Evaluation of $\iint_{x^2-xy+y^2\leq1}(x-y)^2\mathrm{d}x\mathrm{d}y$ $$
I:=\iint_{x^2-xy+y^2\leq1}(x-y)^2\mathrm{d}x\mathrm{d}y
$$
I've got the following 2 colored equations but neither seems useful for evaluating this double integral.
$$\begin{align}
x^2-xy+y^2&=  \color{blue}{\left(x- {1 \over 2 }y \right)^2+ {3 \over  4}y^2}  \\
&= \color{green}{(x-y)^2+xy} 
\end{align}$$
I want your wisdom.
 A: 
The region in question is a slanted ellipse. Using change of variables for ellipses we have
$$
u=x-\frac{y}{2}\\
v=\frac{\sqrt 3y}{2}
$$
Our region is thus mapped into the unit circle $u^2+v^2=1$ with Jacobian $J$ being
$$
J=\begin{bmatrix}
1&-0.5\\0&\sqrt 3/2
\end{bmatrix}=\sqrt 3/2
$$
Use another change of variables into polar coordinates which is trivial. This should make the integral much easier.
A: Observe that $a(x+y)^2+b(x-y)^2=x^2+y^2-xy$ if and only if $a=\frac1{4}$ and $b=\frac{3}{4}$. Now let the change of variable $T(x,y):=\frac1{2}(x+y,\sqrt{3}(x-y))=(z,t)$ and notice that $T=\left[\begin{smallmatrix}1/2&1/2\\\sqrt{3}/2&-\sqrt{3}/2\end{smallmatrix}\right]$ is an invertible linear map, therefore
$$
\begin{align*}
\int_{\left\{(x,y):\frac1{4}(x+y)^2+\frac{3}{4}(x-y)^2\leqslant 1\right\}}(x-y)^2 d(x,y)&=\int_{\{(z,t):z^2+t^2\leqslant 1\}}\frac{4}{3}t^2 d(T^{-1}(z,t))\\
&=\int_{\{(z,t):z^2+t^2\leqslant 1\}}\frac{4}{3}t^2|\det T^{-1}| d(z,t)\\
&=\frac4{3|\det T|}\int_{[0,1]\times [0,2\pi]}r^3(\sin \theta)^2\,d (r,\theta)\\
&=\frac2{3\sqrt{3}}\int_{0}^{2\pi}(\sin \theta )^2 \,d \theta \\
&=\frac{2\pi}{3\sqrt{3}}
\end{align*}
$$
where we used the change of variable $(z,t)=(r\cos \theta ,r \sin \theta )$.∎
