Answer
$$\int_C y^2 \ dx+x^2 \ dy=0$$
Work Step by Step
Green's Theorem states that: $\int_C fdx+g dy=\iint_D (\dfrac{\partial g}{\partial x}-\dfrac{\partial f}{\partial y})dA$
Here, we have $f=y^2 ; g =x^2$
Now, $\dfrac{\partial g}{\partial x} =2x$ and $\dfrac{\partial f}{\partial y}=2y$
Therefore,
$ \iint_D (\dfrac{\partial g}{\partial x}-\dfrac{\partial f}{\partial y})dA=\iint_D (2x-2y) \ dx \ dy = 2 \iint_D x \ dx \ dy -2 \iint_D y \ dx \ dy $
But by symmetry $ \iint_D x \ dx \ dy = \iint_D y \ dx \ dy=0 $
Thus, $\int_C y^2 \ dx+x^2 \ dy=0$