# How do you find the length of cardioid r = 1 - cos theta?

Jun 23, 2016

8

#### Explanation:

arc length in polar is

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

so $\mathrm{ds} = \sqrt{{\sin}^{2} \theta + {\left(1 - \cos \theta\right)}^{2}} \setminus d \theta$

$= \sqrt{{\sin}^{2} \theta + 1 - 2 \cos \theta + {\cos}^{2} \theta} \setminus d \theta$

$= \sqrt{2} \setminus \sqrt{1 - \cos \theta} \setminus d \theta$

 = sqrt(2) \ sqrt{ 1- (1 - 2 sin^2 (theta/2)) \ d theta

$= \sqrt{2} \setminus \sqrt{2 {\sin}^{2} \left(\frac{\theta}{2}\right)} \setminus d \theta$

$= 2 \setminus \sin \left(\frac{\theta}{2}\right) \setminus d \theta$

So, assuming length is of one full revolution.....
$S = 2 \setminus \setminus {\int}_{0}^{\setminus \textcolor{red}{2 \pi}} \setminus \sin \left(\frac{\theta}{2}\right) \setminus d \theta$

$= 2 \setminus {\left[- 2 \cos \left(\frac{\theta}{2}\right)\right]}_{0}^{2 \pi}$

$= 4 {\left[\cos \left(\frac{\theta}{2}\right)\right]}_{2 \pi}^{0}$

$= 4 \left[1 - \left(- 1\right)\right] = 8$