# 2016年考研数二第18题解析：二重积分、二重积分的化简、极坐标系下二重积分的计算

## 题目

$$\iint_{D} \frac{x^{2} – xy – y^{2}}{x^{2} + y^{2}}.$$

## 解析

$$\iint_{D} xy = 0 \Rightarrow$$

$$\iint_{D} \frac{x^{2} – xy – y^{2}}{x^{2} + y^{2}} \rightrightarrows$$

$$\iint_{D} \frac{x^{2} – y^{2}}{x^{2} + y^{2}}. ②$$

$$\theta \in (45^{\circ}, 180^{\circ}-45^{\circ}) \Rightarrow$$

$$\theta \in (\frac{\pi}{4}, \frac{3 \pi}{4}).$$

$$\sin \theta = \frac{1}{r} \Rightarrow$$

$$r = \frac{1}{\sin \theta} \Rightarrow$$

$$r \in (0, \frac{1}{\sin \theta}).$$

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

$$\iint_{D} \frac{x^{2} – y^{2}}{x^{2} + y^{2}} \Rightarrow$$

$$\iint_{D} \frac{r^{2} \cos ^{2} \theta – r^{2} \sin ^{2} \theta}{r^{2} \cos ^{2} \theta + r^{2} \sin ^{2} \theta} r \mathrm{d}r \mathrm{d} \theta \Rightarrow$$

$$\iint_{D} \frac{r^{2} (\cos ^{2} \theta – \sin ^{2} \theta)}{r^{2} (\cos ^{2} \theta + \sin ^{2} \theta)} r \mathrm{d}r \mathrm{d} \theta \Rightarrow$$

$$\iint_{D} \frac{(\cos ^{2} \theta – \sin ^{2} \theta)}{(\cos ^{2} \theta + \sin ^{2} \theta)} r \mathrm{d}r \mathrm{d} \theta \Rightarrow$$

$$\iint_{D} \frac{(\cos ^{2} \theta – \sin ^{2} \theta)}{1} r \mathrm{d}r \mathrm{d} \theta \Rightarrow$$

$$\iint_{D} r(\cos ^{2} \theta – \sin ^{2} \theta) \mathrm{d}r \mathrm{d} \theta \Rightarrow$$

$$\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} (\cos ^{2} \theta – \sin ^{2} \theta) \mathrm{d} \theta \int_{0}^{\frac{1}{\sin \theta}} r \mathrm{d} r \Rightarrow$$

$$\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} (\cos ^{2} \theta – \sin ^{2} \theta) (\frac{1}{2} r^{2}|_{0}^{\frac{1}{\sin \theta}}) \mathrm{d} \theta \Rightarrow$$

$$\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} (\cos ^{2} \theta – \sin ^{2} \theta) (\frac{1}{2} \cdot \frac{1}{\sin ^{2} \theta}) \mathrm{d} \theta \Rightarrow$$

$$\frac{1}{2} \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} (\frac{\cos ^{2} \theta – \sin ^{2} \theta}{\sin ^{2} \theta}) \mathrm{d} \theta \Rightarrow$$

$$\frac{1}{2} \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} (\frac{\cos ^{2} \theta + \sin ^{2} \theta – 2 \sin ^{2} \theta}{\sin ^{2} \theta}) \mathrm{d} \theta \Rightarrow$$

$$\frac{1}{2} \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} (\frac{1}{\sin ^{2} \theta} – 2) \mathrm{d} \theta \Rightarrow$$

$$\frac{1}{2}[ \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{1}{\sin ^{2} \theta} \mathrm{d} \theta – \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} 2 \mathrm{d} \theta] \Rightarrow$$

$$\frac{1}{2}[ \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{1}{\sin ^{2} \theta} \mathrm{d} \theta – \pi] \Rightarrow$$

$$\frac{1}{2}[ \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \csc ^{2} \theta \mathrm{d} \theta – \pi] \Rightarrow$$

$$\frac{1}{2}[ – \cot ^{2} \theta |_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} – \pi] \Rightarrow$$

[1]. $(\cot \theta)^{‘} =$ $-\csc ^{2} \theta$.

$$\frac{1}{2}[\frac{-1}{\tan ^{2} \theta} |_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} – \pi] \Rightarrow$$

[1]. $\tan \frac{\pi}{4} = 1$;

[2]. $\tan \frac{3 \pi}{4} = – 1$.

$$\frac{1}{2} [-1(-1-1) – \pi] = \frac{1}{2} [2- \pi] = 1 – \frac{\pi}{2}.$$