Integral over the unit ball This question has been asked before, but I did not understand it, so I worked on it on my own and got stuck. Any help would be appreciated.
Let $A$ be the region in $\Bbb R^2$ bounded by the curve $x^2-xy+2y^2=1$. Express the integral $\int _A xy$ as an integral over the unit ball in $\Bbb R^2$ centered at $0$. Hint: Complete the square.
This is how far I could go and stuck:
$x^2-xy-2y^2=\frac 78 x^2+(\frac18 x^2-xy+2y^2)=(\sqrt \frac 78 x)^2+(\frac1{2\sqrt2 x}-\sqrt 2 y)^2$
Then, I set $u=\frac 78 x, v=\frac1{2\sqrt2 x}-\sqrt 2 y$
I do not know how from this I can proceed to use the change of variables theorem. 
Thanks in advance!
 A: You should set up a linear change of variable. The curve bounding the region $A$ can be written as $$(x-\frac y2)^2 + \frac 74 y^2 = x^2 - xy + \frac 14 y^2 + \frac 74 y^2 = 1$$
Thus define new coordinates $u = x - \dfrac y2$ and $v = \dfrac{\sqrt 7}{2} y$.  The transformation $T : (x,y) \to (u,v)$ transforms $A$ into the unit disk $D$. The change of variable theorem is more convenient to use the inverse transformation $S = T^{-1} : (u,v) \to (x,y)$ given by
$$  x = u + \frac{1}{\sqrt{7}} v\quad y = \frac{2}{\sqrt 7}v$$
that transforms the unit disk $D$ into $A$.
Since $SD = A$ the change of variables theorem states that $$\int_{A} f(x,y) \,dxdy = \int_{SD} f(x,y) \,dxdy = \int_D f(S(u,v))|JS| \, dudv$$ where $|JS|$ is the Jacobian of $S$. 
The last integral isn't hard to handle: using $f(x,y) = xy$ you get $$f(S(u,v)) = \left(u + \frac{1}{\sqrt{7}} v \right) \left(\frac{2}{\sqrt 7}v \right) = \frac{2}{\sqrt 7} uv + v^2$$ and $$|JS| = \frac{2}{\sqrt 7}.$$
So, what you need to evaluate is $$\int_D \frac{4}{7} uv + \frac{2}{\sqrt 7} v^2 \, dudv.$$
Polar coordinates should do the job nicely.
A: $$x^2-xy+2y^2 = (x-y/2)^2 + \frac{7}{4}y^2 $$
so by setting $u=x-\frac{y}{2},v=y$ we have:
$$ I = \int_{u^2+\frac{7}{4}v^2\leq 1}v\left(u+\frac{v}{2}\right)\,du\,dv=\frac{4}{7}\int_{u^2+w^2\leq 1}w\left(u+\frac{w}{\sqrt{7}}\right)\,du\,dv$$
and exploiting symmetry:
$$ I = \frac{4}{7\sqrt{7}}\int_{u^2+w^2\leq 1}w^2\,du\,dw =\frac{16}{7\sqrt{7}}\int_{0}^{1}(1-t^2)\,dt=\color{red}{\frac{32}{21\sqrt{7}}}.$$
