Answer
$-18\pi $ ; not conservative
Work Step by Step
For a vector field to be Conservative, $\dfrac{\partial f}{\partial y}=\dfrac{\partial g}{\partial x}$
We are given that $F(x,y)=(y, \ -x)$
Also, the curve is $r(t)= (3 \cos t, 3 \sin t)$
This implies that $r'(t) = (-3 \sin t, 3 \cos t)$
Therefore, the integral is:
$\oint F \ dr=\int_0^{2 \pi} F[r(t)] r'(t) \ dt\\=\int_0^{2 \pi} (3 \sin t, -3 \cos t) (-3 \sin t, 3 \cos t)\\=\int_0^{2 \pi} (-9) \ dt \\=-9 \times (2\pi-0) \\=-18\pi $
Since, the integral is not zero,this means that the given vector field is not conservative.
