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)

\begin{align} \mathbf{Area} = \int_a^b y \: dx \\ \mathbf{Area} = \int_{\alpha}^{\beta} g(t) \: f'(t) \: dt \quad \mathrm{or} \quad \mathbf{Area} = \int_{\beta}^{\alpha} g(t) \: f'(t) \: dt \end{align}

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)

\begin{align} A = \int_0^{2\pi} 6(1 - \cos \theta) \cdot 6(1 - \cos \theta) \: d \theta \\ A = 36 \int_0^{2\pi} [ 1- 2\cos \theta + \cos ^2 \theta ] \: d \theta \\ \quad A = 36 \int_0^{2\pi} [1 - 2\cos \theta + \frac{1}{2} + \frac{1}{2} \cos {2 \theta}) ] \: d \theta \\ \quad A = 36 \int_0^{2\pi} [ \frac{3}{2} - 2 \cos \theta + \frac{1}{4} \sin {2 \theta} ] \: d \theta \\ A = 36 \left ( \frac{3}{2}\theta - 2\sin \theta + \frac{1}{4} \sin {2 \theta} \right) \bigg |_0^{2\pi} \\ A = 36(3 \pi) \\ A = 108 \pi \end{align}

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)

\begin{align} A = \int_0^{2\pi} \sin t \cdot - \sin t \: dt \\ A = - \int_0^{2\pi} \sin ^2 t \: dt \\ A = - \int_0^{2\pi} \frac{1}{2}(1 - \cos {2t}) \: dt \\ A = -\frac{1}{2} \left ( t - \frac{\sin {2t}}{2} \right ) \bigg |_0^{2\pi} \\ A = -\frac{1}{2} \cdot 2\pi A = - \pi \end{align}

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