Find the area of $\gamma(t)=(at-a\sin t,a-a\cos t)$ where $t\in[0,2\pi], a>0$
I have bought to use Green's theorem $A=\frac{1}{2}\oint_\gamma(-ydx+xdy)$
$\gamma(t)=(at-a\sin t,a-a\cos t)$
$\gamma'(t)=(a-a\cos t,a\sin t)$
$A=\frac{1}{2}\oint_\gamma(-ydx+xdy)=\frac{1}{2}\int_{0}^{2\pi}(-(a-a\cos t)^2+a^2t\sin t-a^2\sin^2 t )dt=\\=\frac{1}{2}\int_{0}^{2\pi}(-2a^2+2a^2\cos t+a^2\sin t)dt=\frac{1}{2}(-2a^2t+2a^2\sin t-a^2\cos t)|_{0}^{2\pi}=-4a^2\pi$
Which can not be as it is negative, where did it went wrong? I did not use the fact that $a>0$ maybe I had to try other integral
The red line is your curve.
If you don't close the curve, you don't have an enclosed area. The line y=0 will suffice to close the curve.
As for your integration: You can think $\frac 12 (x\ dy - y\ dx)$ as the green area. And the integral will be the sum of triangles with small changes in $(x,y)$ Since we are traversing clockwise, that area is $\frac 12 (x\ dy - y\ dx)$ which is of course the negative of the more familiar $\frac 12 (y\ dx - x\ dy)$ associated with counter-clockwise paths.
As far what does the line y=0 do to your integration? As all triangles connecting the origin to this path are degenerate triangles of 0 area, it has 0 impact.