Computing $\iint_{\Omega} (x+y^2) \ dx \ dy$ where $\Omega$ is a parallelogram 
Compute $$\iint_{\Omega} (x+y^2) \ dx \ dy$$ where $\Omega$ is a parallelogram bounded by $x+y=0$, $x+y=1$, $2x-y=0$ and $2x-y=3.$

Graphing this in Desmos produces the following plot

I was instructed to use change of variables here, but I'm not familiar with that at all. How can I compute this kind of integral? Any tips would be much appreciated.
 A: In single variable calculus, you used a change of variables called $u$-substitution to make some integrands nicer. In multivariable calculus, you can use a change of variables for that purpose, or for making the region over which you are integrating nicer. Here, the region is not a Type 1 or Type 2 or polar region. But we can change variables to $u$ and $v$ in a clever way to make the region much nicer:$$u=x+y,\quad v=2x-y$$
With this change of variables, the region we integrate over is simply a rectangle: $0\leq u\leq 1, \ 0\leq v\leq 3$.
We have to solve these equations for $x$ and $y$ in terms of $u$ and $v$ so that we can substitute in the integrand as well as find the Jacobian of this change of variables. Here, if we add the equations, notice that $y$ is eliminated, so $x=\frac{1}{3}(u+v)$. Alternatively, if we take twice the first equation and subtract the second equation, $x$ is eliminated, so $y=\frac{1}{3}(2u-v)$.
The Jacobian $\frac{\partial (x,y)}{\partial (u,v)}$ of a change of variables is the determinant of the 2x2 matrix $\begin{pmatrix} \frac{\partial{x}}{\partial{u}}&\frac{\partial{x}}{\partial{v}}\\\frac{\partial{y}}{\partial{u}}&\frac{\partial{y}}{\partial{v}}\end{pmatrix}$.
Finally, the change of variables theorem says that $$\iint_\Omega f(x,y)dA = \int_0^3\int_0^1 f\left(\frac{1}{3}u+\frac{1}{3}v,\frac{2}{3}u-\frac{1}{3}v\right)\cdot \frac{\partial (x,y)}{\partial (u,v)}dudv.$$
Can you finish the calculation?
A: If you do $x+y=X$ and $2x-y=Y$, then $x=\frac{X+Y}3$ and $y=\frac{2X-Y}3$. Besides, if you define$$g(X,Y)=\left(\frac{X+Y}3,\frac{2X-Y}3\right),$$then $\Omega=g\bigl([0,1]\times[0,3]\bigr)$ and the absolute value of the Jacobian of $g$ is $\frac13$. So,\begin{align}\iint_\Omega x^2+y\,\mathrm dx\,\mathrm dy&=\iint_{g([0,1]\times[0,3])}x^2+y\,\mathrm dx\,\mathrm dy\\&=\iint_{[0,1]\times[0,3]}\frac13\left(\left(\frac{X+Y}3\right)^2+\frac{2X-Y}3\right)\,\mathrm dX\,\mathrm dY\\&=\frac13\int_0^1\int_0^3\left(\frac{X+Y}3\right)^2+\frac{2X-Y}3\,\mathrm dY\,\mathrm dX.\end{align}Can you take it from here?
