1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Change variables

  1. Nov 8, 2009 #1
    1. The problem statement, all variables and given/known data
    let f be continuous on [0,1] and let R be triangular region with vertices (0,0),(1,0) and (0,1) show that[tex]\int\intf(x+y) dA[/tex]= [tex]\intu*f(u)du[/tex] from 0 to 1


    2. Relevant equations



    3. The attempt at a solution
    i dont know how to start this question. can anyone give me a hint to help me start this question?
     
  2. jcsd
  3. Nov 8, 2009 #2
    i think i didnt type the question correctly, its supposed to be a double integral of f(x+y) over the region R
    and that value is equal to a single integral of u*f(u) du from u=0 to u=1
     
  4. Nov 8, 2009 #3

    Mark44

    Staff: Mentor

    From your two posts, I think this is what you're trying to prove:
    [tex]\int \int_R f(x + y) dA~=~ \int_{u = 0}^1 u f(u) du[/tex]
    where f is continuous on [0, 1] and R is the triangular region defined by the points (0, 0), (1, 0), and (0, 1).

    Here are some thoughts that might be of help to you. Assuming for the moment that f is positive on its domain, the double integral represents the volume of the region above the triangular region. Normally, to calculate a volume like this, we might use either vertical strips (vertical in relation to the x-axis) of width [itex]\Delta x[/itex] and length 1 - x (since y = 1 - x), and height f(x). This would be our incremental volume, or [itex]\Delta V[/itex]. To get the total volume we would integrate on y, as y runs from 0 to 1.

    Alternatively, we might use horizontal strips (in relation to the x-axis) that are [itex]\Delta y[/itex] by 1 - y by f(y) to get [itex]\Delta V[/itex]. To get the total volume, we would integrate on x, as x runs from 0 to 1.

    In this problem, I think that the idea is that they want you to use diagonal strips. Instead of working directly with x or y, I think you want to work with u = x + y. All of the line segments parallel to the hypotenuse of your triangle are such that at every point on the line segment, u is constant. For example, on the segment between (1, 0) and (0, 1), u = 1, where u = x + y. On the segment that joins (1/2, 0) and (0, 1/2), u = 1/2. And so on.

    Using this line of reasoning, the incremental volume element has dimensions (length of diagonal from x-axis to y-axis) X (width of strip = [itex]\Delta u[/itex]) X (height = f(u) = f(x + y)). It's not too big a jump to recognize that you can get the total volume by integrating the incremental volume element from u = 0 to u = 1. If you can convince yourself that the length of each of these diagonal strips is u, then you're pretty well on your way with this problem.

    One other thing: I assumed for convenience that f > 0 so I could think about the double integral as a volume. You won't be able to make this assumption, since we don't know this about f.
     
  5. Nov 8, 2009 #4
    hi, i think i kinda get it. but i dont get where the extra u( infront of f(u)) come from
     
  6. Nov 8, 2009 #5

    Mark44

    Staff: Mentor

    I believe that the extra u comes from the diagonal length of the base of the approximating incremental volume elements. At a given value of u, say u = x + y = k, with 0 <= k <= 1, the diagonal line segment extends from (0, k) to (k, 0), and has a length of sqrt(2k^2) = k*sqrt(2) = (x + y) sqrt(2) = u sqrt(2).

    Can you provide the exact wording of the problem, particularly what you need to prove?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Change variables
  1. Change of Variables (Replies: 0)

  2. Change of Variables (Replies: 4)

  3. Change in variables (Replies: 7)

  4. Change of variable (Replies: 5)

  5. Change of Variables (Replies: 1)

Loading...