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