# 当被积函数可以分离的时候，四重积分就是两个二重积分的积

## 二、解析

$$\textcolor{springgreen}{ f(x, y)=x y+\iint_{D} f(u, v) \mathrm{~ d} u \mathrm{~ d} v }$$

$$\iint_{D} \textcolor{springgreen}{f(x, y)} \mathrm{~ d} x \mathrm{~ d} y=$$

$$\iint_{D} \textcolor{springgreen}{x y} \mathrm{~ d} x \mathrm{~ d} y + \iint_{D}\left[\textcolor{springgreen}{ \iint_{D} f(u, v) \mathrm{~ d} u \mathrm{~ d} v} \right] \mathrm{~ d} x \mathrm{~ d} y \Rightarrow$$

$$\iint_{D} f(x, y) \mathrm{~ d} x \mathrm{~ d} y=$$

$$\iint_{D} x y \mathrm{~ d} x \mathrm{~ d} y + \iint_{D} f(u, v) \mathrm{~ d} u \mathrm{~ d} v \cdot \textcolor{orangered}{\iint_{D} \mathrm{~ d} x \mathrm{~ d} y } \tag{1}$$

$$\iint_{D} x y \mathrm{~ d} x \mathrm{~ d} y=\int_{0}^{1} x \mathrm{~ d} x \int_{0}^{x^{2}} y \mathrm{~ d} y=$$

$$\frac{1}{2} \int_{0}^{1} x^{5} \mathrm{~ d} x=\left.\frac{1}{2} \cdot \frac{1}{6} x^{6}\right|_{0} ^{1}=\frac{1}{12}$$

$$\iint_{D}{ } \mathrm{~ d} x \mathrm{~ d} y=\int_{0}^{1} \mathrm{~ d} x \int_{0}^{x^{2}} \mathrm{~ d} y=\int_{0}^{1} x^{2} \mathrm{~ d} x=$$

$$\left.\frac{1}{3} x^{3}\right|_{0} ^{1}=\frac{1}{3}$$

Tips:

$$\iint_{D} f(x, y) \mathrm{~ d} x \mathrm{~ d} y=\frac{1}{12}+\frac{1}{3} \iint_{D} f(x, y) \mathrm{~ d} x \mathrm{~ d} y \Rightarrow$$

$$\frac{2}{3} \iint_{D} f(x, y) \mathrm{~ d} x \mathrm{~ d} y=\frac{1}{12} \Rightarrow$$

$$\iint_{D} f(x, y) \mathrm{~ d} x \mathrm{~ d} y-\frac{1}{3} \iint_{D} f(x, y) \mathrm{~ d} x \mathrm{~ d} y=\frac{1}{12} \Rightarrow$$

$$\frac{2}{3} \iint_{D} f(x, y) \mathrm{~ d} x \mathrm{~ d} y=\frac{1}{12} \Rightarrow$$

$$\iint_{D} f(x, y)=\frac{1}{12} \times \frac{3}{2}=\frac{1}{8}.$$