# 二重积分中经常使用转变积分区域的形式去根号

## 一、题目

$$I=\iint_{D}(\sqrt{|x|}+\sqrt{|y|}) \mathrm{d} x \mathrm{~d} y = ?$$

## 二、解析

$$(0, 1), \quad (1,0), \quad (\frac{1}{4}, \frac{1}{4}), \quad \cdots$$

$$\iint_{D}(\sqrt{|x|}+\sqrt{|y|}) \mathrm{~d} x \mathrm{~d} y=4 \iint_{D_{1}}(\sqrt{x}+\sqrt{y}) \mathrm{~d} x \mathrm{~d} y$$

$$\int_{D_{1}} \sqrt{x} \mathrm{~d} x \mathrm{~d} y=\int_{D_{1}} \sqrt{y} \mathrm{~d} x \mathrm{~d} y$$

$\iint_{D_{1}}(\sqrt{x}+\sqrt{y}) \mathrm{~d} x \mathrm{~d} y$ $=$ $\frac{1}{2}[\iint_{D_{1}}(\sqrt{x}+\sqrt{y}) \mathrm{~d} x \mathrm{~d} y$ $+$ $\int_{D_{1}}(\sqrt{y}+\sqrt{x}) \mathrm{~d} x \mathrm{~d} y]$

$$\iint_{D}(\sqrt{|x|}+\sqrt{|y|}) \mathrm{~d} x \mathrm{~d} y=8 \iint_{D} \sqrt{x} \mathrm{~d} x \mathrm{~d} y \Rightarrow$$

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

$$8 \int_{0}^{1} \sqrt{x} \mathrm{~d} x \int_{0}^{(1-\sqrt{x})^{2}} 1 \mathrm{~d} y \Rightarrow$$

### 解法 1

$$8 \int_{0}^{1} \sqrt{x}(1-\sqrt{x})^{2} \mathrm{~d} x \Rightarrow$$

$$8 \int_{0}^{1} \sqrt{x}(1+x-2 \sqrt{x}) \mathrm{~d} x \Rightarrow$$

$$8 \int_{0}^{1}\left(x^{\frac{1}{2}}+x^{\frac{3}{2}}-2 x\right) \mathrm{~d} x \Rightarrow$$

$$8\left[\left.\frac{2}{3} x^{\frac{3}{2}}\right|_{0} ^{1}+\left.\frac{2}{5} x^{\frac{5}{2}}\right|_{0} ^{1}-\left.x^{2}\right|_{0} ^{1}\right] \Rightarrow$$

$$8\left[\frac{2}{3}+\frac{2}{5}-1\right]=8\left(\frac{16}{15}-\frac{15}{15}\right)=\frac{8}{15}.$$

### 解法 2

$$8 \int_{0}^{1} \sqrt{x}(1-\sqrt{x})^{2} \mathrm{~d} x.$$

$$\sqrt{x}=\sin ^{2} t \Rightarrow x=\sin ^{4} t \Rightarrow$$

$$x \in(0,1) \Rightarrow t \in\left(0, \frac{\pi}{2}\right) \Rightarrow$$

$$\mathrm{~d} x=4 \sin ^{3} t \cos t \mathrm{~d} t$$

$$8 \int_{0}^{1} \sqrt{x}(1-\sqrt{x})^{2} \mathrm{~d} x \Rightarrow$$

$$8 \int_{0}^{\frac{\pi}{2}} \sin ^{2} t \cos ^{4} t \cdot 4 \sin ^{3} t \cos t \mathrm{~d} t \Rightarrow$$

$$4 \times 8 \int_{0}^{\frac{\pi}{2}} \sin ^{5} t \cos ^{5} t \mathrm{~d} t.$$

$$\sin 2 \alpha=2 \sin \alpha \cos \alpha \Rightarrow$$

$$\frac{1}{2} \sin 2 \alpha=\sin \alpha \cos \alpha \Rightarrow$$

$$\left(\frac{1}{2} \sin 2 \alpha\right)^{5}=(\sin \alpha \cos \alpha)^{5}.$$

$$4 \times 8 \int_{0}^{\frac{\pi}{2}} \sin ^{5} t \cos ^{5} t \mathrm{~d} t \Rightarrow$$

$$4 \times 8 \int_{0}^{\pi} \frac{1}{2^{5}} \sin ^{5} 2 t \mathrm{~d} t \Rightarrow$$

$$\frac{4 \times 8}{2^{5}} \cdot \frac{1}{2} \int_{0}^{\pi} \sin ^{5} 2 t \mathrm{~d} (2 t) \Rightarrow$$

$$\frac{8}{2^{4}} \int_{0}^{\pi} \sin ^{5} \theta \mathrm{~d} \theta \Rightarrow 几何意义 \Rightarrow$$

$$\frac{8}{2^{4}} \times 2 \int_{0}^{\frac{\pi}{2}} \sin ^{5} \theta \mathrm{~d} \theta \Rightarrow 洛必达 \Rightarrow$$

$$\frac{8}{2^{3}} \times \frac{4}{5} \times \frac{2}{3} \times 1=\frac{2^{3} \times 2^{2} \cdot 2^{1}}{2^{3} \times 15}=\frac{8}{15}.$$