integration using change of variables 
find $$\iint_{R}x^2-xy+y^2 dA$$ where $R: x^2-xy^+y^2=2$ using $x=\sqrt{2}u-\sqrt{\frac{2}{3}}v$ and $y=\sqrt{2}u+\sqrt{\frac{2}{3}}v$

To calculate the jacobian I take $$\begin{vmatrix}
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\ 
\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} 
\end{vmatrix}=\begin{vmatrix}
\sqrt{2} &-\sqrt{\frac{2}{3}}\\ 
\sqrt{2} & \sqrt{\frac{2}{3}} 
\end{vmatrix}=\frac{4}{\sqrt{3}}dudv$$
So the integral I have to calculate is now:
$\iint_{R} u^2+v^2\frac{4}{\sqrt{3}}dudv$ or $\iint_{R} u^2+v^2\frac{\sqrt{3}}{4}dudv$
?

Let be $$I=\iint_{R}(x^2-xy+y^2)\, \mathrm dA$$ where $R: x^2-xy+y^2=2$.
Using the change of variables $x=\sqrt{2}u-\sqrt{\frac{2}{3}}v$ and $y=\sqrt{2}u+\sqrt{\frac{2}{3}}v$ the domain of integration $R$ becomes $S:u^2+v^2=1$ and the integrand function $x^2-xy+y^2$ becomes $2(u^2+v^2)$. The Jacobian determinant is
$$\left|\frac{\partial (x,y)}{\partial (u,v)}\right|=\frac{4}{\sqrt 3}$$
Thus we have
$$
I=\iint_{R}(x^2-xy+y^2)\, \mathrm dx\, \mathrm dy=\iint_{S}2(u^2+v^2)\,\frac{4}{\sqrt 3}\, \mathrm du\, \mathrm dv
$$
This integral will be much easier in terms of polar coordinates $x=r\cos\theta$ and $y=r\sin\theta$ and then
\begin{align}
I&=\iint_{S}2(u^2+v^2)\,\frac{4}{\sqrt 3}\, \mathrm du\, \mathrm dv\\
&=
\frac{8}{\sqrt 3}\int_0^{2\pi}\int_0^1 (r^2)\cdot r\, \mathrm dr\, \mathrm d\theta=
\frac{8}{\sqrt 3}\int_0^{2\pi} \left[\frac{r^4}{4}\right]_0^1 \, \mathrm d\theta=\frac{8}{\sqrt 3}\int_0^{2\pi} \frac{1}{4}\,\mathrm d\theta=\frac{4\pi}{\sqrt 3}
\end{align}
