Answer
$\dfrac{\pi}{24} (1-\cos 1)$
Work Step by Step
$Jacobian, J (x,y)=\begin{vmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{vmatrix}=\begin{vmatrix} \dfrac{1}{3}&0\\0&\dfrac{1}{2}\end{vmatrix}=\dfrac{1}{6}$
Let us consider that $u=3x$ and $v=2y$
$\iint_R \sin(9x^2+4y^2) dx dy= \iint_{R} \sin (u^2+v^2) (\dfrac{1}{6}) du dv$
or, $=(\dfrac{1}{6}) \int_0^{\pi/2} \int_0^1 \sin (r^2) r dr d
\theta$
or, $ =(\dfrac{-1}{12}) (\pi/2) |\cos (r^2)|_0^1$
or, $=\dfrac{-1}{12}(\cos (1)-1)$
or, $=\dfrac{\pi}{24} (1-\cos 1)$
