2015年考研数二第06题解析

题目

$$A. \int_{\frac{\pi}{2}}^{\frac{\pi}{3}} d \theta \int_{\frac{1}{2 \sin 2 \theta}}^{\frac{1}{\sin 2 \theta}} f(r \cos \theta, r \sin \theta) r dr$$

$$B. \int_{\frac{\pi}{2}}^{\frac{\pi}{3}} d \theta \int_{\frac{1}{\sqrt{2 \sin 2 \theta}}}^{\frac{1}{\sqrt{\sin 2 \theta}}} f(r \cos \theta, r \sin \theta) r dr$$

$$C. \int_{\frac{\pi}{2}}^{\frac{\pi}{3}} d \theta \int_{\frac{1}{2 \sin 2 \theta}}^{\frac{1}{\sin 2 \theta}} f(r \cos \theta, r \sin \theta) dr$$

$$D. \int_{\frac{\pi}{2}}^{\frac{\pi}{3}} d \theta \int_{\frac{1}{\sqrt{2 \sin 2 \theta}}}^{\frac{1}{\sqrt{\sin 2 \theta}}} f(r \cos \theta, r \sin \theta) dr$$

解析

$$xy = \frac{1}{4} \Rightarrow$$

$$r^{2}\sin \theta \cos \theta = \frac{1}{4} \Rightarrow$$

$$r^{2} \frac{1}{2} \sin 2 \theta = \frac{1}{4} \Rightarrow$$

$$r^{2} \sin 2 \theta = \frac{1}{2} \Rightarrow$$

$$r = \frac{1}{\sqrt{2 \sin 2 \theta}}.$$

$$xy = \frac{1}{2} \Rightarrow$$

$$r^{2} \sin \theta \cos \theta = \frac{1}{2} \Rightarrow$$

$$r^{2} \frac{1}{2} \sin 2 \theta = \frac{1}{2} \Rightarrow$$

$$r = \frac{1}{\sqrt{\sin 2 \theta}}.$$

EOF