## Problem Statement

Calculate the lengths of the arcs of the curves given by the parametric equations.

$$
\begin{aligned}
& \left\{\begin{array}{l}
x=6 \cos ^{3} t \\
y=6 \sin ^{3} t
\end{array}\right. \\
& 0 \leq t \leq \frac{\pi}{3}
\end{aligned}
$$
## Solution

The length of the arc of a curve defined by parametric equations is determined by the formula

$$
L=\int_{t_{1}}^{t_{2}} \sqrt{\left(x_{t}^{\prime}\right)^{2}+\left(y_{t}^{\prime}\right)^{2}} d t
$$

From the equations of the curve, we find:

$$
\begin{aligned}
& x=6 \cos ^{3} t ; x_{t}^{\prime}=6 \cdot 3 \cos ^{2} t(\cos t)^{\prime}=-18 \cos ^{2} t \sin t \\
& y=6 \sin ^{3} t ; y_{t}^{\prime}=6 \cdot 3 \sin ^{2} t(\sin t)^{\prime}=18 \sin ^{2} t \cos t
\end{aligned}
$$

We obtain:

$$
\begin{aligned}
L & =\int_{0}^{\pi / 3} \sqrt{\left(-18 \cos ^{2} t \sin t\right)^{2}+\left(18 \sin ^{2} t \cos t\right)^{2}} d t= \\
& =18 \int_{0}^{\pi / 3} \sqrt{\cos ^{4} t \sin ^{2} t+\sin ^{4} t \cos ^{2} t} d t= \\
& =18 \int_{0}^{\pi / 3} \sqrt{\cos ^{2} t \sin ^{2} t} \cdot \sqrt{\sin ^{2} t+\cos ^{2} t} d t= \\
& =18 \int_{0}^{\pi / 3} \sqrt{\cos ^{2} t \sin ^{2} t} \cdot \sqrt{1} d t=\left|\begin{array}{c}
0 \leq t \leq \pi / 3 \\
\sin (t) \geq 0 \\
\cos (t) \geq 0
\end{array}\right|= \\
& =18 \int_{0}^{\pi / 3} \cos t \sin t d t=\frac{18}{4} \int_{0}^{\pi / 3} \sin (2 t) d(2 t)= \\
& =-\left.\frac{9}{2} \cdot \cos (2 t)\right|_{0} ^{\pi / 3}=-\frac{9}{2} \cdot\left(\cos \frac{2 \pi}{3}-\cos 0\right)=-\frac{9}{2} \cdot\left(-\frac{1}{2}-1\right)=\frac{27}{4}
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/%D0%9A%D1%83%D0%B7%D0%BD%D0%B5%D1%86%D0%BE%D0%B2_%D0%98%D0%BD%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_18-11»

Categories: Kuznetsov's Problem Book Integrals Problem 18 | Integrals

- Last modified: 07:09, 21 May 2010.
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## Problem Kuznetsov Integrals 18-12

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