# 2011年考研数二真题第13题解析：二重积分的计算，三种解法

## 解析

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

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

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

### 解法一

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

$$\iint_{D} f(x, y) d x dy = \iint_{D} f(r \cos \theta, r \sin \theta) r dr d \theta$$

$$\iint_{D} xy d \sigma =$$

$$\iint_{D} r \cos \theta \cdot r \sin \theta \cdot r \cdot dr d \theta =$$

$$\iint_{D} r^{3} \cos \theta \sin \theta dr d \theta =$$

$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos \theta \sin \theta d \theta \int_{0}^{2 \sin \theta} r^{3} dr =$$

$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} [(\cos \theta \sin \theta) \cdot (\frac{1}{4} r^{4} |_{0}^{2 \sin \theta})] d \theta =$$

$$4 \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin^{5} \theta \cos \theta d \theta =$$

$$4 \int_{\frac{\sqrt{2}}{2}}^{1} \sin^{5} \theta d (\sin \theta) =$$

$$4 \cdot \frac{1}{6} (\sin^{6} \theta) |_{\frac{\sqrt{2}}{2}}^{1} =$$

$$\frac{2}{3} \cdot (1 – (\frac{\sqrt{2}}{2})^{6}) =$$

$$\frac{2}{3} \cdot (1 – \frac{1}{8}) =$$

$$\frac{2}{3} \cdot \frac{7}{8} = \frac{7}{12}.$$

$$\because \theta_{1} = 90^{\circ} – \theta_{3}$$

$$\theta_{2} = 180^{\circ} – 90^{\circ} – \theta_{3}$$

$$\therefore \theta_{1} = \theta_{2} = \theta$$

$$\because \sin \theta_{2} = \frac{r}{2}$$

$$\therefore r = 2 \sin \theta_{2} = 2 \sin \theta.$$

### 解法二

$$\iint_{D} xy d \sigma = \iint_{D_{1}} xy d \sigma + \iint_{D_{2}} xy d \sigma.$$

$$\left\{\begin{matrix} x = r \cos \theta,\\ y = r \sin \theta + 1. \end{matrix}\right.$$

$$\iint_{D_{1}} xy d \sigma=$$

$$\iint_{D_{1}} (r \cos \theta) \cdot (r \sin \theta + 1) \cdot r dr d \theta =$$

$$\iint_{D_{1}} r(r^{2} \sin \theta \cos \theta + r \cos \theta) dr d \theta =$$

$$\iint_{D_{1}} (r^{3} \sin \theta \cos \theta + r^{2} \cos \theta) dr d\theta =$$

$$\int_{0}^{1} dr \int_{0}^{\frac{\pi}{2}} (r^{3} \sin \theta \cos \theta + r^{2} \cos \theta) d\theta =$$

$$\int_{0}^{1}[ r^{3}\int_{0}^{\frac{\pi}{2}} \sin \theta \cos \theta d \theta + r^{2} \int_{0}^{\frac{\pi}{2}} \cos \theta d \theta]dr =$$

$$\int_{0}^{1}[r^{3} \int_{0}^{1} \sin \theta d(\sin \theta) d\theta + r^{2} \cdot (1-0)]dr =$$

$$\int_{0}^{1} (\frac{1}{2} r^{3} + r^{2}) dr = \frac{1}{8} + \frac{1}{3}.$$

$$0 < y < 1;$$

$$0<x<y.$$

$$\iint_{D_{2}} xy d\sigma=$$

$$\int_{0}^{1} y dy \int_{0}^{y} x dx =$$

$$\int_{0}^{1}(y \cdot \frac{1}{2} y^{2}) dy =$$

$$\frac{1}{2} \int_{0}^{1} y^{3} dy =$$

$$\frac{1}{2} \cdot \frac{1}{4} = \frac{1}{8}.$$

$$D = D_{1} + D_{2} =$$

$$\frac{1}{3} + \frac{1}{8} + \frac{1}{8} = \frac{7}{12}.$$

### 解法三

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

$$x^{2} + (y – 1)^{2} = 1 \Rightarrow$$

$$(y – 1)^{2} = 1-x^{2} \Rightarrow$$

$$y – 1 = \sqrt{1 – x^{2}} \Rightarrow$$

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

$$0<x<1;$$

$$x<y<1 + \sqrt{1-x^{2}}.$$

$$\iint_{D} xy d \sigma =$$

$$\int_{0}^{1} x \cdot dx\int_{x}^{1+\sqrt{1-x^{2}}} y \cdot dy =$$

$$\int_{0}^{1} x \frac{1}{2}[(1+\sqrt{1-x^{2}})^{2} – x^{2}] dx =$$

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

$$\frac{1}{2} \int_{0}^{1} (2x -2x^{3} +2x\sqrt{1-x^{2}}) dx =$$

$$\int_{0}^{1} (x – x^{3} + x \sqrt{1-x^{2}}) dx =$$

$$\int_{0}^{1} x dx – \int_{0}^{1} x^{3} dx + \int_{0}^{1} x \sqrt{1-x^{2}} dx \Rightarrow$$

$$[(1-x^{2})^{\frac{3}{2}}]^{‘} =$$

$$\frac{3}{2} (1-x^{2})^{\frac{1}{2}} (-2x) =$$

$$-3 x \sqrt{1-x^{2}}.$$

$$[-\frac{1}{3}(1-x^{2})^{\frac{3}{2}}]^{‘} = x \sqrt{1-x^{2}}.$$

$$\int_{0}^{1} x dx – \int_{0}^{1} x^{3} dx + \int_{0}^{1} x \sqrt{1-x^{2}} dx =$$

$$\frac{1}{2} – \frac{1}{4} -[\frac{1}{3}(1-x^{2})^{\frac{3}{2}}|_{0}^{1}] =$$

$$\frac{1}{2} – \frac{1}{4} – (0-\frac{1}{3}) =$$

$$\frac{1}{4} + \frac{1}{3} = \frac{7}{12}.$$