Change of Variables in Multple Integrals

[number0]

Homework Statement

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.

Homework Equations

None

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.

The book answer's is:

v \leq u \leq 1-v and 0 \leq v \leq 1/2Can anyone show me how to figure this problem out without any reference to graphing (just

pure algebraic)?

 

[vela]

Homework Statement

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.

Homework Equations

None

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.

The book answer's is:

v \leq u \leq 1-v and 0 \leq v \leq 1/2Can anyone show me how to figure this problem out without any reference to graphing (just
pure algebraic)?

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

 

[number0]

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

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

book's answer). For example,

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) ?

 

[vela]

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.

 

[number0]

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.

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.

 

[vela]

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?

 

[number0]

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?

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?

 

[vela]

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.