Answer
$a) \ 0$ $ b) \ 0 $ and $c) $ Conservative
Work Step by Step
The two-dimensional curl of the vector field $F=\lt f, g\gt$ can be written as: $\dfrac{\partial g}{\partial x}-\dfrac{\partial f}{\partial y}$.
Here, we have; $f(x, y)=x$ and $g(x, y)=y$
a) The curl of the vector field $F$ can be computed as:
$\ curl=\dfrac{\partial g}{\partial x}-\dfrac{\partial f}{\partial y}=0-0=0$
This implies that the given vector field is irrotational on that region.
b) Next, we will use Green's Theorem - Circulation form such as:
$\oint_C F \cdot \ dr=\oint_C f dx+g \ dy=\iint_R (\dfrac{\partial g}{\partial x} -\dfrac{\partial f}{\partial y}) \ dA=\int_R 0 \ dA=0~~~(1)$
Now, we will compute the integral.
$\oint F \cdot \ dr=\int_0^{2\pi} F[r(t)] r'(t) \ dt=\int_0^{2\pi} \sqrt 2 (\cos t, \sin t) \cdot \sqrt 2 (-\sin t, \cos t) \ dt =0 ~~~~(2)$
We see that both equations (1) and (2) are same.
c) From part (a), we have the vector field $F$ is irrotational. This means that the irrotational vector fields on the connected region in $R^2$ are conservative. So, we conclude that the given vector field is conservative. Also, the curl of $F$ is equal to $0$ which defines the vector field is conservative.
Thus, our results are: $a) \ 0$ $ b) \ 0 $ and $c) $ Conservative
