Find the area enclosed by the curve $x=t-\sin t $, $y=1-\cos t$ from $0$ to $2\pi$

Question

Find the area enclosed by the curve $x=t-\sin t $, $y=1-\cos t$ from $0$ to $2\pi$ and the x-axis

Not sure how to execute. is it just that $.5\int_{0}^{2\pi}x^2+y^2$ ?

and not sure how to account for the x-axis limitation on the area

Answer 1

Reparametrize on $x'$ and $y'$ with $x' = \frac{x}{\pi} - 1$ and $y' = \frac{y}{2}$. You'll get a much clearer drawing, and probably much better formulae. After that, multiply your result by $2\pi$ to account for the initial (linear) transformation.

Answer 2

So, not knowing your level and what you've tried, I'm going to offer this

HINT: You want the area under the curve and above the $x$-axis, so you want $\displaystyle\int_{x=0}^{2\pi} y\,dx$. Rewrite everything in terms of $t$, putting in $y=y(t)$, $dx = \dfrac{dx}{dt}dt$, and you should get a reasonably simple $t$ integral from $t=0$ to $t=2\pi$.