# 当积分区域出现“圆形”时，就要考虑转换为极坐标系求解

## 一、题目

$$I = \int_{0}^{1} \mathrm{~d} x \int_{1-x}^{\sqrt{1-x^{2}}} \frac{x+y}{x^{2}+y^{2}} \mathrm{~d} y = ?$$

## 二、解析

$$x \in (0, 1)$$

$$y=1-x$$

$$y=\sqrt{1-x^{2}} \Rightarrow x^{2}+y^{2}=1$$

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

$$\frac{r \cos \theta+r \sin \theta}{r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta} \Rightarrow \frac{\cos \theta+\sin \theta}{r}$$

$$y=1-x \Rightarrow\left\{\begin{array}{l}x=r \cos \theta \\ y=r \sin \theta\end{array} \Rightarrow\right.$$

$$r \sin \theta=1-r \cos \theta \Rightarrow r(\sin \theta+\cos \theta)=1 \Rightarrow$$

$$r=\frac{1}{\cos \theta+\sin \theta}$$

$$I = \int_{0}^{\frac{\pi}{2}} \mathrm{d} \theta \int_{\frac{\pi}{\cos \theta+\sin \theta}}^{1} \frac{\cos \theta+\sin \theta}{r} \cdot r \mathrm{d} r =$$

$$\int_{0}^{\frac{\pi}{2}} \mathrm{d} \theta \int_{\frac{1}{\cos \theta+\sin \theta}}^{1}(\cos \theta+\sin \theta) \mathrm{d} r =$$

$$\int_{0}^{\frac{\pi}{2}}(\cos \theta+\sin \theta)\left[1-\frac{1}{\cos \theta+\sin \theta}\right] \mathrm{d} \theta =$$

$$\int_{0}^{\frac{\pi}{2}}(\cos \theta+\sin \theta) \times \frac{\cos \theta+\sin \theta-1}{\cos \theta+\sin \theta} \mathrm{d} \theta =$$

$$\int_{0}^{\frac{\pi}{2}}(\cos \theta+\sin \theta-1) \mathrm{d} \theta=$$

$$\left.\sin \theta\right|_{0} ^{\frac{\pi}{2}}-\left.\cos \theta\right|_{0} ^{\frac{\pi}{2}}-\left(\frac{\pi}{2}-0\right)=$$

$$(1-0)-(0-1)-\frac{\pi}{2}=2-\frac{\pi}{2}.$$