Answer
$\dfrac{4 \sqrt 3 \pi}{3}$
Work Step by Step
$\iint_R (x^2-xy+y^2) dA=\iint_{D}(2u^2+2v^2)|J| dA$
We have $|J| =\begin{vmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{vmatrix}=\begin{vmatrix} \sqrt 2&-\sqrt{\dfrac{2}{3}}\\\sqrt 2&\sqrt{\dfrac{2}{3}}\end{vmatrix}=\dfrac{4 \sqrt 3}{3}$
As we are given that $u^2+v^2 \leq 1$
Let us consider that $u =r \cos \theta$ and $v=r \sin \theta$
$\iint_R (x^2-xy+y^2) dA=\iint_{D}(2u^2+2v^2)\dfrac{4 \sqrt 3}{3} dA$
or, $=(2) \int_0^{2 \pi} \int_0^1 r^2 (\dfrac{4 \sqrt 3}{3}) r dr d \theta$
or, $=(\dfrac{8 \sqrt 3}{3}) \int_0^{2 \pi} [ \dfrac{1}{4} r^4]_0^1 d \theta$
or, $=\dfrac{4 \sqrt 3 \pi}{3}$
