# 2017年考研数二第20题解析：二重积分、二重积分的化简、直角坐标系转极坐标系

## 解析

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

$$x^{2} + y^{2} – 2y = 0 \Rightarrow$$

$${\color{Red} x^{2} + (y – 1)^{2} = 1. \quad \quad {\color{White}①} }$$

$$\iint_{D} (x + 1)^{2} \mathrm{d} x \mathrm{d} y \Rightarrow$$

$$\iint_{D} (x ^{2} + 1 + 2x) \mathrm{d} x \mathrm{d} y \Rightarrow$$

$$\iint_{D} x ^{2} \mathrm{d} x \mathrm{d} y + \iint_{D} 1 \mathrm{d} x \mathrm{d} y + \iint_{D} 2x \mathrm{d} x \mathrm{d} y.$$

$$\iint_{D} 2x \mathrm{d} x \mathrm{d} y = 0.$$

$$\iint_{D} 1 \mathrm{d} x \mathrm{d} y = \pi r^{2} = \pi.$$

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

$${\color{Red} \iint_{D} x ^{2} \mathrm{d} x \mathrm{d} y + \pi. \quad \quad {\color{White}②} }$$

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

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

$$r^{2} \cos ^{2} \theta + r^{2} \sin ^{2} \theta \leqslant 2 r \sin \theta \Rightarrow$$

$$r^{2} (\cos ^{2} \theta + \sin ^{2} \theta) \leqslant 2 r \sin \theta \Rightarrow$$

$$r \leqslant 2 \sin \theta.$$

$$\iint_{D} x ^{2} \mathrm{d} x \mathrm{d} y \Rightarrow$$

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

$$\iint_{D} r^{3} \cos ^{2} \theta \mathrm{d} r \mathrm{d} \theta \Rightarrow$$

$${\color{Red} \int_{0}^{\pi} \mathrm{d} \theta \int_{0}^{2 \sin \theta} r^{3} \cos ^{2} \theta \mathrm{d} r}. \quad \quad {\color{White}③} \Rightarrow$$

[1]. 根据前面的图 01 可知，变量 $\theta$ 的取值范围是：$\theta \in (0, 2 \pi)$.

$$\int_{0}^{\pi} \mathrm{d} \theta \cdot \Bigg[ \cos ^{2} \theta \int_{0}^{2 \sin \theta} r^{3} \mathrm{d} r \Bigg] \Rightarrow$$

$$\int_{0}^{\pi} \mathrm{d} \theta \cdot \Bigg[ \cos ^{2} \theta \cdot \frac{1}{4} r^{4} |_{0}^{2 \sin \theta} \Bigg] \Rightarrow$$

$$4 \int_{0}^{\pi} \cos ^{2} \theta \sin ^{4} \theta \mathrm{d} \theta \Rightarrow$$

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

[1]. 在区间 $(0, \pi)$ 上，$\cos^{2} \theta$ 和 $\sin ^{4} \theta$ 的图象都是关于 $x = \frac{\pi}{2}$ 对称的.

$$8 \int_{0}^{\frac{\pi}{2}} (1 – \sin ^{2} \theta) \sin ^{4} \theta \mathrm{d} \theta \Rightarrow$$

$$8 \int_{0}^{\frac{\pi}{2}} (\sin ^{4} \theta – \sin ^{6} \theta) \mathrm{d} \theta \Rightarrow$$

$$8 \Bigg[\int_{0}^{\frac{\pi}{2}} \sin ^{4} \theta \mathrm{d} \theta – \int_{0}^{\frac{\pi}{2}} \sin ^{6} \theta \mathrm{d} \theta \Bigg] \Rightarrow$$

$${\color{White} 华里士点火公式 \Rightarrow }$$

$$8 (\frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} – \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2}) \Rightarrow$$

$$8(\frac{3 \pi}{16} \cdot \frac{1}{6}) = \frac{\pi}{4}.$$

$$\iint_{D} (x + 1)^{2} \mathrm{d} x \mathrm{d} y \Rightarrow$$

$$\iint_{D} x ^{2} \mathrm{d} x \mathrm{d} y + \pi \Rightarrow$$

$$\frac{\pi}{4} + \pi = \frac{5 \pi}{4}.$$