Area with Parametric Equations — Question 9

Question 9

Problem

A designer reverses a parametrization on $0\le t\le1$ by replacing $t$ with $1-t$. What happens to the signed area integral $\int yx'\,dt$?

Question 9 – Solution

Solution

1. Let the original parametrization be $(x(t),y(t))$ on $0\le t\le1$, with signed integral $$I=\int_0^1y(t)x'(t)\,dt.$$

2. Under the reversed parametrization, set $u=1-t$. The new coordinates are $$\widetilde x(t)=x(1-t), \qquad \widetilde y(t)=y(1-t).$$ By the chain rule, $$\widetilde x'(t)=-x'(1-t).$$

3. The reversed signed integral is $$\begin{aligned} \widetilde I &=\int_0^1\widetilde y(t)\widetilde x'(t)\,dt\\ &=-\int_0^1y(1-t)x'(1-t)\,dt. \end{aligned}$$ Substituting $u=1-t$ shows that $\widetilde I=-I$.

4. Therefore, reversing the parametrization changes the sign of the signed area integral but not its magnitude: $$\boxed{\widetilde I=-I, \qquad |\widetilde I|=|I|}.$$ The geometric area is unchanged.