# 2023年考研数二第12题解析：曲线弧长计算、凑微分、挖掘隐含条件

## 二、解析

$$l = \int_{a}^{b} \sqrt{1+ y^{\prime 2}} \mathrm{~d} x$$

$$y^{\prime}=\sqrt{3-x^{2}} \Rightarrow y^{\prime 2}=3-x^{2} \Rightarrow$$

$$y=\sqrt{3-t^{2}} \Rightarrow y^{2}+t^{2}=3 \Rightarrow r=\sqrt{3}$$

$$l=\int_{-\sqrt{3}}^{\sqrt{3}} \sqrt{4-x^{2}} \mathrm{~ d} x =$$

$$2 \int_{-\sqrt{3}}^{\sqrt{3}} \sqrt{1-\left(\frac{x}{2}\right)^{2}} \mathrm{~ d} x=$$

$$4 \int_{0}^{\sqrt{3}} \sqrt{1-\left(\frac{x}{2}\right)^{2}} \mathrm{~ d} x=$$

$$8 \int_{0}^{\sqrt{3}} \sqrt{1-\left(\frac{x}{2}\right)^{2}} \mathrm{~ d} \left(\frac{x}{2}\right)=$$

$$8 \int_{0}^{\frac{\sqrt{3}}{2}} \sqrt{1-x^{2}} \mathrm{~ d} x \Rightarrow$$

$$x=\sin t \Rightarrow t \in\left(0, \frac{\pi}{3}\right) \Rightarrow$$

$$8 \int_{0}^{\frac{\pi}{3}} \cos ^{2} t \mathrm{~ d} t \Rightarrow$$

$$\cos 2 \alpha=2 \cos ^{2} \alpha-1 \Rightarrow$$

$$\cos ^{2} \alpha=\frac{1}{2}(1+\cos 2 \alpha) \Rightarrow$$

$$4 \int_{0}^{\frac{\pi}{3}}(1+\cos 2 t) \mathrm{~ d} t=$$

$$2 \int_{0}^{\frac{\pi}{3}}(1+\cos 2 t) \mathrm{~ d} (2 t)=$$

$$2 \int_{0}^{\frac{2 \pi}{3}}(1+\cos t) \mathrm{~ d} t=$$

$$2\left[\frac{2 \pi}{3} + \sin t \Big|_{0} ^{\frac{2 \pi}{3}}\right]=$$

$$2\left[\frac{2 \pi}{3}+\frac{\sqrt{3}}{2}-0\right] = \frac{4 \pi}{3} + \sqrt{3}$$