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 $\iint_{R} u^2+v^2\frac{\sqrt{3}}{4}dudv$
?
 A: 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}
A: You're sloppy with notation, you're just gluing differentials next to the Jacobian after you get the determinant. Let's settle this question once and for all:
The Jacobian is used in place of the chain rule, so
$$\left|\frac{\partial (x,y)}{\partial (u,v)}\right|=\frac{4}{\sqrt3}$$
Now, just like you can write $dx=\frac{dx}{du}du$ in one dimension, you write
$$dx\,dy=\left|\frac{\partial (x,y)}{\partial (u,v)}\right| du\,dv$$
Now there's no ambiguity how to flip the Jacobian when you do the substitution. It's obvious, dx and dy are on top on both sides, and du and dv are top and bottom on the right, effectively "cancelling out".
I deliberately wrote the Jacobian in compact notation that just records what's on top and bottom, but didn't write out the entire matrix.
