## Problem Statement

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

$$
\begin{aligned}
& \left\{\begin{array}{l}
x=6(\cos t+t \sin t) \\
y=6(\sin t-t \cos t)
\end{array}\right. \\
& 0 \leq t \leq \pi
\end{aligned}
$$
## Solution

We will use the formula for calculating the length of a curve segment:

$l=\int_{t_{0}}^{t_{1}} \sqrt{(d x)^{2}+(d y)^{2}}$

For the given problem, we find:

$d x=x^{\prime} d t=(6(\cos t+t \sin t))^{\prime} d t=6(-\sin t+\sin t+t \cos t) d t=6 t \cos t d t ;$

$d y=y^{\prime} d t=(6(\sin t-t \cos t))^{\prime} d t=6(\cos t-\cos t+t \sin t) d t=6 t \sin t d t$.

According to formula (1), we get:

$l=\int_{0}^{\pi} \sqrt{(6 t \cos t d t)^{2}+(6 t \sin t d t)^{2}}=\int_{0}^{\pi} \sqrt{\left.6^{2}\left(t^{2} \cos ^{2} t+t^{2} \sin ^{2} t\right)\right) d t^{2}}=$

$=\int_{0}^{\pi} 6 \sqrt{\left.t^{2}\left(\cos ^{2} t+\sin ^{2} t\right)\right)} d t=\int_{0}^{\pi} 6 \sqrt{t^{2}} d t=\int_{0}^{\pi} 6 t d t=\left.3 t^{2}\right|_{0} ^{\pi}=3 \pi^{2}$.

In the process of calculation, we used the fundamental trigonometric identity $\cos ^{2} t+\sin ^{2} t=1$.

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� $\% \mathrm{D} 0 \% 9 \mathrm{~A} \% \mathrm{D} 1 \% 83 \% \mathrm{D} 0 \% \mathrm{~B} 7 \% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 1 \% 86 \% \mathrm{D} 0 \% \mathrm{BE} \% \mathrm{D} 0 \% \mathrm{~B} 2 \mathrm{\%} 0 \mathrm{D} 0 \% 98 \% \mathrm{D} 0 \% \mathrm{BD}$ $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+18-15$ »

Categories: Kuznetsov's Problem Book Integrals Problem 18 | Integrals | Problems for Checking

Kuznetsov Integrals Problem 18-15 - PlusPi

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## Kuznetsov's Problem Book Integrals 18-16

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