Answer
$I_x=\iint_{D} \rho y^2 dA$ and $I_y=\iint_{D} \rho x^2 dA$
Work Step by Step
Green's Theorem states that:
$\oint_C M \,dx+ N \,dy=\iint_{D}(\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y})dA$
The moment of inertia about $x$-axis can be given by: $I_x=\dfrac{-\rho}{3} \oint_{C} y^3 dx = \dfrac{-\rho}{3} \iint_{D}(\dfrac{\partial (0)}{\partial x}-\dfrac{\partial (-y^2)}{\partial y})dA =\dfrac{-\rho}{3} \oint_{C} 3y^2 dx = \iint_{D} \rho y^2 dA$
The moment of inertia about $y$-axis can be given by: $I_y= \dfrac{\rho}{3} \oint_{C} x^3 dy =\dfrac{\rho}{3} \iint_{D} \iint_{D}(\dfrac{\partial (x^3)}{\partial x}-\dfrac{\partial (0)}{\partial y})dA =\dfrac{\rho}{3} \iint_{D} 3x^2 dA = \iint_{D} \rho x^2 dA$
