I have to calculate the line integral of the vector field $f$ along the path described
$f(x,y)=(2a-y)i+xj$, along the path described by $\alpha(t)=a(t-\sin(t))i+a(1-\cos(t))j$, $0\leq t \leq 2\pi$.
Now the definition of the line integral is: $\int f \cdot d\alpha=\int_a^bf\big(\alpha(t)\big) \cdot \alpha'(t) \ dt$.
The first thing I did was to find $\alpha'(t).$
We have that $\alpha(t)=(at-a\sin(t))i+(a-a\cos(t))j= ait-ai\sin(t)+aj-aj\cos(t)$
Then $\alpha'(t)=ai-ai\cos(t)+aj\sin(t)=(a-a\cos(t))i+a\sin(t)j$
But how do I find $f\big(\alpha(t)\big)$? Here is what I did:
$y=a\sin(t)$ and $x=(a-a\cos(t))$
Then $f\big(\alpha(t)\big)= (2a-a\sin(t))i+(a-a\cos(t))j$
Now I can calculate the line integral.
$\int_a^bf\big(\alpha(t)\big) \cdot \alpha'(t) \ dt= \int_0^{2\pi}\big((2a-a\sin(t))+(a-a\cos(t))\big) \cdot \big( (a-a\cos(t))+a\sin(t) \big)\ dt=4\pi a^2$
Am I correct?