2015年考研数二第18题解析：二重积分、二重积分的化简、三角函数代换、华里士点火公式

解析

$$\iint_{D} x(x+y) {\rm d} x {\rm d} y \Rightarrow$$

$$\iint_{D} x^{2} {\rm d} x {\rm d} y + \iint_{D} xy {\rm d} x {\rm d} y.$$

$$\iint_{D} x^{2} {\rm d} x {\rm d} y + \iint_{D} xy {\rm d} x {\rm d} y =$$

$$\iint_{D} x^{2} {\rm d} x {\rm d} y + 0 =$$

$$\iint_{D} x^{2} {\rm d} x {\rm d} y = ①$$

$$2 \iint_{D_{1}} x^{2} {\rm d} x {\rm d} y. ②$$

[1]. 注意上面 $①$ 式和 $②$ 式积分区域的不同。

$$x^{2} + y^{2} = 2 \Rightarrow$$

$$y = \pm \sqrt{2-x^{2}}.$$

$$\sqrt{2-x^{2}} = x^{2}.$$

[1]. 由于积分区域 $D$ 全部位于 $y$ 轴的正半轴，因此，我们只需计算 $\sqrt{2-x^{2}} = x^{2}$, 而无需计算 $-\sqrt{2-x^{2}} = x^{2}$.

$$x = \pm 1.$$

$$2 \iint_{D_{1}} x^{2} {\rm d} x {\rm d} y =$$

$$2 \int_{0}^{1} x^{2} {\rm d} x {\rm d} y \int_{x^{2}}^{\sqrt{2-x^{2}}} 1 {\rm d}y =$$

$$2 \int_{0}^{1} x^{2}(\sqrt{2-x^{2}} – x^{2}) {\rm d} x =$$

$$2 [\int_{0}^{1} (x^{2} \sqrt{2-x^{2}}) {\rm d} x – \int_{0}^{1} x^{4} {\rm d} x].$$

$$2 \int_{0}^{1} (x^{2} \sqrt{2-x^{2}}) {\rm d} x – 2 \int_{0}^{1} x^{4} {\rm d} x =$$

$$2 \int_{0}^{1} (x^{2} \sqrt{2-x^{2}}){\rm d} x – 2 \frac{1}{5} x^{5} |_{0}^{1} =$$

$$2 \int_{0}^{1} (x^{2} \sqrt{2-x^{2}}){\rm d} x – \frac{2}{5}.$$

$$0 \leqslant x \leqslant 1 \Rightarrow$$

$$0 \leqslant \sqrt{2} \sin A \leqslant 1 \Rightarrow$$

$$0 \leqslant \sin A \leqslant \frac{\sqrt{2}}{2} \Rightarrow$$

$$0 \leqslant A \leqslant \frac{\pi}{4}.$$

$$2 \int_{0}^{1} x^{2} \sqrt{2-x^{2}} – \frac{2}{5} \Rightarrow$$

$$2 \int_{0}^{\frac{\pi}{4}} 2 \sin^{2}A \cdot \sqrt{2} \cos A {\rm d}(\sqrt{2} \sin A) – \frac{2}{5} =$$

$$2 \int_{0}^{\frac{\pi}{4}} 2 \sin^{2}A \cdot 2 \cos^{2} A {\rm d}(A) – \frac{2}{5} =$$

$$8 \int_{0}^{\frac{\pi}{4}} \sin^{2}A \cdot \cos^{2} A {\rm d}(A) – \frac{2}{5} =$$

$$8 \int_{0}^{\frac{\pi}{4}} (\sin A \cos A) \cdot (\sin A \cos A) {\rm d}(A) – \frac{2}{5} = ③$$

$$8 \cdot \frac{1}{4} \int_{0}^{\frac{\pi}{4}} \sin^{2} (2A) {\rm d}(A) – \frac{2}{5} = ④$$

[1]. $③$ 式到 $④$ 式的变换过程利用了如下公式：

$\sin 2 \alpha = 2 \sin \alpha \cos \alpha.$

$$2 \cdot \frac{1}{2} \int_{0}^{\frac{\pi}{4}} \sin^{2}(2A) {\rm d}(2A) – \frac{2}{5} =$$

[1]. 若令 $B = 2A$, 则由 $0 \leqslant A \leqslant \frac{\pi}{4}$ 可知：

$0 \leqslant B \leqslant \frac{\pi}{2}.$

$$\int_{0}^{\frac{\pi}{2}} \sin^{2} B {\rm d}(B) – \frac{2}{5} = ⑤$$

$$\frac{1}{2} \cdot \frac{\pi}{2} – \frac{2}{5} = ⑥$$

[1]. $⑤$ 式到 $⑥$ 式的变换过程利用了华里士“点火公式”。

$$\frac{\pi}{4} – \frac{2}{5}.$$