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 A\,dx+B \,dy=\iint_{D}(\dfrac{\partial B}{\partial x}-\dfrac{\partial A}{\partial y}) dA $
We need to set up the line integral and compute the integrand of the double integral as follows:
The moment of inertia about the $x$-axis is:
$$I_x=\dfrac{-\rho}{3} \oint_{C} y^3 dx \\= \dfrac{-\rho}{3} \iint_{D}(\dfrac{\partial (0)}{\partial x}-\dfrac{\partial (-y^2)}{\partial y})dA
\\=\dfrac{-\rho}{3} \times \oint_{C} 3y^2 dx \\ = \iint_{D} \rho y^2 dA$$
The moment of inertia about the $y$-axis is:
$I_y= \dfrac{\rho}{3} \oint_{C} x^3 dy \\=\dfrac{\rho}{3} \iint_{D} \iint_{D}(\dfrac{\partial (x^3)}{\partial x}-\dfrac{\partial (0)}{\partial y})dA \\=\dfrac{\rho}{3} \times \iint_{D} 3x^2 dA \\= \iint_{D} \rho x^2 dA$
