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Evaluating A Double Integral over a Rectangle

  1. Nov 17, 2011 #1
    1. The problem statement, all variables and given/known data
    Let R be the rectangle bounded by x - y = 0, x - y = 2, x + y = 0, and x + y = 3. Evaluate

    [itex]\int[/itex][itex]\int[/itex](x + y)ex2-y2dA
    R

    3. 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

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

    [itex]\int[/itex](0≤y≤[itex]\frac{3}{2}[/itex])[itex]\int[/itex](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?
     
  2. jcsd
  3. Nov 17, 2011 #2

    Dick

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    The problem is begging for you to do the change of variables u=x+y, v=x-y.
     
  4. Nov 17, 2011 #3
    Okay. But why should I let v = x - y instead of v = x^2 - y^2?
     
  5. Nov 17, 2011 #4

    Dick

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    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.
     
  6. Nov 18, 2011 #5
    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?
     
  7. Nov 18, 2011 #6

    Dick

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    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.
     
  8. Nov 18, 2011 #7
    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

    Look about right?
     
  9. Nov 18, 2011 #8

    Dick

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    It's 'about' right. I don't think you did the u integration quite right. Check it again.
     
  10. Nov 18, 2011 #9
    Whoops, I meant to type e^(6)/4 - 7/4. How's that?
     
  11. Nov 18, 2011 #10

    Dick

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    That's much better. At least that's what I get. You might notice that you lucked out by picking the order of integration to be first dv and then du. If you'd done it the other way around, it would look impossible. You can sometimes get what looks like an awful integral which gets a lot simpler if you reverse the order of integration. Just file that for future reference.
     
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