Answer
$6 \pi$
Work Step by Step
Since, $R=${$(u,v) | 0 \leq v \leq 1-u, 0\leq u \leq 1$}
$Jacobian, J =\begin{vmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{vmatrix}=\begin{vmatrix} 2&0\\0&3\end{vmatrix}=6-0=6$
Now, we have
$\iint_R x^2 dA=\int_{-1}^1 [\int_{-\sqrt{1-v^2}}^{\sqrt{1-v^2}} 24 u^2 du dv$
Plug $p =r \cos \theta$ and $q=r \sin \theta$
$\iint_R x^2 dA=\int_{-1}^1 [\int_{-\sqrt{1-q^2}}^{\sqrt{1-q^2}} 24 p^2 dp dq$
or, $=\int_0^{2 \pi} \int_0^1 24 (r \cos \theta)^2 r dr d \theta$
or, $=\int_0^{2 \pi} 6 \cos^2 d \theta$
or, $=(3) \int_0^{2 \pi} (1+\cos 2 \theta) d \theta$
Hence, we have $\iint_R x^2 dA=3 [\theta+\dfrac{\sin (2 \theta)}{2}]_0^{2 \pi}=6 \pi$
