Double integral over a parallelogram I understand the general concept behind double integrals but do not understand how to change the coordinates linearly, and what to do from there.
Find $$\int\int_P(x+y)dxdy$$ Where $$P$$is parallelogram with vertices $(0,0), (3,2), (2,4) ,(5,6)$ (Hint: Change the coordinates linearly). 
Thanks
 A: hint: Let $O = (0,0), P = (2,4), Q = (5,6), S = (3,2)$. You can write equations for the lines: $\overline{OP}: 2x-y = 0, \overline{OS}: 2x-3y = 0, \overline{PQ}: 2x-3y = -8, \overline{SQ}: 2x-y = 4$. Now make the substitution: $u = 2x-y, v = 2x-3y$. Can you proceed to the next step?
A: Draw $P$ in the $xy$ plane.
If you want to integrate first with respect to $x$, fix an $\eta$ in $[0,5]$ and draw the vertical line $x=\eta$. From bottom to top, see at what values, depending on $y$, this vertical line goes through $P$. To obtain the first value, you need to find the equation of the line going through the vertices $(0,0)$ and $(3,2)$ of $P$. Once you have this equation, write it in the form $x=L(y)$. Then $L(y)$ is the bottom bound for the first iterated integral (with respect to $x$, expressed in terms of $y$). Obtain the upper bound similarly. Note that $y$ goes from $0$ to $6$, which will be the bounds for the outer iterated integral (with respect to $y$).
The same strategy applies if you want to integrate first with respect to $y$,
