# Change of Variables in Multple Integrals

1. Jul 11, 2010

### number0

1. The problem statement, all variables and given/known data

Let D be the region y $$\leq$$ x $$\leq$$ 1 and 0 $$\leq$$ y $$\leq$$ 1.

Find the boundaries of the integration for the change of variables u and v, if:

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

2. Relevant equations

None

3. The attempt at a solution

I solved for u and for v in terms of x and y.

u = $$\frac{x+y}{2}$$

v = $$\frac{x-y}{2}$$

Then I got stuck.

v $$\leq$$ u $$\leq$$ 1-v and 0 $$\leq$$ v $$\leq$$ 1/2

Can anyone show me how to figure this problem out without any reference to graphing (just

pure algebraic)?

2. Jul 11, 2010

### vela

Staff Emeritus
Why do you guys always like to make things more difficult for yourself by refusing to draw a picture?

Just rewrite the boundaries of D in terms of u and v. For example,

$$y \le x \Rightarrow u-v \le u+v \Rightarrow 0 \le 2v \Rightarrow v \ge 0$$

3. Jul 11, 2010

### number0

Well, I did that part already. However, how is it possible to get v $$\leq$$ 1/2 (the

x $$\leq$$ 1 => u + v $$\leq$$ 1 => v $$\leq$$ 1 - u

Which is supposedly not equivalent to v $$\leq$$ 1/2

How would you go about solving the other ones (v $$\leq$$ u $$\leq$$ v-1) ?

4. Jul 11, 2010

### vela

Staff Emeritus
You need to do the third side of D, y≥0. When you combine it with what you got for x≤1, you can get that upper limit for v.

5. Jul 11, 2010

### number0

Okay, can you explain it a bit more. Sorry about being unknown about what you mean, but I

think if I finally understand what you are trying to say, I can finally get this problem correct.

6. Jul 11, 2010

### vela

Staff Emeritus
y≥0 translates to u≥v, and x≤1 gave you u≤1-v. Combining those, you get v≤u≤1-v. (You almost had this above, but you flipped a sign.) You can't have v=1, for instance, because you'd get 1≤u≤0. What restriction on v must exist for those inequalities to be satisfied?

7. Jul 11, 2010

### number0

Now, I think I am starting to get it. I think I am having a conceptual misunderstanding of the

situation at hand! I am elated that you pointed out how I flipped the sign. Here is the reason

why I flipped the sign:

y $$\leq$$ 1 => u - v $$\leq$$ 1 => u $$\leq$$ 1 + v (correct answer: u $$\leq$$ 1 - v)

Can you tell me why this would not work?

8. Jul 11, 2010

### vela

Staff Emeritus
That is valid, but it doesn't help much. Another constraint gave you v≥0, so if u≤1-v holds, u≤1+v automatically holds as well.