Answer
$\dfrac{-8}{3}$ for the clockwise orientation.
Work Step by Step
Here, we have $f(x, y)=x^2$ and $g (x, y)=2y^2$
This gives: $\dfrac{\partial f}{\partial x}=2x$ and $\dfrac{\partial g}{\partial y}=4y$
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, $=\int_{-1}^{1} \int_0^{\sqrt {1-x^2}} (2x+4y) \ dy \ dx$
or, $=\int_{-1}^{1} (2x\sqrt {1-x^2}-2x^2+2) \ dx$
After computing the above integral, we get $\oint_C (f dy-g dx)=\dfrac{8}{3}$
Also, we have $\oint_C (f dy-g dx)=\dfrac{-8}{3}$ for the clockwise orientation.
