# Evaluating A Double Integral over a Rectangle

## Homework Statement

Let R be the rectangle bounded by x - y = 0, x - y = 2, x + y = 0, and x + y = 3. Evaluate

$\int$$\int$(x + y)ex2-y2dA
R

## The Attempt at a Solution

First I rewrote the boundaries so that I could graph them more easily. I got y = x, y = x - 2, y= -x, and y = -x + 3. I was going to then integrate

$\int$(-1≤y≤0)$\int$(-y≤x≤y+2) ((x + y)ex2-y2) dx dy, and add that to,

$\int$(0≤y≤$\frac{3}{2}$)$\int$(y≤x≤-y+3) ((x + y)ex2-y2) dx dy

But then I realized I didn't even know how to integrate (x + y)ex2-y2. This leads me to believe I'm trying to do the wrong thing here. Suggestions?

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Dick
Homework Helper
The problem is begging for you to do the change of variables u=x+y, v=x-y.

Okay. But why should I let v = x - y instead of v = x^2 - y^2?

Dick
Homework Helper
Okay. But why should I let v = x - y instead of v = x^2 - y^2?
If you try it you'll see why it's a nice choice. One good reason is that you are given the limits in terms x+y and x-y. Factor x^2-y^2.

Okay. Supposing I let x=x+y and v=x-y, then I find that x = (u+v)/2 and y = (u-v)/2.

I think I need to find the Jacobian of this so,

J(u,v) = [x_u x_v]
*******[y_u y_v] (Excuse the ***; I don't know how to correctly format matrices).

I find this to be,

[1/2 1/2]
[1/2 -1/2].

And the absolute value determinant of this is 1/2.

But how do I go about finding new bounds to integrate over?

Dick
Homework Helper
Okay. Supposing I let x=x+y and v=x-y, then I find that x = (u+v)/2 and y = (u-v)/2.

I think I need to find the Jacobian of this so,

J(u,v) = [x_u x_v]
*******[y_u y_v] (Excuse the ***; I don't know how to correctly format matrices).

I find this to be,

[1/2 1/2]
[1/2 -1/2].

And the absolute value determinant of this is 1/2.

But how do I go about finding new bounds to integrate over?
Did you look at the bounds? "bounded by x - y = 0, x - y = 2, x + y = 0, and x + y = 3"? I think that's really easy to translate into u and v.

Do I just plug in? If so, I get v=0 and v=2, and u=0 and u=3. This leads me to believe I can just integrate over a rectangle.

So I have the double integral from u=0 to u=3 and from v=0 to v=2 of (u)e^(vu) dv du. I multiply this by 1/2 because of the Jacobian I calculated earlier.

By "Q-substitution" (since I can't use "u" anymore) I will integrate with respect to Q, where Q=uv. This gives e^(uv)/2 evaluated from v=0 to v=2. So I have,

e^(2u)/2 - 1/2,

which I integrate with respect to u from 0 to 3. This gives

e^(6)/4 - 5/4

Dick
Homework Helper
Do I just plug in? If so, I get v=0 and v=2, and u=0 and u=3. This leads me to believe I can just integrate over a rectangle.

So I have the double integral from u=0 to u=3 and from v=0 to v=2 of (u)e^(vu) dv du. I multiply this by 1/2 because of the Jacobian I calculated earlier.

By "Q-substitution" (since I can't use "u" anymore) I will integrate with respect to Q, where Q=uv. This gives e^(uv)/2 evaluated from v=0 to v=2. So I have,

e^(2u)/2 - 1/2,

which I integrate with respect to u from 0 to 3. This gives

e^(6)/4 - 5/4

It's 'about' right. I don't think you did the u integration quite right. Check it again.

Whoops, I meant to type e^(6)/4 - 7/4. How's that?

Dick