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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

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# Reexpressing double integral using parametric functions

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