1298. Find the length of the astroid

$$
x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}
$$
Solution. Applying formula (1), we will compute the length of the arc $AB$, i.e., $\frac{1}{4}$ of the astroid (Fig. 84). We find the derivative of the implicit function $y$ with respect to $x$, obtaining

$$
\frac{2}{3} x^{-\frac{1}{3}}+\frac{2}{3} y^{-\frac{1}{3}} y_{x}^{\prime}=0
$$

from which

$$
y_{x}^{\prime}=-\frac{y^{\frac{1}{3}}}{x^{\frac{1}{3}}} \text { and } \sqrt{1+y^{\prime 2}}=\int \frac{x^{\frac{2}{3}}+y^{\frac{2}{3}}}{x^{\frac{2}{3}}}=\frac{a^{\frac{1}{3}}}{x^{\frac{1}{3}}}
$$

Therefore,

$$
\frac{1}{4} s=\int_{0}^{a} \sqrt{1+y^{\prime 2}} d x=\int_{0}^{a} a^{\frac{1}{3}} x^{-\frac{1}{3}} d x=\left.a^{\frac{1}{3}} \frac{3 x^{\frac{2}{3}}}{2}\right|_{0} ^{a}=\frac{3 a}{2} \text { and } s=6 a
$$