利用积分区间的对称性和被积函数的奇偶性化简积分运算

一、题目

$$\iint_{D}|x y| \mathrm{d} \sigma = ?$$

二、解析

$$|x| \leqslant 1 \Rightarrow-1 \leqslant x \leqslant 1 \quad|y| \leqslant 1 \Rightarrow-1 \leqslant y \leqslant 1$$

$$x^{2}+y^{2} \geqslant x \Rightarrow x^{2}+y^{2}-x \geqslant 0 \Rightarrow \ \left(x-\frac{1}{2}\right)^{2}+y^{2} \geqslant \frac{1}{4}$$

$$\iint_{D}|x y| \mathrm{~d} \sigma=\iint_{D_{1}}|x y| \mathrm{~d} \sigma-\iint_{D_{2}}|x y| \mathrm{~d} \sigma$$

$$\iint_{D_{1}}|x y| \mathrm{~d} \sigma=4 \int_{0}^{1} \mathrm{~d} x \int_{0}^{1}|x y| \mathrm{~d} y \Rightarrow$$

$$4 \int_{0}^{1} x \mathrm{~d} x \int_{0}^{1} y \mathrm{~d} y=4 \cdot \frac{1}{2} \int_{0}^{1} x \mathrm{~d} x= \ 4 \cdot \frac{1}{2} \cdot \frac{1}{2}=1$$

$$x^{2}+y^{2}-x=0 \Rightarrow\left\{\begin{array}{l}x=r \cos \theta \\ y=r \sin \theta\end{array} \Rightarrow\right.$$

$$r^{2}-r \cos \theta=0 \Rightarrow\left\{\begin{array}{l}r=0 \\ r=\cos \theta .\end{array}\right.$$

$$\iint_{D_{2}}|x y| \mathrm{~d} \sigma = 2 \int_{0}^{\frac{\pi}{2}} \mathrm{~d} \theta \int_{0}^{\cos \theta} r \cdot r \cos \theta \cdot r \sin \theta \mathrm{~d} r \Rightarrow$$

$$2 \int_{0}^{\frac{\pi}{2}} \mathrm{~d} \theta \int_{0}^{\cos \theta} r^{3} \sin \theta \cdot \cos \theta \mathrm{~d} r \Rightarrow$$

$$2 \int_{0}^{\frac{\pi}{2}} \sin \theta \cdot \cos \theta \mathrm{~d} \theta \int_{0}^{\cos \theta} r^{3} \mathrm{~d} r \Rightarrow$$

$$2 \cdot \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \sin \theta \cos ^{5} \theta \mathrm{~d} \theta \Rightarrow$$

$$\frac{-1}{2} \int_{1}^{0} \cos ^{5} \theta d(\cos \theta) \Rightarrow$$

$$\frac{1}{2} \int_{0}^{1} t^{5} \mathrm{~d} t \Rightarrow$$

$$\frac{1}{2} \cdot \frac{1}{6} t \Big|_{0} ^{1}=\frac{1}{12}.$$

$$\iint_{D}|x y| \mathrm{~d} \sigma = 1-\frac{1}{12}=\frac{11}{12}$$