Answer
$\dfrac{8}{3}$
Work Step by Step
Here, we have $f(x, y)=0$ and $g (x, y)=xy$
This gives: $\dfrac{\partial f}{\partial x}=0$ and $\dfrac{\partial g}{\partial y}=x$
Next, we will use the Green's Theorem to compute the vector field.
$\oint_C (f dy-g dx)=\int_R (\dfrac{\partial f}{\partial x}+\dfrac{\partial g}{\partial y}) \ dA$
or, $=\int_0^2 \int_0^{4-2x} x \ dy \ dx$
or, $=\int_0^2 (-2x^2 +4x) \ dx$
or, $=[\dfrac{-2x^3}{3} +\dfrac{4x^2}{2}]_0^2 $
or, $=\dfrac{8}{3}$
