Answer
$I_y=I_x= \dfrac{\pi \rho a^4}{4} $
Work Step by Step
We can parameterize the curve as $x= \cos \theta; y= 2 \sin \theta$ and $ 0 \leq \theta \lt 2 \pi $
Green's Theorem states that:
$\oint_CP\,dx+Q\,dy=\iint_{D}(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y})dA$
We set up the line integral and find out 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} \int_{0}^{2 \pi} ( a\sin^3 \theta) (-a \sin \theta d \theta)
\\=\dfrac{\rho^4}{12} \int_{0}^{2 \pi} (1-\cos \theta)^2 d \theta \\ =\dfrac{\rho^4}{24} [3 \theta-2 \sin 2 \theta+\dfrac{\sin 4 \theta}{4}]_0^{2 \pi} = \dfrac{\pi \rho a^4}{4} $
By symmetry, we can say that the moment of inertia about the $y$-axis is: $I_y=I_x= \dfrac{\pi \rho a^4}{4} $
