- Calculus, Volume 1: One Variable Calculus, with an Introduction to Linear Algebra
- Tom M. Apostol
- Second Edition
- 1967
- 978-1-119-49673-1
14.19 Exercises
- In each of Exercises 8 through 11, make a sketch of the plane curve having the given polar equation and compute its arc length. $$r = e^\theta,\quad 0 \leq \theta \leq \pi.$$
- We know from Exercise 4 that the arc length $s$ of the curve given by the polar equation $r = f(\theta)$ from $\theta = a$ to $\theta = b,$ where $a \leq \theta \leq b \leq a + 2\pi$ is: $$ \begin{align*} \\ s &= \int_a^b \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta. \end{align*} $$
- The arc length $s$ of the curve given by the polar equation $r = f(\theta)$ from $\theta = a$ to $\theta = b,$ where $a \leq \theta \leq b \leq a + 2\pi$ is: $$ \begin{align*} \\ s &= \int_a^b \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta. \end{align*} $$ This means that the arc length of the polar function with radius $r = e^{\theta}$ from $\theta = 0$ to $\theta = \pi$ is $$ \begin{align*} \\ s &= \int_0^{\pi} \sqrt{2e^{2\theta}}\,d\theta \\ \\ &= \sqrt{2}\int_0^{\pi} e^{\theta}\,d\theta \\ \\ &= \sqrt{2}\left(e^{\pi} - 1\right). \quad \blacksquare \end{align*} $$