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