# What is the arclength of the polar curve f(theta) = 2cos(9theta)-3sintheta  over theta in [0,pi/3] ?

Jul 15, 2018

$\approx 13.20973$

#### Explanation:

We use the formula

$s = {\int}_{\alpha}^{\beta} \setminus \sqrt{{r}^{2} + {\left(\frac{\mathrm{dr}}{d \theta}\right)}^{2}} d \theta$
where

$r \left(\theta\right) = 2 \cos \left(9 \theta\right) - 3 \sin \left(\theta\right)$
$r ' \left(\theta\right) = - 18 \sin \left(9 \theta\right) - 3 \cos \left(\theta\right)$

so we have to integrate

${\int}_{0}^{\frac{\pi}{3}} \sqrt{173 - 160 \cos \left(18 \theta\right) + 60 \sin \left(8 \theta\right) + 48 \sin \left(10 \theta\right)} d \theta$
by a numerical method we get

$\approx 13.20973$