# 2023年考研数二第20题解析：极坐标系二重积分

## 二、解析

\begin{aligned} & \begin{cases} & x = r \cos \theta \\ \\ & y = r \sin \theta \end{cases} \\ \\ & \Rightarrow \begin{cases} & x^{2}+y^{2}-x y=1 \\ \\ & x^{2}+y^{2}-x y=2 \end{cases} \\ \\ & \Rightarrow \begin{cases} & r_{1}^{2}-r_{1}^{2} \sin \theta \cos \theta=1 \\ \\ & r_{2}^{2}-r_{2}^{2} \sin \theta \cos \theta=2 \end{cases} \\ \\ & \Rightarrow \begin{cases} & r_{1}^{2}=\frac{1}{1-\sin \theta \cos \theta} \\ \\ & r_{2}^{2}=\frac{2}{1-\sin \theta \cos \theta} \end{cases} \end{aligned}

$$\theta \in \left(0, \frac{\pi}{3}\right)$$

$r$ 的取值范围为：

$$r \in \left( \sqrt{ \frac{1}{1-\sin \theta \cos \theta} }, \ \sqrt{ \frac{2}{1-\sin \theta \cos \theta} } \right)$$

\begin{aligned} I \\ \\ & = \iint_{D} \frac{1}{3 x^{2}+y^{2}} \mathrm{~d} x \mathrm{~d} y \\ \\ & = \int_{0}^{\frac{\pi}{3}} \mathrm{~d} \theta \int_{\sqrt{\frac{1}{1-\sin \theta \cos \theta}}}^{\sqrt{ \frac{2}{1-\sin \theta \cos \theta}}} \frac{1}{r^{2}\left(3 \cos ^{2} \theta+\sin ^{2} \theta\right)} r \mathrm{~d} r \\ \\ & = \int_{0}^{\frac{\pi}{3}} \frac{1}{3 \cos ^{2} \theta+\sin ^{2} \theta} \mathrm{~d} \theta \int_{\sqrt{\frac{1}{1-\sin \theta \cos \theta}}}^{\sqrt{\frac{2}{1-\sin \theta \cos \theta}}} \frac{1}{r} \mathrm{~d} r \\ \\ & = \ln \sqrt{2} \int_{0}^{\frac{\pi}{3}} \frac{1}{3 \cos ^{2} \theta+\sin ^{2} \theta} \mathrm{~d} \theta \\ \\ & = \ln \sqrt{2} \int_{0}^{\frac{\pi}{3}} \frac{\frac{1}{\cos ^{2} \theta}}{3+\tan ^{2} \theta} \mathrm{~d} \theta \\ \\ & = \ln \sqrt{2} \int_{0}^{\frac{\pi}{3}} \frac{1}{3+\tan ^{2} \theta} \mathrm{~d} (\tan \theta) \\ \\ & = \frac{\sqrt{3} \ln \sqrt{2}}{3} \int_{0}^{\frac{\pi}{3}} \frac{1}{1+\left(\frac{\tan \theta}{\sqrt{3}}\right)^{2}} \mathrm{~d} \left(\frac{\tan \theta}{\sqrt{3}}\right) \\ \\ & = \left.\frac{\sqrt{3} \ln \sqrt{2}}{3} \arctan \left(\frac{\tan \theta}{\sqrt{3}}\right)\right|_{0} ^{\frac{\pi}{3}} \\ \\ & = \frac{\sqrt{3} \ln \sqrt{2}}{3}\left(\frac{\pi}{4}-0\right) \\ \\ & = \frac{\sqrt{3} (\frac{1}{2} \ln 2)}{3} \cdot \frac{\pi}{4} = \frac{\sqrt{3} (\ln 2)}{24} \cdot \pi \end{aligned}