Calculating the area between a curve and an axis How can I calculate the area between $p(t) =\begin{pmatrix}a(t-\sin(t)) \\ a(1-\cos(t)) \end{pmatrix}$ ($t\in (0,2\pi)$) and the x-axis.
I thought of a reparameterization to view it as an usual integral, but couldn't work it through. Thank you in advance!
 A: Hint:
WLOG $a=1$ and let us consider $t$ in the first quadrant.
From $y=1-\cos t$ you draw $t=\arccos(1-y)$ and $x=\arccos(1-y)-\sqrt{1-(1-y)^2}.$
Now, rather than $\int y\,dx$, you may consider $\int x\,dy$, which is complementary.
WA gives the antiderivative that follows: http://www.wolframalpha.com/input/?i=integrate+arccos(1-y)-sqrt(1-(1-y)%5E2). Not really nice.
Try with Green's theorem instead.
A: I used Green's theorem - thanks to Yves Daoust for the hint:
$C$ is counterclockwise curve that is the boundary of $D$.
$C = C_1 \cup C_2$ where $C_1$ is the straight line between the points $(0,0)$ and $(2a\pi,0)$ with parametrisation $q(t) =\begin{pmatrix}a.t \\ 0 \end{pmatrix}$ for $t\in (0,2\pi)$ and $C_2$ is the curve that has $p(t)$ as parametrisation. If $C_2$ is traced in the counterclockwise direction, the value of its parameter decreases from $2\pi$ to $0$.  
$$\textrm{area}\:D =\frac{1}{2}\int_{C} xdy-ydx \\
= \frac{1}{2}\int_{C_1} x\,dy-y\,dx + \frac{1}{2}\int_{C_2} x\,dy-y\,dx\\
= \frac{1}{2}\int_{0}^{2\pi} 0\,dt +
\frac{1}{2}\int_{2\pi}^{0} \left[a(t-\sin t).a\sin t - a(1-\cos t).a.(1-\cos t)\right]dt  \\
= \frac{a^2}{2}\int_{2\pi}^{0} \left(t\sin t+2\cos t - 2\right)dt\\
= \frac{a^2}{2}\left(\left. 3\sin t - t\cos t - 2t\,\right|^{0}_{2\pi} \right)= \frac{a^2}{2}\cdot(6\pi)=3a^2\pi   
$$
The parametrisation $p(t)$ has an intresting property, namely $y=\dfrac{dx}{dt}$. Therefore, I tried to solve the problem in another manner:
$$\textrm{area}\:D=\int f(x)dx =\int y\,dx = \int \dfrac{dx}{dt}.dx = \int \dfrac{dx}{dt}\cdot\dfrac{dx}{dt}\cdot dt = \int_0^{2\pi}\left(\dfrac{dx}{dt}\right)^2\cdot dt$$
Let's compute this integral.
$$\textrm{area}\:D=\int_0^{2\pi}\left(\dfrac{dx}{dt}\right)^2 dt=\int_0^{2\pi} a^2(1-\cos t)^2\,dt\\
=\int_0^{2\pi} a^2\left(1-2\cos t +\cos^2 t \right)dt
=a^2\left(\left. \frac{3}{2}t-2\sin t +\frac{1}{4}\sin(2t) \right|^{2\pi}_0 \right)=a^2(3\pi)=3a^2\pi$$
Fortunately, I get the same solution. 
