Evaluate $\iint xy $ over a parallelogram given with $2x-y=1$, $2x-y=3$, $x+y=-2$, $x+y=0$

Question

I have to compute an integral: $$\iint_R xy \; dxdy$$ over a parallelogram given with $2x-y=1$, $2x-y=3$, $x+y=-2$, $x+y=0$, in a way that I use variable substitution:$u=2x-y,v=x+y$. If I substitute x and y and multiply by Jacobian determinant I get the following integral: $$\int_{-2}^0\int_1^3\frac{(1-uv)(2v-u)(v+u)}{9}\;dudv$$ which is not correct. Did I do something wrong?

Answer 1

$\displaystyle x = \frac{u+v}{3}, y = \frac{2v-u}{3}$

$\displaystyle x_u = \frac{1}{3}, y_v = \frac{2}{3}, x_v = \frac{1}{3}, y_u = -\frac{1}{3}$

So your Jacobian for the transformation should be $\, J = \frac{1}{3} \times \frac{2}{3} - \frac{1}{3} \times (-\frac{1}{3}) = \frac{1}{3}$.

So your integral becomes

$I = \frac{1}{3} \displaystyle \int_{-2}^{0}\int_1^3\frac{(2v-u)(u+v)}{9} \, du \, dv$