Areas Under Parametric Curves Examples 1

# Areas Under Parametric Curves

Recall that if we have a parametric curve defined by the equations $x = f(t)$ and $y = g(t)$, then we can calculate the area trapped by the parametric curve using the following formula:

(1)
\begin{align} A = \int_{\alpha}^{\beta} g(t) \: f'(t) \: dt \end{align}

Where $\alpha ≤ t ≤ \beta$ and where $a = f(\alpha)$ and $b = f(\beta)$ OR $b = f(\alpha)$ and $a = f(\beta)$. We will now apply this formula in the following examples.

## Example 1

Determine the area trapped between the x-axis and the parametric curve defined by the equations $x = t^2 -4$ and $y = t$ on the interval [0, 4].

We first note that $\frac{dx}{dt} = 2t$. Furthermore, we also note that our limits of integration are 0 and 4. Hence applying this to the formula we get that:

(2)
\begin{align} A = \int_0^4 t \cdot 2t \: dt \\ A = 2 \int_0^4 t^2 \: dt \\ A = 2 \left ( \frac{t^3}{3} \right ) \bigg |_0^4 \\ A = 2 \left ( \frac{64}{3} \right ) \\ A = \frac{128}{3} \end{align}

The following graph shows the area that we have just computed:

Notice that it appears that we have integrated from -4 to 12. In fact, this is not true. Recall that $a = f(\alpha)$ and $b = f(\beta)$. In fact we can see that $f(0) = -4$, and $f(4) = 12$.

## Example 2

Determine the area trapped between the x-axis and the curve with parametric equations $x = t^3$ and $y = e^t$ on the interval [0, 1].

Once again we note that $\frac{dx}{dt} = 3t^2$. Our limits of integration are from 0 to 1. Using the formula for calculating areas of parametric curves, we obtain that:

(3)
\begin{align} A = \int_0^1 e^t \cdot 3t^2 \: dt \\ A = 3 \int_0^1 t^2 \cdot e^t \: dt \\ A = 3 (( t^2 - 2t + 2)e^t ) \bigg |_0^1 \\ A = 3(e - 2) \\ A \approx 2.155 \end{align}

## Example 3

Determine the area trapped between the x-axis and the curve with parametric equations $x = t$ and $y = -t^2 + 2$.

Applying the formula for area, we get that $\int_{\alpha}^{\beta} g(t) \: f'(t) \: dt$. We note that $y = g(t) = -t^2 + 2$ and $f'(t) = \frac{dx}{dt} = 1$. We now need to find our limits of integration $\alpha$ and $\beta$ which we will then substitute into our equation.

First let's eliminate the parameter $t$ to get $y = -x^2 + 2$. Note that $y = 0$ when $x = \sqrt{2} = b$ or $x = -\sqrt{2} = a$.

Now to get $\alpha$ and $\beta$, plug $a$ into our parametric substitution equation to get $\alpha$, namely $t \rvert_{x = a = -\sqrt{2}} = -\sqrt{2} = \alpha$, and plug $b$ into our parametric substitution equation $t \rvert_{x = b = \sqrt{2}} = -\sqrt{2} = \beta$ to get that $\beta = \sqrt{2}$. Thus:

(4)
\begin{align} A = \int_{-\sqrt{2}}^{\sqrt{2}} (-t^2 + 2)(1) \: dt \\ A = \left [ -\frac{t^3}{3} + 2t \right ]_{-\sqrt{2}}^{\sqrt{2}} \\ A \approx 3.77 \end{align}

Hopefully you can see how integration by parametric equations works with regards to example 3.