Parametric Integration

On Friday, we finished up 2-dimensional integration by looking at how to integrate parametric functions. Noticing how the standard parametric function is denoted as:

we determined that the formula for integration:

can be rewritten as:

by simple substitution.

We proceeded to looking at some problems. One of the more interesting ones was the parametric curve:

which looks like this:

We found that the function crosses the x-axis at t = sqrt 3. After integrating using our new formula to find the area bounded by the curve, we also discovered that the integral from t = -sqrt 3 to t = sqrt 3 was negative. Not only that, but 2x the integral from t = 0 to t = sqrt (3), which we thought would yield the correct answer, was also negative.

We got these results for two reasons, one for each end.

• When we integrated from t = 0 to t = sqrt 3, we were integrating the area denoted by green. As we know, the area under the x-axis is negative, resulting in a negative area when we multiplied by 2.
• When we integrated from t = -sqrt 3 to t = sqrt 3, two things happened:

1. For the first half, from t = -sqrt 3 to 0, we were integrating the function 'backwards', denoted by the red arrow.

2. The second half, from t = 0 to t = sqrt 3, we integrated the green area, which was already established to be negative.

What we gleaned from this problem is that when integrating parametric functions, both the "direction" of integration and the location of the area we are integrating (whether above or below the x-axis) must be taken into consideration.

Happy integrating!