# 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 $\int y \, dx = \int_{a}^{b} y(t) x'(t) \, dt.$

## Derivation Of Formula

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.

## Examples And Problems

What is the area of a circle with radius $r$?

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

cycloidis the parametric curve given by equations $x(t) = t-\sin(t),$ $y(t) = 1-\cos(t).$

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?

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

**Cite as:**Parametric Area.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/parametric-equations-area-basic/