# 计算平面曲线的弧长需要知道积分上下限，但如果这个积分上线限题目中没有给出该怎么办？

## 二、解析

$$\textcolor{springgreen}{ l=\int_{a}^{b} \sqrt{1+y^{\prime 2}(x)} \mathrm{~ d} x } \quad \mathrm{~ d} x \in(a, b)$$

$$y=\int_{-\sqrt{3}}^{x} \sqrt{3-t^{2}} \mathrm{~ d} t \Rightarrow$$

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

$$3-x^{2} \geqslant 0 \Rightarrow x^{2} \leq 3 \Rightarrow$$

$$\textcolor{orange}{ x \in(-\sqrt{3}, \sqrt{3}) }$$

$$l=\int_{-\sqrt{3}}^{\sqrt{3}} d \sqrt{1+3-x^{2}} \mathrm{~ d} x=$$

$$\textcolor{orange}{ \int_{-\sqrt{3}}^{\sqrt{3}} \sqrt{4-x^{2}} \mathrm{~ d} x }$$

Next

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

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

$$\left.x^{2} \cdot \frac{\sqrt{4-x^{2}}}{2 x}\right|_{-\sqrt{3}} ^{\sqrt{3}}-\int_{-\sqrt{3}}^{\sqrt{3}} x^{2} \mathrm{~ d} \left(\frac{\sqrt{4-x^{2}}}{2 x}\right)$$

$$I_{1}=\left.x^{2} \cdot \frac{\sqrt{4-x^{2}}}{2 x}\right|_{-\sqrt{3}} ^{\sqrt{3}}=\left.\frac{1}{2} x \sqrt{4-x^{2}}\right|_{-\sqrt{3}} ^{\sqrt{3}}=\sqrt{3}$$

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

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

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

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

$$\int_{-\sqrt{3}}^{\sqrt{3}} \frac{-8}{4 \sqrt{4-x^{2}}} \mathrm{~ d} x=$$

$$-2 \int_{-\sqrt{3}}^{\sqrt{3}} \frac{1}{\sqrt{4-x^{2}}} \mathrm{~ d} x=$$

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

$$-\left.2 \arcsin \frac{x}{2}\right|_{-\sqrt{3}} ^{\sqrt{3}}=-2\left(\frac{\pi}{3}+\frac{\pi}{3}\right)=\frac{-4 \pi}{3}$$

$$l=I_{1}-I_{2}=\sqrt{3}+\frac{4}{3} \pi$$

Next

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

$$x=2 \sin t \Rightarrow \sin t \in\left(\frac{-\sqrt{3}}{2}, \frac{\sqrt{3}}{2}\right) \Rightarrow$$

$$t \in\left(\frac{-\pi}{3}, \frac{\pi}{3}\right)$$

$$4-x^{2}=4-4 \sin ^{2} t=4 \cos ^{2} t$$

$$\mathrm{~ d} x=2 \cos t \mathrm{~ d} t \Rightarrow$$

$$l=\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}} 2 \cos t \cdot 2 \cos t \mathrm{~ d} t=4 \int_{\frac{-\pi}{3}}^{\frac{\pi}{3}} \cos ^{2} t \mathrm{~ d} t \Rightarrow$$

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

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

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

$$l=2 \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(\cos 2 t+1) \mathrm{~ d} t=\left.2\left(\frac{1}{2} \sin 2 t+t\right)\right|_{\frac{-\pi}{3}} ^{\frac{\pi}{3}}=$$

$$\left.\sin 2 t\right|_{\frac{-\pi}{3}} ^{\frac{\pi}{3}}+\left.2 t\right|_{\frac{-\pi}{3}} ^{\frac{\pi}{3}}=\left(\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right)+\frac{4}{3} \pi \Rightarrow$$

$$l=\sqrt{3}+\frac{4}{3} \pi$$