设曲线 $\gamma$ 由 $y^2=\frac{1}{3} x^2(1-4 x), y \geq 0, x \in\left[0, \frac{1}{4}\right]$ 所定义,计 算 $\gamma$ 的弧长
【答案】 令 $x=\frac{1-t^2}{4}, y=\frac{\sqrt{3} t\left(1-t^2\right)}{12}, t \in[0,1]$ ,则
$$
\begin{aligned}
\left(x^{\prime}(t)\right)^2+\left(y^{\prime}(t)\right)^2 & =\left(-\frac{t}{2}\right)^2+\left(\frac{\sqrt{3}}{12}\left(1-3 t^2\right)\right)^2 \\
& =\frac{t^2}{4}+\frac{1}{48}\left(1-6 t^2+9 t^4\right) \\
& =\frac{1}{48}\left(1+6 t^2+9 t^4\right) \\
& =\frac{1}{48}\left(3 t^2+1\right)^2
\end{aligned}
$$

于是由弧长计算公式,得
$$
\begin{aligned}
L_\gamma & =\int_0^1 \sqrt{\left(x^{\prime}(t)\right)^2+\left(y^{\prime}(t)\right)^2} \mathrm{~d} t=\int_0^1 \frac{3 t^2+1}{4 \sqrt{3}} d t \\
& =\frac{\sqrt{3}}{12}\left(t^3+\left.t\right|_0 ^1\right)=\frac{\sqrt{3}}{12}(2-0)=\frac{\sqrt{3}}{6}
\end{aligned}
$$


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