Chapter 11: Parametric Equations and Polar Coordinates - Section 11.5 - Areas and Lengths in Polar Coordinates - Exercises 11.5 - Page 671: 23

$8$

The length of the curve can be calculated as: $L=\int_{p}^{q}\sqrt{r^2+(\dfrac{dr}{d\theta})^2}d\theta$ Here, we have $L=\int_{0}^{2\pi} \sqrt{(1+\cos \theta)^2+(-\sin \theta)^2} d \theta=\sqrt 2 \int_{0}^{2 \pi}\sqrt {(1+\cos \theta)} d\theta$ $\implies L=2 \int_{0}^{2 \pi} |\cos (\theta/2)| d \theta$ Plug $2k=\theta \implies d\theta=2k dk$ Now, we have $L =4 \int_{0}^{\pi} |\cos k| d k$ $\implies L=4 \int_{0}^{\pi} 2 (\cos k) dk -4 \int_{\pi/2}^{\pi} (\cos k )dk=8$