Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Reexpressing double integral using parametric functions

  1. Jul 31, 2010 #1
    If:

    x = f(t) (continuous and differentiable)
    y = g(t) (continuous)
    x is nondecreasing on [a, b]
    y is nonnegative on [a, b]

    Then when we trace the points (x,y) from t=a to t=b, we can calculate the area bounded above by the traced curve (below by y = 0, left by x = f(a), and right by x = f(b)) by

    [tex]\int_a^b (y \frac{dx}{dt}) dt[/tex]

    I find this fascinating because we don't need to first express y as a function of x, and just integrate over t directly.

    I am now trying to find the same parallel in double integral. Suppose the area defined above is R, and we want to evaluate:

    [tex]\int\int_R h(x,y) dA[/tex]

    If we can express y as a function of x, say y=i(x), then the integral above evaluates to:

    [tex]\int_{x=f(a)}^{f(b)}\int_{y=0}^{i(x)} h(x,y) dy \, dx[/tex]

    What I'm trying to do from now on in certainly not rigor. However, I'm thinking that because as t spans from a to b, x spans from f(a) to f(b), and on a particular t we can evaluate the integral h(x, y) over y (with x = f(t) constant, and moving the parameter u from 0 to g(t)). But we need to factor how fast x changes to t. Thus:

    [tex]\int_{t=a}^{b}\left\{\int_{u=0}^{g(t)} h(f(t),g(u)) du \right\} (\frac{dx}{dt}) dt[/tex]

    However I tried some calculations and the two integrals are certainly different. What is the correct formulation and what is the idea to derive it? Thanks
     
    Last edited: Jul 31, 2010
  2. jcsd
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted