Example 1. Find the length of the arc of the astroid $x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$. What is the length of the astroid when $a=1, a=\frac{2}{3}?$
Solution. Since the astroid

![](https://cdn.mathpix.com/cropped/2024_05_22_7ef2114a321cf223ba99g-214.jpg?height=389&width=478&top_left_y=515&top_left_x=12)

Fig. 5.13 is symmetric with respect to the coordinate axes (Fig. 5.13), it is sufficient to compute the length of the arc $AB$ and multiply the result by 4.

Differentiating the function $x^{\frac{2}{3}} + y^{\frac{2}{3}} = a^{\frac{2}{3}}$ implicitly, we get

$$
\frac{2}{3} x^{-\frac{1}{3}} + \frac{2}{3} y^{-\frac{1}{3}} y' = 0, \quad y' = -\frac{y^{\frac{1}{3}}}{x^{\frac{1}{3}}}
$$

We find the expression for the integrand in formula (2.18). We have

$$
\sqrt{1 + y'^2} = \sqrt{1 + \frac{y^{\frac{2}{3}}}{x^{\frac{2}{3}}}} = \sqrt{\frac{x^{\frac{2}{3}} + y^{\frac{2}{3}}}{x^{\frac{2}{3}}}} = \sqrt{\frac{a^{\frac{2}{3}}}{x^{\frac{2}{3}}}} = \frac{a^{\frac{1}{3}}}{x^{\frac{1}{3}}}
$$

Using formula (5.18), we find

$$
\begin{aligned}
\frac{1}{4} l & = \int_{0}^{a} \sqrt{1 + y'^2} \, dx = \int_{0}^{a} \frac{a^{\frac{1}{3}}}{x^{\frac{1}{3}}} \, dx = a^{\frac{1}{3}} \int_{0}^{a} x^{-\frac{1}{3}} \, dx = \\
& = \left. a^{\frac{1}{3}} \frac{x^{\frac{2}{3}}}{\frac{2}{3}} \right|_{0}^{a} = \left. \frac{3}{2} a^{\frac{1}{3}} x^{\frac{2}{3}} \right|_{0}^{a} = \frac{3}{2} a, \quad l = 6a
\end{aligned}
$$

For $a = 1$, we get $l = 6$, and for $a = \frac{2}{3}$, we get $l = 4$.