How can I calculate the moment of inertia of an ellipse ($\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$) using polar coordinates and the formula: $$I_x=\int_\Omega y^2dx dy$$
I know that the result is supposed to be $I_x=\frac{\pi a b^3}{4}$ ($b$ corresponding to the semi-axis in the $y$ coordinate), but I just can't seem to derive it.
PS: I've read other posts on this topic but none I've found so far uses polar coordinates to get to the answer.
You can parametrize your ellipse by $$ x=ar\cos t,\ \ \ y=br\sin t,\ \ \ t\in[0,2\pi],\ \ \ r\in[0,1]. $$ The Jacobian is $$ \begin{vmatrix} a\cos t& b\sin t\\ -ar\sin t& br\cos t\end{vmatrix}=abr. $$ So $$ I_x=\int_{\Omega}y^2\,dx\,dx=\int_0^1\int_0^{2\pi}b^2r^2\sin^2t\,abr\,dt\,dr=\pi ab^3\int_0^1r^3\,dr=\frac{\pi a b^3}4. $$