# What is the arclength of r=3/4theta  on theta in [-pi,pi]?

May 19, 2018

$L = \frac{3}{4} \pi \sqrt{{\pi}^{2} + 1} + \frac{3}{4} \ln \left(\pi + \sqrt{{\pi}^{2} + 1}\right)$ units.

#### Explanation:

$r = \frac{3}{4} \theta$
${r}^{2} = \frac{9}{16} {\theta}^{2}$

$r ' = \frac{3}{4}$
${\left(r '\right)}^{2} = \frac{9}{16}$

Arclength is given by:

$L = {\int}_{-} {\pi}^{\pi} \sqrt{\frac{9}{16} {\theta}^{2} + \frac{9}{16}} d \theta$

Simplify:

$L = \frac{3}{4} {\int}_{-} {\pi}^{\pi} \sqrt{{\theta}^{2} + 1} d \theta$

From symmetry:

$L = \frac{3}{2} {\int}_{0}^{\pi} \sqrt{{\theta}^{2} + 1} d \theta$

Apply the substitution $\theta = \tan \phi$:

$L = \frac{3}{2} \int {\sec}^{3} \phi \mathrm{dp} h i$

This is a known integral:

$L = \frac{3}{4} \left[\sec \phi \tan \phi + \ln | \sec \phi + \tan \phi |\right]$

Reverse the substitution:

$L = \frac{3}{4} {\left[\theta \sqrt{{\theta}^{2} + 1} + \ln | \theta + \sqrt{{\theta}^{2} + 1} |\right]}_{0}^{\pi}$

Insert the limits of integration:

$L = \frac{3}{4} \pi \sqrt{{\pi}^{2} + 1} + \frac{3}{4} \ln \left(\pi + \sqrt{{\pi}^{2} + 1}\right)$