Area with Parametric Equations — Question 3

Question 3 – Solution

Solution

1. Locate the parameter values at which the loop closes. Since $$x=t^2-1, \qquad y=t(t^2-1),$$ both $t=-1$ and $t=1$ produce the point $(0,0)$. The interval $-1\le t\le1$ traces the loop once.

2. Differentiate $x$: $$x'(t)=2t.$$ The signed area integral is $$\int_{-1}^{1}y(t)x'(t)\,dt =\int_{-1}^{1}(t^3-t)(2t)\,dt.$$

3. Simplify and integrate: $$\begin{aligned} \int_{-1}^{1}(2t^4-2t^2)\,dt &=\left[\frac{2}{5}t^5-\frac{2}{3}t^3\right]_{-1}^{1}\\ &=\frac45-\frac43=-\frac{8}{15}. \end{aligned}$$

4. The negative sign records the orientation of traversal; geometric area is nonnegative. Hence $$\boxed{A=\left|-\frac{8}{15}\right|=\frac{8}{15}}.$$