Change of Variable finding simple bounds

Question

I have a very simple problem that I've solved before (and it gave me a lot of trouble then) and I somehow can not come up with the right bound again.

In Michael Corral's book he uses a change of variable transformation to make $x = \frac{1}{2}(u+v)$ and $y = \frac{1}{2}(v-u)$.

The original substitutions are: $x - y = u$

$x + y = v$

The bounds on the $x,y$ plane are $(1,0)$, $(0,1)$ and $x = 1-y$ Setting x to $0$ I can come up with the first bound of $u = -v$ I did so by plugging in $0$ for $x$ and $\frac{1}{2}(v-u)$ for $y$ and set it equal to $u$ i.e. $u = -\frac{1}{2}(v-u)$ It also works simply pluggin in the numbers.

But the second bound for when $x = 1-y$ should be $u = v$ When plugging in I always get $0$ (i.e. the u's go away) or I get $v = 1$ which is a correct line but I have no bound for u on this? I'm doing something wrong and I'd be so grateful if you could help me out. I have a total black out here. I weirdly am super comfortable with polar and spherical substitution bounds but they don't apply here.

Thank you very much!

Edit: also if someone could recommend a ressource that has lot's of practice for finding these bounds that'd be great!

Answer 1

Your region on $xy-$plane is bounded by a right angled triangle with boundaries $x=0, y=0 $ and $x+y=1$. You just need to know the boundaries of the transformed region in $uv-$plane using the given transformations:

$x-y=u$ and $x+y=v$

• $x=0$ maps to $u=-y, v=y$ or equivalently $v=-u$

• $y=0$ maps to $u=x, v=x$ or equivalently $v=u$

• $x+y=1$ maps to $v=1$

So the new transformed region is a triangle in $uv-$plane bounded by the edges $v=-u, v=u$ and $v=1$.