How to calculate the line integral with respect to the circle in counterclockwise direction Consider the vector field $F=<y,-x>$.
Compute the line integral $$\int_CF\cdot dr$$ where $C$ is the circle of radius $3$ centered at the origin counterclockwise.
My Try:
The circle is $x^2+y^2=9$
$$\cases{x=3\cos t \\ y=3\sin t} \text{ for } 0\le t\le2\pi$$
Now how do I calculate $\int_CF\cdot dr$?
Can anyone explain how to solve this?

The radial vector is 
$\vec r = \begin{pmatrix} x \\ y \end{pmatrix}; \tag 1$
also,
$F = \begin{pmatrix} y \\ - x \end{pmatrix}; \tag 2$
with
$x = 3 \cos t, \tag 3$
$y = 3 \sin t, \tag 4$
we have
$\vec r = \begin{pmatrix} 3 \cos t \\ 3\sin t \end{pmatrix}, \tag 5$
$d \vec r = \dfrac{d \vec r}{dt} dt = \begin{pmatrix} -3\sin t \\ 3\cos t \end{pmatrix} dt, \tag 5$
and 
$F = \begin{pmatrix} 3\sin t \\ -3\cos t \end{pmatrix}; \tag 6$
then
$F \cdot d \vec r = \begin{pmatrix} 3\sin t \\ -3\cos t \end{pmatrix} \cdot \begin{pmatrix} -3\sin t \\ 3\cos t \end{pmatrix} dt$
$= (-9 \sin^2 t - 9 \cos^2 t)dt = -9(\sin^2 t + \cos^2 t)dt = -9(1)dt = -9dt; \tag 7$
finally,
$\displaystyle \int_C F \cdot d \vec r = \int_C -9 dt = -9 \int_C dt = -9(2\pi) = -18\pi. \tag 8$
