Chapter H.2, Problem 38E

To determine

To find: the exact length of the polar curve.

Answer to Problem 38E

Theexact length of the curve is $\pi \sqrt{4{\pi }^2+1}+\frac{1}{2}\left[{\sinh }^{-1}\left(2\pi \right)-{\sinh }^{-1}\left(0\right)\right]$

Explanation of Solution

Given:

Given that the curve $r=\theta ,0\le \theta \le 2\pi$

Calculation:

Theexactlength of the curve $r=\theta$ is

  $\begin{array}{c}L={\displaystyle \underset{0}{\overset{2\pi }{\int }}\sqrt{r^2+{\left(\frac{dr}{d\theta }\right)}^2}}d\theta \\ ={\displaystyle \underset{0}{\overset{2\pi }{\int }}\sqrt{{\left(\theta \right)}^2+{\left(1\right)}^2}d\theta }\\ ={\displaystyle \underset{0}{\overset{2\pi }{\int }}\sqrt{{\theta }^4+1}d\theta }\\ ={\left[\frac{\theta \sqrt{{\theta }^2+1}}{2}\frac{1}{2}{\sinh }^{-1}\left(\theta \right)\right]}_0^{2\pi }\left[\because {\displaystyle \int \sqrt{a^2+x^2}}dx=\frac{x}{2}\sqrt{a^2+x^2}+\frac{a^2}{2}\ln \left(x+\sqrt{a^2+x^2}\right)\right]\\ =\left[\frac{2\pi \sqrt{4{\pi }^2+1}}{2}+\frac{1}{2}{\sinh }^{-1}\left(2\pi \right)\right]-\frac{1}{2}\left[{\sinh }^{-1}\left(0\right)\right]\\ =\pi \sqrt{4{\pi }^2+1}+\frac{1}{2}\left[{\sinh }^{-1}\left(2\pi \right)-{\sinh }^{-1}\left(0\right)\right]\end{array}$

Conclusion:

Therefore, theexact length of the curve is $\pi \sqrt{4{\pi }^2+1}+\frac{1}{2}\left[{\sinh }^{-1}\left(2\pi \right)-{\sinh }^{-1}\left(0\right)\right]$