Area with Parametric Equations — Question 2

Question 2 – Solution

Solution

1. The given interval traces one arch from $(0,0)$ to $(2\pi,0)$. Since $$x'(t)=1-\cos t\ge0,$$ the curve moves from left to right, so the area under it is $\int y\,dx$ with the stated bounds.

2. Substitute $y(t)=1-\cos t$ and $dx=x'(t)dt$: $$A=\int_0^{2\pi}y(t)x'(t)\,dt =\int_0^{2\pi}(1-\cos t)^2\,dt.$$

3. Expand the integrand: $$(1-\cos t)^2=1-2\cos t+\cos^2t.$$ Therefore, $$A=\int_0^{2\pi}1\,dt -2\int_0^{2\pi}\cos t\,dt +\int_0^{2\pi}\cos^2t\,dt.$$

4. These three integrals equal $2\pi$, $0$, and $\pi$, respectively. Thus $$\boxed{A=3\pi}.$$