Parametric Area

Parametric Area is the area under a parametric curve. For instance, in the graph to the right, we have a curve for the parametric equations \(x(t) = t^2 + t\) and \(y(t) = 2t - 1\). We could want to find the area under the curve between \(t=-\frac{1}{2}\) and \(t=1\). This would be called the parametric area and is represented by the area in blue to the right.

Generalizing, to find the parametric areas means to calculate the area under a parametric curve of real numbers in two-dimensional space, \(\mathbb{R}^2\). Given some parametric equations, \(x(t)\), \(y(t)\). Where the parameter \(t\) ranges over some given interval. This curve may be bound by a region of the Cartesian plane (in particular, when the curve is closed and has no self-intersections), and this region is the area to be calculated. The formula for this generalized form is:

If \(x(t)\), \(y(t)\), \(t\in [a,b]\) describes a parametric curve bounding a region of the Cartesian plane, the area of this region is \[\int y \, dx = \int_{a}^{b} y(t) x'(t) \, dt.\]

Suppose \(x(t)\), \(y(t)\), \(t\in [0,1]\) describes a parametric curve that bounds a region \(R \subset \mathbb{R}^2\). Intuitively, one can approximate the area of \(R\) using rectangles; for an infinitesimal change in \(x\), from \(x\) to \(x+dx\), the portion of \(R\) contained between the "fenceposts" \(x\) and \(x+dx\) is approximately a rectangle, with height \(y\) and width \(dx\).

Since the area of this rectangle is \(y \, dx\), integrating this element over all possible \(x\)-values should give the area of \(R\). Thus, the area is precisely \[\int y \, dx = \int_{0}^{1} y(t) x'(t) \, dt.\] This proof can be made rigorous using Green's theorem.

If the argument above had been carried out swapping the roles of \(x\) and \(y\), one would obtain the formula \[\int x \, dy = \int_{0}^{1} x(t) y'(t) \, dt.\] This makes sense because the curve must be closed in order for it to bound a region, i.e. one must have \((x(0), y(0)) = (x(1), y(1))\). Then, one may compute \[\int x \, dy + \int y \, dx = \int_{0}^{1} [x(t) y'(t) + y(t) x'(t)] \, dt = \int_{0}^{1} (x(t) y(t))' \, dt = x(1) y(1) - x(0) y(0) = 0.\] Thus, the two formulas for area are equal up to a sign.

This illustrates the idea that this formula actually gives signed area, meaning that the value of the integral \(\int y\, dx\) is positive or negative depending on whether the bounded region lies to the left or right of the curve. For example, note that the regions bounded by closed curves \((x(t), y(t))\) and \((y(t), x(t))\) have the same unsigned area (which equals absolute value of the signed area), since one curve is just the reflection of the other over the line \(y = x\); but their signed areas sum to zero.

What is the area of a circle with radius \(r\)?

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Parametrize the circle as \((x(t), y(t)) = (r\cos(t), r\sin(t))\), where \(t\in [0,2\pi]\). From the formula for area bounded by a parametric curve, we have \[\text{Area}(\text{circle of radius } r) = \int_{0}^{2\pi} r \cos(t) \cdot (r\sin(t))' \, dt = \int_{0}^{2\pi} r^2 \cos^2 (t) \, dt = \pi r^2. \]

A cycloid is the parametric curve given by equations \[x(t) = t-\sin(t),\] \[y(t) = 1-\cos(t).\]

Our cycloid is traced by a marked point on a rolling circle of radius 1.

From the image, one notes that the cycloid consists of many congruent arches, traced as \(t\) ranges over the real numbers. What is the area bounded by one arch of the cycloid?

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One arch is traced out in its entirety as \(t\) ranges over \([0,2\pi]\). We compute \[\int y \, dx = \int_{0}^{2\pi} (1-\cos(t))^2 \, dt = \left[ \frac{3}{2} x - 2\sin(x) + \frac{1}{4} \sin(2x) \right] \Big\vert_{0}^{2\pi} = 3\pi.\]