Most of the time, we use y = f(x) to integrate curves. However, there are instances where this isn't convenient or even possible. This is where parametric coordinates come into play.
In parametric coordinates, a 'dummy' variable (usually denoted as t) assigns a value to either the x or y coordinate. Unlike a function in the form of y = f(x), a curve is represented by y(t) = g(t) and x(t) = h(t), where h and g are functions describing the change in x and y coordinates respectively.
For instance, a curve can be described by y(t) = 2t and x(t) = 2t, where 0 < t < 2π. By expressing it as (x(t))^2 + (y(t))^2 = (2cos(t))^2 + (2sin(t))^2 = 4(cos^2(t) + sin^2(t)) = 4, we can see that it forms a circle with a radius of 4, or x^2 + y^2 = 4.
The key to parametric integration lies in switching the limits. When we have an integral in the form of ∫f(x)dx, we must also switch the limits, resulting in the integral being ∫f(x(t))x'(t)dt.
For example, if we have an integral ∫y(x)dx from x = a to x = b, switching to t coordinates requires using the inverse formula of x(t) to find the t limits from the x limits, giving us ∫y(t)x'(t)dt from t = x(a) to t = x(b).
Parametric integration may seem daunting, so here are a couple of examples to better understand it.
Example 1: Find the area of the region bounded by the parabola y = x^2 and the line y = x from (0,0) to (1,1).
To represent the curve parametrically, we can let x(t) = t and y(t) = t^2, where 0 <= t <= 1. The area can then be found by integrating y(t)x'(t)dt from t = 0 to t = 1, giving us an answer of 1/3.
Example 2: Find the length of the curve y = x^2 from (0,0) to (1,1).
Using the same parametric representation as before, we can calculate the length of the curve by integrating √(x'(t)^2 + y'(t)^2)dt from t = 0 to t = 1, which gives us a length of √2/3.
