# Question #f6b13

Mar 27, 2017

$= 8 \pi$

#### Explanation:

The simplest thing to do here is to de-parameterise.

From $x \left(t\right) = 2 t$ we have $t = \frac{x}{2}$.

From $y \left(t\right) = 2 \left(1 - \cos t\right)$ we have $y \left(x\right) = 2 \left(1 - \cos \left(\frac{x}{2}\right)\right)$

For $t \in \left[0 , 2 \pi\right]$, we have $x \in \left[0 , 4 \pi\right]$

So we are looking at:

$2 {\int}_{x = 0}^{4 \pi} 1 - \cos \left(\frac{x}{2}\right) \setminus \mathrm{dx} = 8 \pi$

Have assumed that you know how to work with that integrand.