Find \iint_R \frac{3x+y}{-4x-3y} dA, where R is the parallelogram enclosed by the lines
$3x + y = 0$, $3x + y = 2$, $-4x - 3y = 1$, $-4x - 3y = 5$.
Round your answer to four decimal places.
This can be done directly with a tedious computation, or can be done with a change of variables to transform
the parallelogram into a rectangle.
Hint: Let $u = 3x + y$ and $v = -4x - 3y$.