Change of Variables: which order of integration limits?

Question

An exercise in Larson/Edwards Calculus (10th ed.) asks the reader to evaluate the double integral

$$\int_R\int 4(x^2+y^2)\,\mathrm{d}A$$ using a given change of variables $$x=\frac{1}{2}(u+v)\mathrm{,}\quad y=\frac{1}{2}(u-v)$$ over the region pictured below.

Since, with the given change of variables, $\,v=x-y\,$ and $\,u=x+y$, it looks like my $v$ limits of integration are going to be $-1$ and $1$, and my $u$ limits of integration are going to be $-1$ and $1$ as well. My question is: how do I know which number should be the upper limit of integration and which number should be the lower limit of integration? Going off the picture, it looks like I should be integrating from $v=1$ to $v=-1$, and from $u=-1$ to $u=1$, so my integral would look like this:

$$\int_{1}^{-1}\int_{-1}^{1}f(u,v)\,\mathrm{d}u\,\mathrm{d}v$$

This gives me the opposite of the correct answer (the correct answer multiplied by $-1$). I know I get the correct answer by integrating $$\int_{-1}^{1}\int_{-1}^{1}f(u,v)\,\mathrm{d}u\,\mathrm{d}v$$ instead, but I don't see any reason why I ought to put the limits in that order. I suppose it might always be correct to choose the least of the two values to be the lower limit and the greatest of the two values to be the upper limit (thought I'm not sure), but even if that were so I still wouldn't really understand why that's the case.

Edit: I tried writing the original integral without changing the variables, like this: $$\int_{-1}^{0}\int_{-x-1}^{x+1}g(x,y)\,\mathrm{d}y\,\mathrm{d}x+\int_0^1\int_{x-1}^{-x+1}g(x,y)\,\mathrm{d}y\,\mathrm{d}x$$ and then manipulating the inequalities $$-x-1\leq y\leq x+1\quad\mathrm{and}\quad x-1\leq y \leq -x+1$$ to find $-1\leq x+y\mathrm{,}$ $-1\leq x-y\mathrm{,}$ $x-y\leq 1\mathrm{,}$ and $x+y\leq 1$. Substituting $v$ for $x-y$ and $u$ for $x+y$, I get $-1\leq v\leq 1$ and $-1\leq u\leq 1$. Basing my limits of integration off these inequalities seems to work, but I'm not sure this procedure for finding limits of integration works in general.

Answer 1

No, you could not always just use the least value as the lower limit: See what would happen if you had chosen the substitution $v = y-x$ instead of $x-y$.

The correct reasoning is the following: You want the area "above" (in $u$) the line $x+y = -1$ yet "below" (in $u$) the line $x+y = +1$. Here "above" and "below" mean greater that $u$ is greater than (less than) $-1$ (or $+1$). Now on $x+y = -1$ we have $u=-1$ so the lower limit in $u$ is $-1$. Similarly the upper limit is $+1$.

The best way to see that is that the area you want has $-1 \leq x+y \leq +1$ so the lower limit is $-1$ and the upper limit is $+1$.

Now consider $v = x-y$: You want the area with $-1 \leq x-y \leq +1$ so again the integral goes from $-1$ to $+1$.

To see if you understand it, try using a dfferent $v = y-x$ and see that you get the correct limits then.