# Prove that the arc length of the polar curve r = a(1-cos theta) is 8a?

Sep 15, 2017

The polar arc length of a curve is given by:

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

We have:

$r = a \left(1 - \cos \theta\right)$
$\setminus \setminus = a - a \cos \theta$

Thus:

$\frac{\mathrm{dr}}{d \theta} = a \sin \theta$

So, the arc length is:

$L = {\int}_{0}^{2 \pi} \setminus \sqrt{{\left(a - a \cos \theta\right)}^{2} + {\left(a \sin \theta\right)}^{2}} \setminus d \theta$
$\setminus \setminus \setminus = {\int}_{0}^{2 \pi} \setminus \sqrt{\left({a}^{2} - 2 {a}^{2} \cos \theta + {a}^{2} {\cos}^{2} \theta\right) + {\left(a \sin \theta\right)}^{2}} \setminus d \theta$
$\setminus \setminus \setminus = {\int}_{0}^{2 \pi} \setminus a \sqrt{1 - 2 \cos \theta + {\cos}^{2} \theta + {\sin}^{2} \theta} \setminus d \theta$
$\setminus \setminus \setminus = a \setminus {\int}_{0}^{2 \pi} \sqrt{2 - 2 \cos \theta} \setminus d \theta$

$\setminus \setminus \setminus = a \setminus {\int}_{0}^{2 \pi} \sqrt{2} \sqrt{1 - \cos \theta} \setminus d \theta$

Using the trig identity:

$\cos 2 x \equiv 1 - 2 {\sin}^{2} x \implies 2 {\sin}^{2} x \equiv 1 - \cos 2 x$

We then have:

$L = a \setminus {\int}_{0}^{2 \pi} \sqrt{2} \sqrt{2 {\sin}^{2} \left(\frac{\theta}{2}\right)} \setminus d \theta$
$\setminus \setminus \setminus = a \setminus {\int}_{0}^{2 \pi} 2 \sin \left(\frac{\theta}{2}\right) \setminus d \theta$
$\setminus \setminus \setminus = 2 a \setminus {\left[- 2 \cos \left(\frac{\theta}{2}\right)\right]}_{0}^{2 \pi}$
$\setminus \setminus \setminus = - 4 a \setminus \left(\cos \pi - \cos 0\right)$
$\setminus \setminus \setminus = - 4 a \setminus \left(- 1 - 1\right)$
$\setminus \setminus \setminus = \left(- 4 a\right) \left(- 2\right)$
$\setminus \setminus \setminus = 8 a$ QED