# 三角函数套进其反三角函数——湮灭为一个变量

## 一、题目

$$\iint_{D} \arctan \frac{y}{x} \mathrm{~d} \sigma = ?$$

## 二、解析

$$\left\{\begin{array}{l}y=\sqrt{3} x \Rightarrow \theta=\frac{\pi}{3} \\ x=\sqrt{3} y \Rightarrow \theta=\frac{\pi}{6}\end{array}\right.$$

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

$$I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \mathrm{~d} \theta \int_{1}^{3} \arctan \frac{r \cos \theta}{r \sin \theta} \cdot r \mathrm{~d} r \Rightarrow$$

$$I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \mathrm{~d} \theta \int_{1}^{3} \theta \cdot r \cdot \mathrm{~d} r \Rightarrow$$

$$I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \theta \mathrm{~d} \theta \int_{1}^{3} r \mathrm{~d} r \Rightarrow$$

$$I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\left(\left.\frac{1}{2} r^{2}\right|_{1} ^{3}\right) \theta \mathrm{~d} \theta \Rightarrow$$

$$I=\frac{1}{2}(9-1) \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \theta \mathrm{~d} \theta \Rightarrow$$

$$I=\left.4 \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \theta \mathrm{~d} \theta = 4 \cdot \frac{1}{2} \theta^{2}\right|_{\frac{\pi}{6}} ^{\frac{\pi}{3}} \Rightarrow$$

$$I=2\left(\frac{\pi^{2}}{9}-\frac{\pi^{2}}{36}\right)=2 \cdot \frac{3 \pi^{2}}{36}=\frac{\pi^{2}}{6}.$$