Evaluate $$I=\iint_A xy\,\mathrm{d}x\,\mathrm{d}y$$ where $$A=\{(x,y): ax^2+2hxy+by^2\le r^2, a>0, ab-h^2>0\}$$
I tried diagonalising the quadratic form $h(x,y)=ax^2+2hxy+by^2$ to some
$$g(u,v)=u^2+v^2$$
where
$$\begin{pmatrix}u\\v\end{pmatrix}=\begin{pmatrix} \sqrt{a} & \frac{h}{\sqrt{a}}\\\ 0 & \sqrt{\frac{ab-h^2}{a}} \\\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$$
The Jacobian of the transformation $(x,y)\to(u,v)$ is
$$\det(J) = \frac{1}{\sqrt{ab-h^2}}$$
The integral then reduces to
$$I=\iint_{u^2+v^2\le r^2}\left(\frac{uv}{\sqrt{ab-h^2}}-\frac{hv^2}{ab-h^2}\right)\frac{1}{\sqrt{ab-h^2}}\,\mathrm{d}u\,\mathrm{d}v$$
Now I could apply a polar transformation to get the answer. But this method seems a bit cumbersome, especially solving for $x$ and $y$ from $u,v$. Is there an obvious alternate solution?
The problem can also be seen as finding $\mathbb{E}(XY)$ where $(X,Y)$ is jointly uniform over $A$.
EDIT.
Proceeding in the way I have done so far, I get the final answer as
$$I=-\frac{\pi r^4h}{4(ab-h^2)^{3/2}}$$
Is this indeed the correct answer?
The final answer in my post is correct, as I was able to verify using Mathematica here.