使用极坐标系简化二重积分的运算：升级版例题

一、题目

$$
I=
$$

$$
\int_{0}^{\frac{\sqrt{2}}{2} R} \mathrm{e}^{-y^{2}} \mathrm{~d} y \int_{0}^{y} \mathrm{e}^{-x^{2}} \mathrm{~d} x+\int_{\frac{\sqrt{2}}{2} R}^{R} \mathrm{e}^{-y^{2}} \mathrm{~d} y \int_{0}^{\sqrt{R^{2}-y^{2}}} \mathrm{e}^{-x^{2}} \mathrm{~d} x = ?
$$

难度评级：

二、解析

由题知：

$$
y \in\left(0, \frac{\sqrt{2}}{2} R\right) \Rightarrow x=y
$$

$$
y \in\left(\frac{\sqrt{2}}{2} R, R\right) \Rightarrow x=\sqrt{R^{2}-y^{2}} \Rightarrow x^{2}+y^{2}=R^{2}
$$

于是，我们可以绘制出该二重积分的积分区域：

又：

$$
e^{-x^{2}} \times e^{-y^{2}} \Rightarrow e^{-\left(x^{2}+y^{2}\right)} \Rightarrow
$$

$$
\left\{\begin{array}{l}x=r \cos \theta \\ y=r \sin \theta\end{array} \Rightarrow e^{-\left(x^{2}+y^{2}\right)}=e^{-r^{2}}\right.
$$

于是：

$$
I=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \mathrm{d} \theta \int_{0}^{R} e^{-r^{2}} \cdot r \mathrm{d} r \Rightarrow
$$

$$
I = \theta \Bigg|_{\frac{\pi}{4}} ^{\frac{\pi}{2}} \times \int_{0}^{R} e^{-r^{2}} \cdot r \mathrm{d} r \Rightarrow
$$

$$
I=\frac{\pi}{4} \int_{0}^{R} e^{-r^{2}} \cdot r \mathrm{d} r \Rightarrow
$$

又：

$$
\left(e^{-r^{2}}\right)_{r}^{\prime}=-2 r e^{-r^{2}} \Rightarrow
$$

$$
I=\frac{\pi}{4} \times \frac{-1}{2} \times\left. e^{-r^{2}}\right|_{0} ^{R}=\frac{-\pi}{8}\left(e^{-R^{2}}-1\right) \Rightarrow
$$

$$
I=\frac{\pi}{8}\left(1-e^{-R^{2}}\right).
$$

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