Express integral over curve as integral over unit ball Let $A$ be the region in $\mathbb{R}^2$ bounded by the curve $x^2-xy+2y^2=1$. Express the integral $\int_Axy$ as an integral over the unit ball in $\mathbb{R}^2$ centered at $0$.
So I rearranged $x^2-xy+2y^2=(x-\sqrt{2}y)^2+(2\sqrt{2}-1)xy$. It might be helpful to change the variables from $x$ and $y$ to $x-\sqrt{2}y$ and $xy$ or something similar. But I can't see how I would be able to get an integral over the unit ball.
 A: You should know that $A$ is an ellipse rotated at an angle of $\pi/8$ with respect to the $x$ axis.  You can show this by making the transformation 
$$x'=x \cos{\theta} - y \sin{\theta}$$
$$y'=x \sin{\theta}+y \cos{\theta}$$
You then substitute into the equation for the curve bounding $A$ above and find $\theta$ such that the coefficient of $x'y'$ vanishes.  Details here.  The equation of the ellipse in rotated coordinates is
$$\frac{x'^2}{a^2} + \frac{y'^2}{b^2} = 1$$
where $a=(2+\sqrt{2})/\sqrt{7}$ and $b=(2-\sqrt{2})/\sqrt{7}$.  You may then transform the integral to an integral over the unit ball by writing $x'=a \rho \cos{\phi}$ and $y'=b \rho \sin{\phi}$, $\rho \in [0,1]$, $\phi \in [0,2 \pi]$.  Note that $dx dy = dx' dy'$ because the transformation is a rotation and is therefore unitary.  The integral then becomes
$$\int_A dx dy \, x y = \frac{\sqrt{2}}{4} a b \int_0^{2 \pi} d\phi \, \int_0^1 d\rho \, \rho^3 (a^2 \cos^2{\phi} - b^2 \sin^2{\phi} + 2 a b \cos{\phi} \sin{\phi})$$
which may be evaluated easily.
