Example 2.58. Compute the length of the arc of the astroid $x=\cos ^{3} t$, $y=\sin ^{3} t, 0 \leqslant t \leqslant 2 \pi$.
Solution. From Fig. 2.9, it is clear that due to symmetry, it is sufficient to compute only a quarter of the astroid for

![](https://cdn.mathpix.com/cropped/2024_05_22_0f84e17422d7c0df8d68g-084.jpg?height=307&width=449&top_left_y=128&top_left_x=388)

Fig. 2.9.

$0 \leqslant t \leqslant \pi / 2$. We will use formula (2.26):

$$
\begin{gathered}
L=\int_{0}^{\pi / 2} \sqrt{9 \sin ^{2} t \cos ^{4} t+9 \sin ^{4} t \cos ^{2} t} d t= \\
=\int_{0}^{\pi / 2} \sqrt{9 \sin ^{2} t \cos ^{2} t\left(\sin ^{2} t+\cos ^{2} t\right)} d t= \\
=12 \int_{0}^{\pi / 2}|\sin t \cos t| d t=6 \int_{0}^{\pi / 2}|\sin 2 t| d t=\left.6(-\cos (2 t))\right|_{0} ^{\pi / 2}=12
\end{gathered}
$$

Under the integral, it is necessary to write the arithmetic value of the root. If this is not done, an incorrect result can be obtained. In the example considered here, if instead of $3 \int_{0}^{\pi / 2}|\sin t \cos t| d t$ the integral $3 \int_{0}^{\pi / 2} \sin t \cos t d t$ were written, the result would be $\left.\frac{3}{2} \sin ^{2} t\right|_{0} ^{2 \pi}=0-$ a zero arc length.