Area with Parametric Equations — Question 8

Question 8 – Solution

Solution

1. The interval $0\le t\le\pi/2$ traces the first-quadrant arc from $(1,0)$ to $(0,1)$. By symmetry, the enclosed area is four times the first-quadrant area.

2. It is convenient to integrate with respect to $y$: $$A_1=\int x\,dy=\int_0^{\pi/2}x(t)y'(t)\,dt.$$ Since $y=\sin^3t$, $$y'(t)=3\sin^2t\cos t.$$

3. Substitute $x=\cos^3t$: $$A_1=3\int_0^{\pi/2}\sin^2t\cos^4t\,dt.$$ Using the even-power integral (or power-reduction identities), $$\int_0^{\pi/2}\sin^2t\cos^4t\,dt=\frac{\pi}{32}.$$ Thus $A_1=3\pi/32$.

4. Multiply by four quadrants: $$A=4A_1=4\left(\frac{3\pi}{32}\right) =\boxed{\frac{3\pi}{8}}.$$