Region of double integration change under co-ordinate transformation

Question

If we have the region $0<x<\frac{1}{2}$,$\\ \ \ $ $x<y<1-x$ for a double integral on the $xy$ plane then it will appear as such:

Now say we want to find the region of integration under the new co-ordinates $r=x+y$ and $s=y-x$ - what will the upper and lower bounds on these be?

I thought of subbing $r$ and $s$ into the original boundaries (and multiplying everything by 2 to give):

$\rightarrow 0<r-s<1$ & $r-s<r+s<2-(r-s)$

adding these together...

$\rightarrow r-s<2r<3-r+s$

but found no way to obtain an inequality in only $r$ and an inequality with only $s$ in the centre.

I also thought of trying to draw an $x+y$ and $y-x$ axis onto the plane but didn't really know if these were just the lines $y=x$ and $y=-x$ or something more weird and nonlinear perhaps (if it is indeed these two equations, I would appreciate a clear explanation).

Many thanks.

Answer 1

One approach is purely mechanistic: you express the boundary of the region of integration in terms of a system of equations, then transform that system.

The boundary is defined in $(x,y)$ coordinates as $$x = 0, \quad y = x, \quad x+y = 1.$$ So under the transformation $$r = x + y, \quad s = y - x,$$ or equivalently $$x = \frac{r - s}{2}, \quad y = \frac{r + s}{2},$$ we obtain $$r - s = 0, \quad r - s = r + s, \quad r = 1,$$ and simplifying yields $$r = s, \quad s = 0, \quad r = 1.$$ Now we plot these in the $(r,s)$ plane, and the interior region is just a triangle with vertices are $(r,s) \in \{(0,0), (1,0), (1,1)\}$. Therefore, we can set up an iterated integral as $$\int_{r=0}^1 \int_{s=0}^r f(r,s) \, ds, \, dr,$$ or if we want to integrate with respect to $r$ first, $$\int_{s=0}^1 \int_{r=s}^1 f(r,s) \, dr \, ds.$$

Another approach is to look at how the transformation maps the vertices of the boundary, knowing that this particular transformation preserves lines. So for instance, $(x,y) = (0,0)$ becomes $(r,s) = (0+0, 0-0) = (0,0)$; and $(x,y) = (0,1)$ becomes $(r,s) = (0+1,1-0) = (1,1)$. This gives us the same list of transformed vertices we got above.

Answer 2

To me, the easiest way is to use the fact that "boundary will map to the boundary". This is because a change of variables is related a to continuous function, in this case, $$F:\mathbb{R}^2 \to \mathbb{R}^2$$ $$(x,y) \mapsto F(x,y) = (\underbrace{x+y}_{r},\underbrace{y-x}_{s})$$

Set $A(0,0),B(\frac{1}{2},\frac{1}{2}),C(0,1)$ the vertices of the region,

Thus, $$\overline{AB}=\{(x,x)/x\in [0,\frac{1}{2}]\} \implies F(\overline{AB}) = (2x,0), x\in [0,\frac{1}{2}]$$ representing a line over the $r-$axis, joining $(0,0)\rightarrow (1,0)$. Similarly, $$\overline{BC}=\{(x,1-x)/x\in [0,\frac{1}{2}]\} \implies F(\overline{AB}) = (1,1-2x), x\in [0,\frac{1}{2}]$$ a line over the $s-$axis joining $(1,0)\rightarrow (1,1)$

$$\overline{CA}=\{(0,y)/y\in [0,1]\} \implies F(\overline{AB}) = (y,y), y\in [0,1]$$ which is the diagonal $r=s$: Putting all together:

Thus, $r\in [0,1], 0\leq s \leq r$ is the desired region.