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Double Integrals

  1. Jul 23, 2010 #1
    1. The problem statement, all variables and given/known data

    f(x,y) = x2 + y2 over the triangular region with vertices (0,0), (1,0), (0,1)

    2. Relevant equations



    3. The attempt at a solution

    [tex]\int[/tex]from 0 to 1-x[tex]\int[/tex]from 0 to 1 of x2 + y2 dydx = [tex]\int[/tex]from 0 to 1-x of (1/3)y3 evaluated from 0 to 1 dx
    = [tex]\int[/tex] from 0 to 1-x of (1/3) dx
    = (1/3)x evaluated from 0 to 1-x
    = (1/3)(1-x)

    According to the book, the answer is 1/6... How do I know what the value of x is?
     
  2. jcsd
  3. Jul 23, 2010 #2
    Is there another way you can define the line 1-x? Maybe in terms of y :wink:.

    Oh, and if you choose to do so you might need to change the order in which you're integrating :wink:.
     
  4. Jul 23, 2010 #3
    You've written the equivalent of this:

    [tex] \int^{1-x}_0 f(x) dx [/tex].

    You can't "let x go from 0 to (1-x)." That doesn't make any sense.

    Try writing down your double integral again, but pay more careful attention to which direction you're integrating and what order your dy and dx are in.
     
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