2014年考研数二第17题解析：二重积分、极坐标系

题目

$$\iint_{D} \frac{x \sin (\pi \sqrt{x^{2}+y^{2}})}{x+y} dxdy.$$

解析

方法一

$$I = \iint_{D} \frac{x \sin (\pi \sqrt{x^{2}+y^{2}})}{x+y} dxdy.$$

$$I = \iint_{D} \frac{y \sin (\pi \sqrt{x^{2}+y^{2}})}{x+y} dxdy.$$

$$I=$$

$$\frac{1}{2}[\iint_{D} \frac{x \sin (\pi \sqrt{x^{2}+y^{2}})}{x+y} dxdy + \iint_{D} \frac{y \sin (\pi \sqrt{x^{2}+y^{2}})}{x+y} dxdy] \Rightarrow$$

$$I = \frac{1}{2} \iint_{D} \frac{(x+y) \sin (\pi \sqrt{x^{2} + y^{2}})}{x+y} \Rightarrow$$

$$I = \frac{1}{2} \iint_{D} \sin (\pi \sqrt{x^{2} + y^{2}}) \Rightarrow 转化为极坐标系 \Rightarrow$$

$$I = \frac{1}{2} \iint_{D} \sin (\pi r) \cdot r dr d \theta \Rightarrow$$

$$I = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} d \theta \int_{1}^{2} r \sin (\pi r) dr.$$

$$\frac{1}{2} \int_{0}^{\frac{\pi}{2}} d \theta = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}.$$

$$\int_{1}^{2} r \sin (\pi r) dr =$$

$$\frac{1}{\pi^{2}} \int_{\pi}^{2\pi} (\pi r) \sin (\pi r) d(\pi r) =$$

$$\frac{1}{\pi^{2}} \int_{\pi}^{2\pi} A \sin A d A =$$

$$\frac{-1}{\pi^{2}} \int_{\pi}^{2 \pi} A d(\cos A) \Rightarrow$$

$$\frac{-1}{\pi^{2}} [A \cos A |_{\pi}^{2 \pi} – \int_{\pi}^{2 \pi} \cos A dA] \Rightarrow$$

$$\frac{-1}{\pi^{2}} [A \cos A |_{\pi}^{2 \pi} – \sin A|_{\pi}^{2 \pi}] \Rightarrow$$

$$\frac{-1}{\pi^{2}} [2 \pi + \pi – 0] = \frac{-3}{\pi}.$$

$$I = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} d \theta \int_{1}^{2} r \sin (\pi r) dr \Rightarrow$$

$$I = \frac{\pi}{4} \cdot \frac{-3}{\pi} = \frac{-3}{4}.$$

方法二

$$\iint_{D} \frac{x \sin (\pi \sqrt{x^{2}+y^{2}})}{x+y} r d \theta dr \Rightarrow 转化为极坐标系 \Rightarrow$$

$$\iint_{D} \frac{r \cos \theta \sin (\pi r)}{r \cos \theta + r \sin \theta} r d \theta dr \Rightarrow$$

$$\iint_{D} \frac{\cos \theta \sin (\pi r)}{ \sin \theta + \cos \theta} r d \theta d r \Rightarrow$$

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

$$\int_{1}^{2} r \sin (\pi r) dr = \frac{-3}{\pi}.$$

$$\int_{0}^{\frac{\pi}{2}} \frac{\cos \theta}{\sin \theta + \cos \theta} d \theta = \int_{0}^{\frac{\pi}{2}} \frac{\sin \theta}{\sin \theta + \cos \theta} d \theta$$

$$\int_{0}^{\frac{\pi}{2}} \frac{\cos \theta}{\sin \theta + \cos \theta} d \theta =$$

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

$$\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{\sin \theta + \cos \theta}{\sin \theta + \cos \theta} d \theta =$$

$$\frac{1}{2} \int_{0}^{\frac{\pi}{2}} 1 d \theta = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}.$$

$$\iint_{D} \frac{x \sin (\pi \sqrt{x^{2}+y^{2}})}{x+y} r d \theta dr =$$

$$\frac{-3}{\pi} \cdot \frac{\pi}{4} = \frac{-3}{4}.$$

补充

$$\int_{0}^{\frac{\pi}{2}} \frac{\cos \theta}{\sin \theta + \cos \theta} d \theta =$$

$$\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{\cos \theta + \sin \theta + \cos \theta – \sin \theta}{\cos \theta + \sin \theta} d \theta =$$

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

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

$$\frac{1}{2} [\int_{0}^{\frac{\pi}{2}} 1 d \theta + \ln |\cos \theta + \sin \theta| |_{0}^{\frac{\pi}{2}}] =$$

$$\frac{1}{2} [\frac{\pi}{2} + \ln(0+1) – \ln (1+0)] =$$

$$\frac{1}{2} [\frac{\pi}{2} + 0 – 0] = \frac{\pi}{4}.$$