Evaluate $\iint_{R} 2 y-8 x d A$ where $R$ is the region formed by the quadrilateral with vertices $(6,0),(8,4),(6,8) \text { and }(4,4)$.
My try: The region is a parallelogram bounded by $2x+y=12$,$2x+y=20$,$2x-y=4$ and $2x-y=12$ Now i used the transformation as: $$2x+y=u, 2x-y=v$$ $\implies$ $$x=\frac{u+v}{4}. y=\frac{u-v}{2}$$ Now the integral is: $$\iint_{R}(2y-8x)dA=\iint_{R^*}(-u-3v)|J|\:dudv$$
Now should i choose $u$ goes from $12$ to $20$ and $v$ goes from $4$ to $12$?
In that case i am getting answer as $-640$?
But volume cannot be negative?
So how should we choose the range of values taken by $u$ and $v$?