos r \mathrm{~d} r \\ \\
& = -2 \cos \sqrt{2} + 2 \int_{0}^{\sqrt{2}} r \cdot \cos r \mathrm{~d} r \\ \\
& = -2 \cos \sqrt{2} + 2 \int_{0}^{\sqrt{2}} r \mathrm{~d} (\sin r) \\ \\
& = -2 \cos \sqrt{2} + 2 \left[ r \sin r \Bigg|_{0}^{\sqrt{2}} – \int_{0}^{\sqrt{2}} \sin r \mathrm{~d} r \right] \\ \\
& = -2 \cos \sqrt{2} + 2 \left[ \sqrt{2} \sin \sqrt{2} – \cos r \Bigg|_{0}^{\sqrt{2}} \right] \\ \\
& = -2 \cos \sqrt{2} + 2 \sqrt{2} \sin \sqrt{2} + 2 \cos \sqrt{2} – 2 \\ \\
& = 2 \sqrt{2} \sin \sqrt{2} – 2
\end{aligned}
$$

综上：

$$
I = \sqrt{2} \cdot (2\sqrt{2} \sin \sqrt{2} – 2) = 4 \sin \sqrt{2} – 2\sqrt{2}
$$

解法 2

首先，根据直角坐标系转极坐标系的方法，令 $x = r \cos \theta$, $y = r \sin \theta$, 则：

$$
\sin \sqrt{x^{2} + y^{2}} = \sin \sqrt{r^{2} \cos^{2} x + r^{2} \sin^{2} x}
$$

又因为积分区域 $D$ 关于 $Y$ 轴对称，被积函数 $y \sin \sqrt{x^{2} + y^{2}}$ 关于 $x$ 是偶函数，所以，被积函数在整个积分区域 $D$ 上的积分，就相当于被积函数在积分区域位于第一象限（如图 02 所示）的部分 $D_{1}$ 积分的 $2$ 倍：

于是：

$$
\begin{aligned}
I & = \iint_{D} y \sin \sqrt{x^{2} + y^{2}} \mathrm{~d} x \mathrm{d} y \\ \\
& = 2 \iint_{D_1} y \sin \sqrt{x^{2} + y^{2}} \mathrm{~d} x \mathrm{d} y \\ \\
& = 2 \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \mathrm{~d} \theta \int_{0}^{\sqrt{2}} r^{2} \sin \theta \sin r \mathrm{~d} r \\ \\
& = 2 \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin \theta \mathrm{~d} \theta \int_{0}^{\sqrt{2}} r^{2} \sin r \mathrm{~d} r \\ \\
& \textcolor{lightgreen}{ \leadsto } \textcolor{gray}{2 \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin \theta \mathrm{~d} \theta = – 2 \cos \theta \Bigg|_{\frac{\pi}{4}}^{\frac{\pi}{2}} = – 2 \left( 0-\frac{\sqrt{2}}{2} \right) = \sqrt{2}} \\ \\
& = \sqrt{2} \int_{0}^{\sqrt{2}} r^{2} \sin r \mathrm{~d} r \\ \\
& = -\sqrt{2} \int_{0}^{\sqrt{2}} r^{2} \mathrm{~d} \cos r \\ \\
& = – \sqrt{2} \left( r^{2} \cos r \Bigg|_{0}^{\sqrt{2}} – 2 \int_{0}^{\sqrt{2}} r \cos r \mathrm{~d} r \right) \\ \\
& = – \sqrt{2} \left( r^{2} \cos r \Bigg|_{0}^{\sqrt{2}} – 2 \int_{0}^{\sqrt{2}} r \mathrm{~d} \sin r \right) \\ \\
& = – \sqrt{2} \left( r^{2} \cos r \Bigg|_{0}^{\sqrt{2}} – 2 r \sin r \Bigg|_{0}^{\sqrt{2}} – 2\int_{0}^{\sqrt{2}} \sin r \mathrm{~d} r \right) \\ \\
& = – \sqrt{2} \left( r^{2} \cos r \Bigg|_{0}^{\sqrt{2}} – 2 r \sin r \Bigg|_{0}^{\sqrt{2}} – 2 \cos r \Bigg|_{0}^{\sqrt{2}} \right) \\ \\
& = – \sqrt{2} \left( r^{2} \cos r – 2r \sin r – 2 \cos r \right) \bigg|_{0}^{\sqrt{2}} \\ \\
& = 4 \sin \sqrt{2} – 2\sqrt{2}
\end{aligned}
$$

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