Answer
$\dfrac{4 \sqrt 3 \pi}{3}$
Work Step by Step
Here, we have: $Jacobian =\begin{vmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{vmatrix}=\begin{vmatrix} \sqrt 2&-\sqrt{\dfrac{2}{3}}\\\sqrt 2&\sqrt{\dfrac{2}{3}}\end{vmatrix}=\dfrac{4 \sqrt 3}{3}$
$\iint_R (x^2-xy+y^2) dA=\iint_{D}(2u^2+2v^2)\dfrac{4 \sqrt 3}{3} dA$
Here, $u^2+v^2 \leq 1$
Plug $u =r \cos \theta$ and $v=r \sin \theta$
$\iint_R (x^2-xy+y^2) dA=\iint_{D}(2u^2+2v^2)\dfrac{4 \sqrt 3}{3} dA=\int_0^{2 \pi} \int_0^1 2r^2 (\dfrac{4 \sqrt 3}{3}) r dr d \theta$
Thus, $\iint_R (x^2-xy+y^2) dA=\dfrac{8 \sqrt 3}{3}\int_0^{2 \pi} \dfrac{1}{4} r^4|_0^1 d \theta=\dfrac{4 \sqrt 3 \pi}{3}$
