# Thread: Change of Variables...final tomorrow!

1. ## Change of Variables...final tomorrow!

Use the transformation:
x = (u+v)/2
y = (v−u)/2
double integral bounded by R of: (x − y)(x + y)dxdy

where R is the region bounded by the lines y = x + 2, y = x, y = −x + 2 and
y = −x + 4.

Again...its the bounds of the integral that are killing me. Thanks ahead of time my book is just terrible.

2. Note from your transformation that $\displaystyle u=x-y$ and $\displaystyle v=x+y$. Now, note that our region is a rectangle at 45° in the xy axes, and out bounding lines can be rewritten as: $\displaystyle x-y=-2$, $\displaystyle x-y=0$, $\displaystyle x+y=2$, and $\displaystyle x+y=4$, so we have bounds in uv of $\displaystyle u=-2$, $\displaystyle u=0$, $\displaystyle v=2$ and $\displaystyle v=4$. Does that help?

--Kevin C.

3. how did you find out what u and v were equal to

4. I found u and v by the same methods one uses to solve a system of equations;
since you have:
$\displaystyle x=\frac{u+v}{2}=\tfrac{1}{2}u+\tfrac{1}{2}v$
$\displaystyle y=\frac{v-u}{2}=-\tfrac{1}{2}u+\tfrac{1}{2}v$
adding the two equations together gives:
$\displaystyle x+y=\tfrac{1}{2}u+\tfrac{1}{2}v-\tfrac{1}{2}u+\tfrac{1}{2}v=v$
and subtracting the second equation from the first gives:
$\displaystyle x-y=\tfrac{1}{2}u+\tfrac{1}{2}v-(-\tfrac{1}{2}u+\tfrac{1}{2}v)=u$
Does that make sense?

--Kevin C.