# Areas Under Parametric Curves

Recall that the area under a curve $y =f(x)$ for $f(x) ≥ 0$ can be computed with the following integral: $\int_a^b f(x) \: dx$. Now suppose that a curve $C$ is traced out by the parametric equations $x = f(t)$ and $y = g(t)$. Note that $\frac{dy}{dt} = g'(t)$, or rather, $dx = f'(t) \: dt$. Making these substitutions into our area integral above, we get that:

(1)Where $a = f(\alpha)$ and $b = f(\beta)$ or $a = f(\beta)$ and $b = f(\alpha)$ respectively. We can now use this newly derived formula to determine the area under a parametric curve.

## Example 1

**Determine the area under the parametric curve defined by $x = 6(\theta - \sin \theta)$ and $y = 6(1 - \cos \theta)$ on the interval from $[0, 2\pi ]$.**

We first note that our limits of integration will be from $0$ to $2 \pi$. Additionally, over this interval we trace the curve out only once and notice that it is a cycloid as shown below:

Now let's first differentiate $x$. We thus get that $\frac{dx}{d \theta} = 6(1 - \cos \theta)$. Applying this into our formula we get that:

(2)## Example 2

**Determine the area inside the curve defined by the parametric equations $x = \cos t$ and $y = \sin t$.**

We first notice that the curve defined by the parametric equations in the problem form a unit circle, that is $x^2 + y^2 = \cos ^2 t + \sin ^2 t = 1$:

This circle has radius 1, and hence we should obtain an area of π. We can now verify this using parametric equations.

First we note that $\frac{dx}{dt} = - \sin t$. Additionally, our interval will be from $0$ to $2 \pi$ as it traces our the circle exactly once. We can now plug this into our formula for calculating areas of parametric equations:

(3)Clearly the area cannot be negative, so we take the absolute value of $A$ to get that the area is $\pi$.